The Cantor diagonal construction is an algorithm for computing an incomputable number.
But if there is an algorithm for computing the number, then it is by definition a computable number.
The Cantor diagonal construction is an algorithm for computing an incomputable number.
But if there is an algorithm for computing the number, then it is by definition a computable number.
The Cantor diagonal construction is an algorithm for computing an incomputable number.
But if there is an algorithm for computing the number, then it is by definition a computable number.
On 03/04/2025 23:18, Lawrence D'Oliveiro wrote:
The Cantor diagonal construction is an algorithm for computing an
incomputable number.
But if there is an algorithm for computing the number, then it is by
definition a computable number.
I invite you to present an algorithm (a finite sequence of
mathematically rigorous instructions) for computing the nth Cantor
diagonal.
On 03/04/2025 23:18, Lawrence D'Oliveiro wrote:
The Cantor diagonal construction is an algorithm for computing an
incomputable number.
It is not an algorithm for computing something. Algorithms are
instructions that operate on finite inputs and must terminate with an
answer at some point for every input.
On 4/3/25 6:18 PM, Lawrence D'Oliveiro wrote:
The Cantor diagonal construction is an algorithm for computing an
incomputable number.
But if there is an algorithm for computing the number, then it is by
definition a computable number.
He shows a METHOD to generate that number, but it uses an infinite
number of steps, and shows that the number couldn't have been any of the numbers in the original set.
On 04/04/2025 03:29, Lawrence D'Oliveiro wrote:
On Fri, 4 Apr 2025 02:41:09 +0100, Mike Terry wrote:
On 03/04/2025 23:18, Lawrence D'Oliveiro wrote:
The Cantor diagonal construction is an algorithm for
computing an
incomputable number.
It is not an algorithm for computing something. Algorithms are
instructions that operate on finite inputs and must terminate
with an
answer at some point for every input.
The definition of a “computable number” is that for any integer
N, there
is an algorithm that will compute digit N of the number in a
finite
sequence of steps.
Does the Cantor diagonal construction fit this definition? Yes
it does.
No it doesn't, because for a computable number the algorithm
cannot have an infinite amount of input data. Typically we would
have a TM running with a tape containing just the number N
finitely encoded somehow and blank elsewhere.
The Cantor diagonal construction takes an infinite list as input...
On Fri, 4 Apr 2025 00:49:15 +0100, Richard Heathfield wrote:
On 03/04/2025 23:18, Lawrence D'Oliveiro wrote:
The Cantor diagonal construction is an algorithm for computing an
incomputable number.
But if there is an algorithm for computing the number, then it is by
definition a computable number.
I invite you to present an algorithm (a finite sequence of
mathematically rigorous instructions) for computing the nth Cantor
diagonal.
Simply repeat the diagonal construction n times.
On Fri, 4 Apr 2025 02:41:09 +0100, Mike Terry wrote:
On 03/04/2025 23:18, Lawrence D'Oliveiro wrote:
The Cantor diagonal construction is an algorithm for computing an
incomputable number.
It is not an algorithm for computing something. Algorithms are
instructions that operate on finite inputs and must terminate with an
answer at some point for every input.
The definition of a computable number is that for any integer N, there
is an algorithm that will compute digit N of the number in a finite
sequence of steps.
Does the Cantor diagonal construction fit this definition? Yes it does.
On 04/04/2025 03:30, Lawrence D'Oliveiro wrote:
On Fri, 4 Apr 2025 00:49:15 +0100, Richard Heathfield wrote:
On 03/04/2025 23:18, Lawrence D'Oliveiro wrote:
The Cantor diagonal construction is an algorithm for computing an
incomputable number.
But if there is an algorithm for computing the number, then it is by
definition a computable number.
I invite you to present an algorithm (a finite sequence of
mathematically rigorous instructions) for computing the nth Cantor
diagonal.
Simply repeat the diagonal construction n times.
The diagonal construction starts with a set, and the result you get
depends on the set you start with, so you end up not with /the/ nth
Cantor diagonal but one of infinitely many nth Cantor diagonals.
On 04/04/2025 03:29, Lawrence D'Oliveiro wrote:
No it doesn't, because for a computable number the algorithm cannot have
On Fri, 4 Apr 2025 02:41:09 +0100, Mike Terry wrote:
On 03/04/2025 23:18, Lawrence D'Oliveiro wrote:
The Cantor diagonal construction is an algorithm for computing an
incomputable number.
It is not an algorithm for computing something. Algorithms are
instructions that operate on finite inputs and must terminate with an
answer at some point for every input.
The definition of a “computable number” is that for any integer N,
there is an algorithm that will compute digit N of the number in a
finite sequence of steps.
Does the Cantor diagonal construction fit this definition? Yes it does.
an infinite amount of input data.
On Fri, 4 Apr 2025 04:35:59 +0100, Richard Heathfield wrote:
The Cantor argument constructs a number that is not
in the input list and thus proves that the input list, no matter how
large, is incomplete.
But that proof takes an infinite number of steps.
At every point, the
probability that the N digits computed so far match some number later in
the list is 1.
On 04/04/2025 05:24, Lawrence D'Oliveiro wrote:
At every point, the probability that the N digits computed so far match
some number later in the list is 1.
Counter-example follows.
Input list:
1111
2222
3333
4444
On Fri, 4 Apr 2025 04:47:13 +0100, Richard Heathfield wrote:
On 04/04/2025 03:30, Lawrence D'Oliveiro wrote:
On Fri, 4 Apr 2025 00:49:15 +0100, Richard Heathfield wrote:
On 03/04/2025 23:18, Lawrence D'Oliveiro wrote:
The Cantor diagonal construction is an algorithm for computing an
incomputable number.
But if there is an algorithm for computing the number, then it is by >>>>> definition a computable number.
I invite you to present an algorithm (a finite sequence of
mathematically rigorous instructions) for computing the nth Cantor
diagonal.
Simply repeat the diagonal construction n times.
The diagonal construction starts with a set, and the result you get
depends on the set you start with, so you end up not with /the/ nth
Cantor diagonal but one of infinitely many nth Cantor diagonals.
But for every n, the number of possibilities so far is always finite.
On 04/04/2025 05:24, Lawrence D'Oliveiro wrote:
On Fri, 4 Apr 2025 04:47:13 +0100, Richard Heathfield wrote:
On 04/04/2025 03:30, Lawrence D'Oliveiro wrote:
On Fri, 4 Apr 2025 00:49:15 +0100, Richard Heathfield wrote:
On 03/04/2025 23:18, Lawrence D'Oliveiro wrote:
The Cantor diagonal construction is an algorithm for computing an
incomputable number.
But if there is an algorithm for computing the number, then it is
by definition a computable number.
I invite you to present an algorithm (a finite sequence of
mathematically rigorous instructions) for computing the nth Cantor
diagonal.
Simply repeat the diagonal construction n times.
The diagonal construction starts with a set, and the result you get
depends on the set you start with, so you end up not with /the/ nth
Cantor diagonal but one of infinitely many nth Cantor diagonals.
But for every n, the number of possibilities so far is always finite.
For which input list, specifically?
On Fri, 4 Apr 2025 05:46:46 +0100, Richard Heathfield wrote:
On 04/04/2025 05:24, Lawrence D'Oliveiro wrote:
At every point, the probability that the N digits computed so far match
some number later in the list is 1.
Counter-example follows.
Input list:
1111
2222
3333
4444
Is that all? Just 4 numbers?
On Fri, 4 Apr 2025 06:16:20 +0100, Richard Heathfield wrote:
The Cantor diagonal argument shows that *any* list, finite or infinite,
is incomplete.
But it takes an infinite number of steps to show that for an infinite
list. And at every point, the probability that the N digits computed so
far match some number later in the list is 1.
The Cantor diagonal argument shows that *any* list, finite or infinite,
is incomplete.
If we say the 'real number' in the diagonal construction is rational
number, it will also fits the description.
On 04/04/2025 07:15, Lawrence D'Oliveiro wrote:
On Fri, 4 Apr 2025 06:16:20 +0100, Richard Heathfield wrote:
The Cantor diagonal argument shows that *any* list, finite or
infinite,
is incomplete.
But it takes an infinite number of steps to show that for an infinite
list. And at every point, the probability that the N digits computed so
far match some number later in the list is 1.
Depends on the list.
On Fri, 4 Apr 2025 07:24:35 +0100, Richard Heathfield wrote:
On 04/04/2025 07:15, Lawrence D'Oliveiro wrote:
On Fri, 4 Apr 2025 06:16:20 +0100, Richard Heathfield wrote:
The Cantor diagonal argument shows that *any* list, finite or
infinite,
is incomplete.
But it takes an infinite number of steps to show that for an infinite
list. And at every point, the probability that the N digits computed so
far match some number later in the list is 1.
Depends on the list.
No it doesn’t. At every point N, we have the first N digits of our hypothetical number-that-is-not-in-the-list. But we have an infinitude of remaining numbers in the list we haven’t looked at, among which all possible combinations of those N digits will occur.
Therefore there is
guaranteed to be some number we haven’t looked at yet with all those first N digits the same.
On 04/04/2025 08:21, Lawrence D'Oliveiro wrote:
At every point N, we have the first N digits of our
hypothetical number-that-is-not-in-the-list. But we have an infinitude
of remaining numbers in the list we haven’t looked at, among which all
possible combinations of those N digits will occur.
Show me your first N digits, and I'll show you a counterexample.
Therefore there is guaranteed to be some number we haven’t looked at
yet with all those first N digits the same.
And yet you still won't post those first N digits.
The Cantor diagonal construction is an algorithm for computing an incomputable number.
On 2025-04-03 22:18:38 +0000, Lawrence D'Oliveiro said:
The Cantor diagonal construction is an algorithm for computing an
incomputable number.
Can you prove that it computes an incomputable number?
On Fri, 4 Apr 2025 11:00:36 +0300, Mikko wrote:
On 2025-04-03 22:18:38 +0000, Lawrence D'Oliveiro said:
The Cantor diagonal construction is an algorithm for computing an
incomputable number.
Can you prove that it computes an incomputable number?
It’s trying to come up with a number that cannot fit into a set with cardinality ℵ₀. The cardinality of the computable numbers is the same as that of the integers, which is ℵ₀.
On Fri, 4 Apr 2025 08:41:35 +0100, Richard Heathfield wrote:
On 04/04/2025 08:21, Lawrence D'Oliveiro wrote:
At every point N, we have the first N digits of our
hypothetical number-that-is-not-in-the-list. But we have an infinitude
of remaining numbers in the list we haven’t looked at, among which all >>> possible combinations of those N digits will occur.
Show me your first N digits, and I'll show you a counterexample.
Counterexample to what?
At every point N, we have the first N digits of our
hypothetical number-that-is-not-in-the-list. But we have
an infinitude of remaining numbers in the list we haven’t
looked at, among which all possible combinations of those
N digits will occur.
Therefore there is guaranteed to be some number we haven’t looked at
yet with all those first N digits the same.
And yet you still won't post those first N digits.
Digit 1 is the first digit of entry 1 in the list.
Digit 2 is the second digit of entry 2 in the list.
.
.
.
Digit N is the Nth digit of entry N in the list.
=3: 3*10^n
The Cantor diagonal construction is an algorithm for computing an incomputable number.
But if there is an algorithm for computing the number, then it is by definition a computable number.
On Fri, 4 Apr 2025 04:15:39 +0100, Mike Terry wrote:
On 04/04/2025 03:29, Lawrence D'Oliveiro wrote:
No it doesn't, because for a computable number the algorithm cannot have
On Fri, 4 Apr 2025 02:41:09 +0100, Mike Terry wrote:
On 03/04/2025 23:18, Lawrence D'Oliveiro wrote:
The Cantor diagonal construction is an algorithm for computing an
incomputable number.
It is not an algorithm for computing something. Algorithms are
instructions that operate on finite inputs and must terminate with an
answer at some point for every input.
The definition of a computable number is that for any integer N,
there is an algorithm that will compute digit N of the number in a
finite sequence of steps.
Does the Cantor diagonal construction fit this definition? Yes it does.
an infinite amount of input data.
It doesnt need to. To compute the Nth digit of the diagonal, it only
needs N * (N + 1) / 2 digits of the numbers in the first N entries of the table.
On 2025-04-04 08:03:44 +0000, Lawrence D'Oliveiro said:
On Fri, 4 Apr 2025 11:00:36 +0300, Mikko wrote:
On 2025-04-03 22:18:38 +0000, Lawrence D'Oliveiro said:
The Cantor diagonal construction is an algorithm for computing an
incomputable number.
Can you prove that it computes an incomputable number?
It’s trying to come up with a number that cannot fit into a set with
cardinality ℵ₀. The cardinality of the computable numbers is the same
as that of the integers, which is ℵ₀.
It is also the cardinality of rationals but not of reals. Cantor's proof
is that for each list of reals one can construct a real that is not in
the list.
On 04/04/2025 05:22, Lawrence D'Oliveiro wrote:
Yes it does. The missing number has infinitely many digits ...
On Fri, 4 Apr 2025 04:15:39 +0100, Mike Terry wrote:
On 04/04/2025 03:29, Lawrence D'Oliveiro wrote:
No it doesn't, because for a computable number the algorithm cannot
On Fri, 4 Apr 2025 02:41:09 +0100, Mike Terry wrote:
On 03/04/2025 23:18, Lawrence D'Oliveiro wrote:The definition of a “computable number” is that for any integer N, >>>> there is an algorithm that will compute digit N of the number in a
finite sequence of steps.
Does the Cantor diagonal construction fit this definition? Yes it
does.
have an infinite amount of input data.
It doesn’t need to. To compute the Nth digit of the diagonal, it only
needs N * (N + 1) / 2 digits of the numbers in the first N entries of
the table.
On Fri, 4 Apr 2025 11:20:28 +0300, Mikko wrote:<snip>
On 2025-04-04 08:03:44 +0000, Lawrence D'Oliveiro said:
On Fri, 4 Apr 2025 11:00:36 +0300, Mikko wrote:
Can you prove that it computes an incomputable number?
It’s trying to come up with a number that cannot fit into a set with
cardinality ℵ₀. The cardinality of the computable numbers is the same >>> as that of the integers, which is ℵ₀.
It is also the cardinality of rationals but not of reals. Cantor's proof
is that for each list of reals one can construct a real that is not in
the list.
That proof doesn’t quite work, because at any point in the construction, the number can be shown to match some later item in the list.
The mismatch
with all items in the list is only demonstrated when the proof completes.
But the proof never completes. Therefore there is no mismatch. QED.
On 04/04/2025 20:13, Lawrence D'Oliveiro wrote:
That proof doesn’t quite work, because at any point in the
construction, the number can be shown to match some later item in the
list.
Then show it.
The anti-diagonal is *as computable as the list is*: and the argument in
fact proves there can be no such list, computable or otherwise...
On 04/04/2025 00:18, Lawrence D'Oliveiro wrote:
The Cantor diagonal construction is an algorithm for computing an
incomputable number.
To be more precise, it is an argument, not an algorithm, although it
is a fact that "diagonalisation" as a method has become ubiquitous.
BTW, the argument has indeed been proved constructively: or so they
say, I find that debatable as it needs *extensionality*.
But if there is an algorithm for computing the number, then it is byThe anti-diagonal is *as computable as the list is*: and the argument
definition a computable number.
in fact proves there can be no such list, computable or otherwise...
(no complete list of all such *sequences*: the connection to the real *numbers* is a further step and problem).
On Fri, 4 Apr 2025 20:53:31 +0100, Richard Heathfield wrote:
On 04/04/2025 20:13, Lawrence D'Oliveiro wrote:
That proof doesn’t quite work, because at any point in the
construction, the number can be shown to match some later item in the
list.
Then show it.
Simple: you have N digits of the supposed-incomputable number so far, and there are only B**N (where B is the base of the number system, e.g. 10) possible combinations for those digits, compared to an infinite number of remaining items in the rest of the list that you haven’t looked at. Therefore
every possible one of those B**N possibilities will occur
somewhere in that remaining list.
Therefore the partial number constructed
so far will always match some later item in the list. QED.
Not all lists are random.
-- The construction of a computable number that differs from every
number in the list shows that there is no way to construct a list
of all computable numbers ...
On Sat, 5 Apr 2025 00:04:23 +0100, Richard Heathfield wrote:
Not all lists are random.
I didn’t say they were.
But if you like, it is possible to construct a list where the property of finding matches later in the list is guaranteed.
In fact, such a list also
provably leaves out whole swathes of known computable numbers.
So you’d
think the Cantor construction would have an easier time finding omissions, given that you know they exist on other grounds. But it doesn’t.
On Fri, 4 Apr 2025 17:02:57 +0100, Mike Terry wrote:
On 04/04/2025 05:22, Lawrence D'Oliveiro wrote:
Yes it does. The missing number has infinitely many digits ...
On Fri, 4 Apr 2025 04:15:39 +0100, Mike Terry wrote:
On 04/04/2025 03:29, Lawrence D'Oliveiro wrote:
No it doesn't, because for a computable number the algorithm cannot
On Fri, 4 Apr 2025 02:41:09 +0100, Mike Terry wrote:
On 03/04/2025 23:18, Lawrence D'Oliveiro wrote:The definition of a computable number is that for any integer N,
there is an algorithm that will compute digit N of the number in a
finite sequence of steps.
Does the Cantor diagonal construction fit this definition? Yes it
does.
have an infinite amount of input data.
It doesnt need to. To compute the Nth digit of the diagonal, it only
needs N * (N + 1) / 2 digits of the numbers in the first N entries of
the table.
Any given one of which can be computed in a finite time, with a finite
amount of data. Thats the definition of computable.
On Fri, 4 Apr 2025 11:20:28 +0300, Mikko wrote:
On 2025-04-04 08:03:44 +0000, Lawrence D'Oliveiro said:
On Fri, 4 Apr 2025 11:00:36 +0300, Mikko wrote:
On 2025-04-03 22:18:38 +0000, Lawrence D'Oliveiro said:
The Cantor diagonal construction is an algorithm for computing an
incomputable number.
Can you prove that it computes an incomputable number?
It’s trying to come up with a number that cannot fit into a set with
cardinality ℵ₀. The cardinality of the computable numbers is the same >>> as that of the integers, which is ℵ₀.
It is also the cardinality of rationals but not of reals. Cantor's proof
is that for each list of reals one can construct a real that is not in
the list.
That proof doesn’t quite work, because at any point in the construction, the number can be shown to match some later item in the list. The mismatch with all items in the list is only demonstrated when the proof completes.
But the proof never completes. Therefore there is no mismatch. QED.
You don't understand what a computable number is ...
Um, right, the individual digits are computable. :)
But the missing (real) number from the Cantor list is not necessarilycomputable.
Since all elements (except your two openers) begin with a 3, none of
them start 12, and so after just two iterations we have already
constructed a number that's not in the infinite list.
On Fri, 4 Apr 2025 09:16:17 +0100, Richard Heathfield wrote:
Since all elements (except your two openers) begin with a 3, none of
them start 12, and so after just two iterations we have already
constructed a number that's not in the infinite list.
Remember that the hypothesis of the Cantor “proof” is that the list is already supposed to contain every computable number.
The fact that the
contruction succeeds for your list examples does not mean it will succeed with mine. Remember, the “proof” depends on it succeeding in the general case, with every possible list.
The proof is finite and complete.
On Sat, 5 Apr 2025 00:09:13 +0100, Andy Walker wrote:
-- The construction of a computable number that differs from every... which in turn must mean that there is no way to construct a list of
number in the list shows that there is no way to construct a list
of all computable numbers ...
all integers, which is clearly nonsense.
On 05/04/2025 11:05, Andy Walker wrote:
ask yourself what nature of integer has8, right?
an infinite number of digits.
On Fri, 4 Apr 2025 09:16:17 +0100, Richard Heathfield wrote:
Since all elements (except your two openers) begin with a 3, none ofRemember that the hypothesis of the Cantor “proof” is that the list is already supposed to contain every computable number.
them start 12, and so after just two iterations we have already
constructed a number that's not in the infinite list.
The fact that the contruction succeeds for your list examples does not mean it will succeed with mine. Remember, the “proof” depends on it succeeding in the general case, with every possible list.
On Fri, 4 Apr 2025 13:47:29 +0200, Julio Di Egidio wrote:
The anti-diagonal is *as computable as the list is*: and the argument in
fact proves there can be no such list, computable or otherwise...
No it doesn’t. The cardinality of the computable numbers is ℵ₀, same as that of the integers. And the integers can in fact be arranged in a list, therefore so can the computable numbers. QED.
ask yourself what nature of integer has
an infinite number of digits.
On 05/04/2025 03:20, Lawrence D'Oliveiro wrote:
On Sat, 5 Apr 2025 00:09:13 +0100, Andy Walker wrote:
-- The construction of a computable number that differs from every
number in the list shows that there is no way to construct a list
of all computable numbers ...
... which in turn must mean that there is no way to construct a list of
all integers, which is clearly nonsense.
No it doesn't.
Think about what it means to construct a list of the computable numbers.
You can’t, of course, physically write out a number of infinite digits. So what you can do instead is write (or at least try to write) a list of the computer programs, in the form of callable functions, for calculating each computable number, ordered according to some Gödel numbering. Each
function takes a single positive integer parameter N, and is supposed to return the Nth digit of its computable number.
On Sat, 5 Apr 2025 04:26:29 +0100, Mike Terry wrote:
You don't understand what a computable number is ...
A number which can be computed to any number of digits desired by a
Turing machine.
<https://mathworld.wolfram.com/ComputableNumber.html>
But to be computable, numbers must be computed in a finite number of
steps.
On 04/04/2025 23:49, Lawrence D'Oliveiro wrote:
On Fri, 4 Apr 2025 13:47:29 +0200, Julio Di Egidio wrote:
The anti-diagonal is *as computable as the list is*: and the argument
in fact proves there can be no such list, computable or otherwise...
No it doesn’t. The cardinality of the computable numbers is ℵ₀, same as
that of the integers. And the integers can in fact be arranged in a
list, therefore so can the computable numbers. QED.
You should try and prove it formally ...
The problem is that to calculate say the first 100 digits of the Cantor "anti-diagonal" of Cantor's list requires input of 100 digits-worth of information out of the list. More properly, the TM cannot receive 100 digits-worth on its tape, since by definition the tape contains only the number 100 and is otherwise empty.
It does succeed with every possible list.
Which is more likely, that you've found a flaw in a proof that's been accepted by mathematicians for over a century, or that you've reached an incorrect conclusion?
On 05/04/2025 03:19, Lawrence D'Oliveiro wrote:
But if you like, it is possible to construct a list where the property
of finding matches later in the list is guaranteed.
Yes, of course. To construct such a list is trivially easy, because you
can just add a match yourself. But so what?
On Sat, 5 Apr 2025 09:07:22 +0100, Richard Heathfield wrote:
But to be computable, numbers must be computed in a finite number of
steps.
“Computable Number: A number which can be computed to any number of digits desired by a Turing machine.”
<https://mathworld.wolfram.com/ComputableNumber.html>
On 06/04/2025 06:40, Lawrence D'Oliveiro wrote:
On Sat, 5 Apr 2025 09:07:22 +0100, Richard Heathfield wrote:
But to be computable, numbers must be computed in a finite number of
steps.
“Computable Number: A number which can be computed to any number of
digits desired by a Turing machine.”
<https://mathworld.wolfram.com/ComputableNumber.html>
"The “computable” numbers may be described briefly as the real numbers whose expressions as a decimal are calculable by finite means." - Alan Turing.
And therefore, to be computable, numbers must be computed in a finite
number of steps.
On Sat, 5 Apr 2025 11:40:14 +0100, Andy Walker wrote:
It does succeed with every possible list.
Here’s a counterexample list: write out the whole numbers (non-negative integers) from 0 in increasing order, and flip the digits of each one so
that the digit from the 10⁰ place goes to the 10¯¹ place, 10¹ to 10¯² etc:
0.0000000000000...
0.1000000000000...
0.2000000000000...
0.3000000000000...
...
0.9000000000000...
0.0100000000000...
0.1100000000000...
0.2100000000000...
0.3100000000000...
...
0.9100000000000...
0.0200000000000...
...
This is not a *complete* list of all computable numbers, let alone all the reals (the original point of the Cantor construction). You’d think it
would make things easier for the Cantor construction, but it doesn’t.
Step 1 of the Cantor construction: choose a digit in the 10¯¹ place different from that of the first item in the list. There are 9
possibilities we could pick. But all the 10 possibilities for that first digit occur in the following 10 numbers, so our pick will definitely match one of them.
Step 2: choose the next digit, in the 10¯² place, different from that of the second item in the list. There are, again, 9 possibilities we could
pick. But all the 100 combinations of possibilities for the first two
digits occur in the following 100 numbers, so our picks so far will definitely match one of them.
And so on: at step N, we pick a digit in the Nth decimal place, to be different from that of the Nth number in the list. But all the 10**N possibilities for the digits we have picked so far occur in the following 10**N numbers, so the number we have constructed so far will provably
match one of them.
Note this is a proof by induction: if our choices at step N match some existing entry in the list, then so will the addition of our next choice
at step N + 1. Since our first choice already matches some existing entry
in the list, it follows that, however many digits we choose, the result
will always match some existing entry in the list.
So even in a list which we already know does not contain every possible computable number, or every real number, the Cantor construction fails to find one of the missing ones.
On Sun, 6 Apr 2025 07:27:43 +0100, Richard Heathfield wrote:
On 06/04/2025 06:40, Lawrence D'Oliveiro wrote:
On Sat, 5 Apr 2025 09:07:22 +0100, Richard Heathfield wrote:
But to be computable, numbers must be computed in a finite number of
steps.
“Computable Number: A number which can be computed to any number of
digits desired by a Turing machine.”
<https://mathworld.wolfram.com/ComputableNumber.html>
"The “computable” numbers may be described briefly as the real numbers >> whose expressions as a decimal are calculable by finite means." - Alan
Turing.
And therefore, to be computable, numbers must be computed in a finite
number of steps.
I would say you are quoting Turing out of context.
By your
(mis)interpretation of his words, even something like 1/3 is an
incomputable number, since its “expressions as a decimal are not
calculable by finite means”.
Turing would not have been so dumb as to have his definition of
computability depend on something as trivial as the choice of number base.
You can slice and dice as much as you like, but the Cantor
process applied to any list of reals or to any computable list of
computable reals generates a real or computable real not in the list. So
any such list is incomplete.
On Sat, 5 Apr 2025 12:36:16 +0200, Julio Di Egidio wrote:
On 04/04/2025 23:49, Lawrence D'Oliveiro wrote:
On Fri, 4 Apr 2025 13:47:29 +0200, Julio Di Egidio wrote:
The anti-diagonal is *as computable as the list is*: and the argument
in fact proves there can be no such list, computable or otherwise...
No it doesn’t. The cardinality of the computable numbers is ℵ₀, same as
that of the integers. And the integers can in fact be arranged in a
list, therefore so can the computable numbers. QED.
You should try and prove it formally ...
Look up “Gödel numbering”.
Simply put, repeating decimals are irrational.
On Sun, 2025-04-06 at 08:36 +0100, Richard Heathfield wrote:
On 06/04/2025 08:15, wij wrote:
<snip>
Simply put, repeating decimals are irrational.
Simply put, an irrational number is one that can't be expressed
as the ratio of two integers.
Prove it.
On Sun, 2025-04-06 at 09:11 +0100, Richard Heathfield wrote:
On 06/04/2025 08:49, wij wrote:
On Sun, 2025-04-06 at 08:36 +0100, Richard Heathfield wrote:
On 06/04/2025 08:15, wij wrote:
<snip>
Simply put, repeating decimals are irrational.
Simply put, an irrational number is one that can't be expressed
as the ratio of two integers.
Prove it.
The clue is in the name.
Thus quoth Wolfram: "An irrational number is a number that cannot
be expressed as a fraction p/q for any integers p and q."
<https://mathworld.wolfram.com/search/?query=irrational+number>
When you've persuaded Wolfram that they're wrong, call.
What is this, Give A Crank A Home Week?
Obviously, you don't even understand what 'definition' is, neither.
On 06/04/2025 07:52, Lawrence D'Oliveiro wrote:
On Sat, 5 Apr 2025 12:36:16 +0200, Julio Di Egidio wrote:
On 04/04/2025 23:49, Lawrence D'Oliveiro wrote:
On Fri, 4 Apr 2025 13:47:29 +0200, Julio Di Egidio wrote:
The anti-diagonal is *as computable as the list is*: and the argument >>>>> in fact proves there can be no such list, computable or otherwise...
No it doesn’t. The cardinality of the computable numbers is ℵ₀, same as
that of the integers. And the integers can in fact be arranged in a
list, therefore so can the computable numbers. QED.
You should try and prove it formally ...
Look up “Gödel numbering”.
Fuck you and the whole indecent band-wagon.
*Plonk*
On Fri, 4 Apr 2025 09:16:17 +0100, Richard Heathfield wrote:
Since all elements (except your two openers) begin with a 3, none of
them start 12, and so after just two iterations we have already
constructed a number that's not in the infinite list.
Remember that the hypothesis of the Cantor “proof” is that the list is already supposed to contain every computable number. The fact that the contruction succeeds for your list examples does not mean it will succeed with mine.
On Sun, 2025-04-06 at 06:43 +0000, Lawrence D'Oliveiro wrote:
On Sun, 6 Apr 2025 07:27:43 +0100, Richard Heathfield wrote:
On 06/04/2025 06:40, Lawrence D'Oliveiro wrote:
On Sat, 5 Apr 2025 09:07:22 +0100, Richard Heathfield wrote:
But to be computable, numbers must be computed in a finite number of >>>>> steps.
“Computable Number: A number which can be computed to any number of
digits desired by a Turing machine.”
<https://mathworld.wolfram.com/ComputableNumber.html>
"The “computable” numbers may be described briefly as the real numbers >>> whose expressions as a decimal are calculable by finite means." - Alan
Turing.
And therefore, to be computable, numbers must be computed in a finite
number of steps.
I would say you are quoting Turing out of context. By your>
(mis)interpretation of his words, even something like 1/3 is an>
incomputable number, since its “expressions as a decimal are not>
calculable by finite means”.
Simply put, repeating decimals are irrational. https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber2-en.txt/download
All your doubt should be addressed in the file of the link.
Turing would not have been so dumb as to have his definition of>
computability depend on something as trivial as the choice of number
base.
On 2025-04-05 07:38:19 +0000, Lawrence D'Oliveiro said:
On Fri, 4 Apr 2025 09:16:17 +0100, Richard Heathfield wrote:
Since all elements (except your two openers) begin with a 3,
none of
them start 12, and so after just two iterations we have already
constructed a number that's not in the infinite list.
Remember that the hypothesis of the Cantor “proof” is that the
list is
already supposed to contain every computable number. The fact
that the
contruction succeeds for your list examples does not mean it
will succeed
with mine.
How can Cantor's construction fail to succeed on a list?
On Sat, 5 Apr 2025 10:10:44 +0300, Mikko wrote:
The proof is finite and complete.
It requires showing that there is no complete match among an infinity of digits.
On 06/04/2025 11:29, Mikko wrote:
On 2025-04-05 07:38:19 +0000, Lawrence D'Oliveiro said:
On Fri, 4 Apr 2025 09:16:17 +0100, Richard Heathfield wrote:
Since all elements (except your two openers) begin with a 3, none of
them start 12, and so after just two iterations we have already
constructed a number that's not in the infinite list.
Remember that the hypothesis of the Cantor proof is that the list is
already supposed to contain every computable number. The fact that the
contruction succeeds for your list examples does not mean it will succeed >>> with mine.
How can Cantor's construction fail to succeed on a list?
As I understand it, his argument can be summarised as follows:
1. Let C[inf][inf] be a list of all the digits of all the computable numbers.
2. Let D be the Cantor diagonal, eg via
for(n = 0; n <= inf; n++)
{
D[n] = (C[n][n] + 1) % 10;
}
3, Because we have computed D, it is a computable number, and therefore it must have an entry in C[,
so the construction of D must somehow be in error.
The flaw, of course, is in overlooking that we required infinitely many steps to derive D. for(n =
0; n <= inf; n++){whatever} is not an algorithm, because by definition algorithms must have at most
finitely many steps.
On Sat, 5 Apr 2025 23:10:35 +0100, Andy Walker wrote:
You can slice and dice as much as you like, but the CantorApply it to the list I posted elsewhere, then.
process applied to any list of reals or to any computable list of
computable reals generates a real or computable real not in the list. So
any such list is incomplete.
On Sat, 5 Apr 2025 11:40:14 +0100, Andy Walker wrote:[... snippage ...]
It does succeed with every possible list.Here’s a counterexample list: write out the whole numbers (non-negative integers) from 0 in increasing order, and flip the digits of each one so
that the digit from the 10⁰ place goes to the 10¯¹ place, 10¹ to 10¯² etc:
0.0000000000000...
0.1000000000000...
0.2000000000000...
0.3000000000000...
And so on: at step N, we pick a digit in the Nth decimal place, to be different from that of the Nth number in the list. But all the 10**N possibilities for the digits we have picked so far occur in the following 10**N numbers, so the number we have constructed so far will provably
match one of them.
So even in a list which we already know does not contain every possible computable number, or every real number, the Cantor construction fails to find one of the missing ones.
On 06/04/2025 07:41, Lawrence D'Oliveiro wrote:
On Sat, 5 Apr 2025 23:10:35 +0100, Andy Walker wrote:
You can slice and dice as much as you like, but the CantorApply it to the list I posted elsewhere, then.
process applied to any list of reals or to any computable list of
computable reals generates a real or computable real not in
the list. So
any such list is incomplete.
What did your last servant die of?
More specifically, if you are trolling then I strongly suggestIt seems to be a trend: find a well-established proof and bend
that you follow Julio's [rather intemperate] advice. If on the
other hand you are a crank, then you have just joined Peter
and Wij on the Naughty Step.
If, on the third hand, you have confused yourself
[and perhaps Richard]
On 06/04/2025 11:52, Richard Heathfield wrote:
On 06/04/2025 11:29, Mikko wrote:
On 2025-04-05 07:38:19 +0000, Lawrence D'Oliveiro said:
On Fri, 4 Apr 2025 09:16:17 +0100, Richard Heathfield wrote:
Since all elements (except your two openers) begin with a 3,
none of
them start 12, and so after just two iterations we have already
constructed a number that's not in the infinite list.
Remember that the hypothesis of the Cantor “proof” is that
the list is
already supposed to contain every computable number. The fact
that the
contruction succeeds for your list examples does not mean it
will succeed
with mine.
How can Cantor's construction fail to succeed on a list?
As I understand it, his argument can be summarised as follows:
1. Let C[inf][inf] be a list of all the digits of all the
computable numbers.
Cantor made no reference to computability or computable numbers.
2. Let D be the Cantor diagonal, eg via
for(n = 0; n <= inf; n++)
{
D[n] = (C[n][n] + 1) % 10;
}
You are writing the above as though it is some kind of
step-by-step program to be executed.
That's a source of
confusion amongst certain non-mathematicians, particularly those
with a programming perspective on things. (Not that you're
confused - I'd just avoid such notation in case others are
confused.)
The diagonal definition is just that: a definition, not a step by
step sequence of anything.
3, Because we have computed D, it is a computable number, and
therefore it must have an entry in C[, so the construction of D
must somehow be in error.
Cantor was not concerned with computability.
His proof assumes
we have the list as in (1), and constructs (defines) a new real
number which is manifestly not in the list using the well known
diagonal argument.
There is no error with the construction of D, given the list.
Depending on exact wording of the proof, it either shows that
every list of reals misses at least one real number [the
"anti-diagonal"], or that a contradiction is reached from the
assumption that the original list was complete [the new
anti-diagonal being a real number both in the list and not in the
list] and so the assumption of completeness of the list was false.
The flaw, of course, is in overlooking that we required
infinitely many steps to derive D. for(n = 0; n <= inf;
n++){whatever} is not an algorithm, because by definition
algorithms must have at most finitely many steps.
Hmmm, maybe you're talking about applying Cantor's argument to
the list of computable numbers? Cantor never did that.
If we do
this, we can certainly start with a list of computable numbers
which is complete, since there are only countably many such
numbers. Then the anti-diagonal argument produces a new
(non-computable) number that is not in the list.
It might seem that the number produced must be computable because
the anti-diagonal "computes" it, but the anti-diagonal
"computation" would only work given infinitely many digits of
data out of the list. (E.g. the whole list that you've called
C[inf][inf] could be represented on a tape, or perhaps just the
diagonal etc. but in any case the job can't be done with only a
finite amount of data.
Or perhaps we might think that a "computable list" of computable
numbers could be constructed where a single TM can somehow
generate all the C[inf][inf] on request (no extra data being
input other than the row/col indices of the required entry).
Then we could have one single TM that calculates all the
anti-diagonal digits with no further data input from the expanded
list since it can just calculate those digits as required. That
would present a contradiction since the new anti-diagonal number
would then be computable! But such a "computable list" is just
wishful thinking and does not exist...
On 06/04/2025 07:41, Lawrence D'Oliveiro wrote:
On Sat, 5 Apr 2025 23:10:35 +0100, Andy Walker wrote:
You can slice and dice as much as you like, but the CantorApply it to the list I posted elsewhere, then.
process applied to any list of reals or to any computable list of
computable reals generates a real or computable real not in the list. So >>> any such list is incomplete.
What did your last servant die of?
More specifically, if you are trolling then I strongly suggest
that you follow Julio's [rather intemperate] advice. If on the other
hand you are a crank, then you have just joined Peter and Wij on the
Naughty Step. If, on the third hand, you have confused yourself [and perhaps Richard], then I have done my best to enlighten you, and you
need to learn more about the differences between maths and CS, and more specifically about the historical context of Cantor's and Turing's work. Whichever, I shall not be responding further in this thread unless you
show signs of learning and of being a student rather than a belligerent.
On 06/04/2025 17:24, Mike Terry wrote:
On 06/04/2025 11:52, Richard Heathfield wrote:
On 06/04/2025 11:29, Mikko wrote:
On 2025-04-05 07:38:19 +0000, Lawrence D'Oliveiro said:
On Fri, 4 Apr 2025 09:16:17 +0100, Richard Heathfield wrote:
Since all elements (except your two openers) begin with a 3, none of >>>>>> them start 12, and so after just two iterations we have already
constructed a number that's not in the infinite list.
Remember that the hypothesis of the Cantor proof is that the list is >>>>> already supposed to contain every computable number. The fact that the >>>>> contruction succeeds for your list examples does not mean it will succeed >>>>> with mine.
How can Cantor's construction fail to succeed on a list?
As I understand it, his argument can be summarised as follows:
1. Let C[inf][inf] be a list of all the digits of all the computable numbers.
Cantor made no reference to computability or computable numbers.
Nevertheless, that's precisely what Mr D'Oliveiro is talking about. From the start of the thread:
"The Cantor diagonal construction is an algorithm for computing an incomputable number. But if there
is an algorithm for computing the number, then it is by definition a computable number."
<snip>
2. Let D be the Cantor diagonal, eg via
for(n = 0; n <= inf; n++)
{
D[n] = (C[n][n] + 1) % 10;
}
You are writing the above as though it is some kind of step-by-step program to be executed.
It is an attempt to formalise what I believe to be what the OP considers to be the relevant
algorithm for arriving at the diagonal.
That's a source of confusion amongst certain non-mathematicians, particularly those with a
programming perspective on things. (Not that you're confused - I'd just avoid such notation in
case others are confused.)
It's just shorthand. I'm very willing to consider suggestions for alternative ways to convey meaning
so succinctly.
The diagonal definition is just that: a definition, not a step by step sequence of anything.
Yes. We don't have to run alongside Achilles.
3, Because we have computed D, it is a computable number, and therefore it must have an entry in
C[, so the construction of D must somehow be in error.
Cantor was not concerned with computability.
Mr D'Oliveiro, however, is.
His proof assumes we have the list as in (1), and constructs (defines) a new real number which is
manifestly not in the list using the well known diagonal argument.
There is no error with the construction of D, given the list. Depending on exact wording of the
proof, it either shows that every list of reals misses at least one real number [the
"anti-diagonal"], or that a contradiction is reached from the assumption that the original list
was complete [the new anti-diagonal being a real number both in the list and not in the list] and
so the assumption of completeness of the list was false.
Yes. Because of the way "computable" is defined, the same proof works for computable numbers and
suffices to debunk Mr D'Oliveiro.
The flaw, of course, is in overlooking that we required infinitely many steps to derive D. for(n
= 0; n <= inf; n++){whatever} is not an algorithm, because by definition algorithms must have at
most finitely many steps.
Hmmm, maybe you're talking about applying Cantor's argument to the list of computable numbers?
Cantor never did that.
No, but Mr D'Oliveiro did.
If we do this, we can certainly start with a list of computable numbers which is complete, since
there are only countably many such numbers. Then the anti-diagonal argument produces a new
(non-computable) number that is not in the list.
Yes.
It might seem that the number produced must be computable because the anti-diagonal "computes" it,
but the anti-diagonal "computation" would only work given infinitely many digits of data out of
the list. (E.g. the whole list that you've called C[inf][inf] could be represented on a tape, or
perhaps just the diagonal etc. but in any case the job can't be done with only a finite amount of
data.
It's okay; we're playing with the whole set.
Or perhaps we might think that a "computable list" of computable numbers could be constructed
where a single TM can somehow generate all the C[inf][inf] on request (no extra data being input
other than the row/col indices of the required entry). Then we could have one single TM that
calculates all the anti-diagonal digits with no further data input from the expanded list since it
can just calculate those digits as required. That would present a contradiction since the new
anti-diagonal number would then be computable! But such a "computable list" is just wishful
thinking and does not exist...
Actually, it does. I got it for Christmas in 1996. Last time I looked it was somewhere in the loft.
On 06/04/2025 18:04, Richard Heathfield wrote:
On 06/04/2025 17:24, Mike Terry wrote:
On 06/04/2025 11:52, Richard Heathfield wrote:
On 06/04/2025 11:29, Mikko wrote:
On 2025-04-05 07:38:19 +0000, Lawrence D'Oliveiro said:
On Fri, 4 Apr 2025 09:16:17 +0100, Richard Heathfield wrote:
Since all elements (except your two openers) begin with a
3, none of
them start 12, and so after just two iterations we have
already
constructed a number that's not in the infinite list.
Remember that the hypothesis of the Cantor “proof” is that
the list is
already supposed to contain every computable number. The
fact that the
contruction succeeds for your list examples does not mean
it will succeed
with mine.
How can Cantor's construction fail to succeed on a list?
As I understand it, his argument can be summarised as follows:
1. Let C[inf][inf] be a list of all the digits of all the
computable numbers.
Cantor made no reference to computability or computable numbers.
Nevertheless, that's precisely what Mr D'Oliveiro is talking
about. From the start of the thread:
"The Cantor diagonal construction is an algorithm for computing
an incomputable number. But if there is an algorithm for
computing the number, then it is by definition a computable
number."
<snip>
Sure, but... I'm still not clear on whether OP is
expecting/requiring the Cantor list to be a list of computable
numbers. I don't believe he has stated that anywhere... He just
keeps saying the missing number is computable.
[...]2. Let D be the Cantor diagonal, eg via
for(n = 0; n <= inf; n++)
{
D[n] = (C[n][n] + 1) % 10;
}
It's just shorthand. I'm very willing to consider suggestions
for alternative ways to convey meaning so succinctly.
The "D[n] = (C[n][n] + 1) % 10" bit is no problem, but the for
loop is definitely encouraging people to think of the whole thing
as some kind of supertask
I'd have just said that the D was the unique real number whose
n'th digit is
D[n] := 5 [if C[n][n] != 5]
6 [otherwise]
(no need for procedural loops... also sidesteps the whole
0.999999... discussion we've thankfully not had)
Hmmm, maybe you're talking about applying Cantor's argument to
the list of computable numbers? Cantor never did that.
No, but Mr D'Oliveiro did.
I don't see anywhere that he says the list is a list of
computable numbers - just where he says the missing number must
be computable.
OP's main problem seems to be that he doesn't understand that a
computable real has a /finite/ definition via an algorithm/TM,
not a sequence of ever larger TMs that can each compute just
finite digit prefixes. Computing the anti-diag requires the
Cantor list as input, which cannot be packaged into a finite
amount of data. Maybe OP has something else in mind but simply
hasn't expressed it anywhere so people can't tell what he's
actually getting at. (A bit like assuming the numbers in the
list are computable without bothering to actually say that. :) )
Or maybe OP is trolling or a just confused about things, hard to
say.
The constructed number will not continue to match any particular member
of the list indefinitely.
On 06/04/2025 07:43, Lawrence D'Oliveiro wrote:
On Sun, 6 Apr 2025 07:27:43 +0100, Richard Heathfield wrote:
On 06/04/2025 06:40, Lawrence D'Oliveiro wrote:
On Sat, 5 Apr 2025 09:07:22 +0100, Richard Heathfield wrote:
But to be computable, numbers must be computed in a finite number of >>>>> steps.
“Computable Number: A number which can be computed to any number of
digits desired by a Turing machine.”
<https://mathworld.wolfram.com/ComputableNumber.html>
"The “computable” numbers may be described briefly as the real numbers >>> whose expressions as a decimal are calculable by finite means." - Alan
Turing.
And therefore, to be computable, numbers must be computed in a finite
number of steps.
I would say you are quoting Turing out of context.
There is no context before his words because they are the paper's
opening words.
Cantor made no reference to computability or computable numbers.
In fact, I don't think Cantor proposed the argument for uncountability
of real numbers in this form ...
Cantor was not concerned with computability.
His proof assumes we have the list as in (1), and constructs (defines) a
new real number which is manifestly not in the list using the well known diagonal argument.
Hmmm, maybe you're talking about applying Cantor's argument to the list
of computable numbers? ...
It might seem that the number produced must be computable because the anti-diagonal "computes" it, but the anti-diagonal "computation" would
only work given infinitely many digits of data out of the list.
On Sun, 6 Apr 2025 08:05:42 +0100, Richard Heathfield wrote:
On 06/04/2025 07:43, Lawrence D'Oliveiro wrote:
On Sun, 6 Apr 2025 07:27:43 +0100, Richard Heathfield wrote:
On 06/04/2025 06:40, Lawrence D'Oliveiro wrote:
On Sat, 5 Apr 2025 09:07:22 +0100, Richard Heathfield wrote:
But to be computable, numbers must be computed in a finite number of >>>>>> steps.
“Computable Number: A number which can be computed to any number of >>>>> digits desired by a Turing machine.”
<https://mathworld.wolfram.com/ComputableNumber.html>
"The “computable” numbers may be described briefly as the real numbers >>>> whose expressions as a decimal are calculable by finite means." - Alan >>>> Turing.
And therefore, to be computable, numbers must be computed in a finite
number of steps.
I would say you are quoting Turing out of context.
There is no context before his words because they are the paper's
opening words.
The paper is “On Computable Numbers, With An Application To The Entscheidungsproblem” <https://www.cs.virginia.edu/~robins/Turing_Paper_1936.pdf>.
Go on, then. Show how his conclusions are at odds with the generally- accepted (and more concise) definition of “computable number” quoted above.
On Sun, 6 Apr 2025 17:22:28 +0100, Andy Walker wrote:
The constructed number will not continue to match any particular member
of the list indefinitely.
Congratulations, you got the point of my proof.
Isn’t the Cantor construction supposed to come up with a number not in the list, for *any* list?
The drawback with this construction is it fails to converge. That is, at
any point, we only have some finite number of digits of the number that is supposedly not in the list, yet it can always match some later entry in
the list.
It is an infinite list, after all.
Hmmm, maybe you're talking about applying Cantor's argument to the list
of computable numbers? ...
It might seem that the number produced must be computable because the
anti-diagonal "computes" it, but the anti-diagonal "computation" would
only work given infinitely many digits of data out of the list.
Actually, at any point, it only needs a finite number of digits out of a finite number of entries in the list.
The definition of “computability”
doesn’t mean “I can produce infinitely many digits”, but “I can produce as
many digits as you ask for”.
After infinitely many steps ...
First, Cantor's Diagonal Argument wasn't about construable
numbers, but about numbers being countable.
On 06/04/2025 17:24, Mike Terry wrote:
On 06/04/2025 11:52, Richard Heathfield wrote:
On 06/04/2025 11:29, Mikko wrote:
On 2025-04-05 07:38:19 +0000, Lawrence D'Oliveiro said:
On Fri, 4 Apr 2025 09:16:17 +0100, Richard Heathfield wrote:
Since all elements (except your two openers) begin with a 3, none of >>>>>> them start 12, and so after just two iterations we have already
constructed a number that's not in the infinite list.
Remember that the hypothesis of the Cantor “proof” is that the list is
already supposed to contain every computable number. The fact that the >>>>> contruction succeeds for your list examples does not mean it will
succeed
with mine.
How can Cantor's construction fail to succeed on a list?
As I understand it, his argument can be summarised as follows:
1. Let C[inf][inf] be a list of all the digits of all the computable
numbers.
Cantor made no reference to computability or computable numbers.
Nevertheless, that's precisely what Mr D'Oliveiro is talking about. From
the start of the thread:
"The Cantor diagonal construction is an algorithm for computing an incomputable number. But if there is an algorithm for computing the
number, then it is by definition a computable number."
<snip>
2. Let D be the Cantor diagonal, eg via
for(n = 0; n <= inf; n++)
{
D[n] = (C[n][n] + 1) % 10;
}
You are writing the above as though it is some kind of step-by-step
program to be executed.
It is an attempt to formalise what I believe to be what the OP considers
to be the relevant algorithm for arriving at the diagonal.
That's a source of confusion amongst certain non-mathematicians,
particularly those with a programming perspective on things. (Not
that you're confused - I'd just avoid such notation in case others are
confused.)
It's just shorthand. I'm very willing to consider suggestions for
alternative ways to convey meaning so succinctly.
The diagonal definition is just that: a definition, not a step by step
sequence of anything.
Yes. We don't have to run alongside Achilles.
3, Because we have computed D, it is a computable number, and
therefore it must have an entry in C[, so the construction of D must
somehow be in error.
Cantor was not concerned with computability.
Mr D'Oliveiro, however, is.
His proof assumes we have the list as in (1), and constructs (defines)
a new real number which is manifestly not in the list using the well
known diagonal argument.
There is no error with the construction of D, given the list.
Depending on exact wording of the proof, it either shows that every
list of reals misses at least one real number [the "anti-diagonal"],
or that a contradiction is reached from the assumption that the
original list was complete [the new anti-diagonal being a real number
both in the list and not in the list] and so the assumption of
completeness of the list was false.
Yes. Because of the way "computable" is defined, the same proof works
for computable numbers and suffices to debunk Mr D'Oliveiro.
The flaw, of course, is in overlooking that we required infinitely
many steps to derive D. for(n = 0; n <= inf; n++){whatever} is not an
algorithm, because by definition algorithms must have at most
finitely many steps.
Hmmm, maybe you're talking about applying Cantor's argument to the
list of computable numbers? Cantor never did that.
No, but Mr D'Oliveiro did.
If we do this, we can certainly start with a list of computable
numbers which is complete, since there are only countably many such
numbers. Then the anti-diagonal argument produces a new (non-
computable) number that is not in the list.
Yes.
It might seem that the number produced must be computable because the
anti-diagonal "computes" it, but the anti-diagonal "computation" would
only work given infinitely many digits of data out of the list. (E.g.
the whole list that you've called C[inf][inf] could be represented on
a tape, or perhaps just the diagonal etc. but in any case the job
can't be done with only a finite amount of data.
It's okay; we're playing with the whole set.
Or perhaps we might think that a "computable list" of computable
numbers could be constructed where a single TM can somehow generate
all the C[inf][inf] on request (no extra data being input other than
the row/col indices of the required entry). Then we could have one
single TM that calculates all the anti-diagonal digits with no further
data input from the expanded list since it can just calculate those
digits as required. That would present a contradiction since the new
anti-diagonal number would then be computable! But such a "computable
list" is just wishful thinking and does not exist...
Actually, it does. I got it for Christmas in 1996. Last time I looked it
was somewhere in the loft.
On Sun, 6 Apr 2025 07:53:06 +0100, Richard Heathfield wrote:
After infinitely many steps ...
I.e. never.
On Sun, 6 Apr 2025 17:24:33 +0100, Mike Terry wrote:
Cantor made no reference to computability or computable numbers.
Didnt Turing apply his argument to them?
In fact, I don't think Cantor proposed the argument for uncountability
of real numbers in this form ...
What form did he give for his proof, then?
Cantor was not concerned with computability.
Did the concept exist in his time?
He just keeps saying the missing number is computable.
... there doesn't need to be a master algorithm, that given k and n
gives the first n digits of the kth number, that can be shown to cover
the full set of [computable] numbers ...
On 4/6/2025 2:18 PM, Lawrence D'Oliveiro wrote:
On Sun, 6 Apr 2025 17:22:28 +0100, Andy Walker wrote:
The constructed number will not continue to match any particular member
of the list indefinitely.
Congratulations, you got the point of my proof.
Isn’t the Cantor construction supposed to come up with a number not in the >> list, for *any* list?
Hell no! It "comes up with a number" that is not in the list being
discussed ...
On 06/04/2025 23:01, Lawrence D'Oliveiro wrote:
On Sun, 6 Apr 2025 07:53:06 +0100, Richard Heathfield wrote:
After infinitely many steps ...
I.e. never.
If you mean you can never know all the digits, hey, you're right.
No algorithm can derive the number. It's incomputable.
But "never" is a strong word.
On 2025-04-05 07:26:38 +0000, Lawrence D'Oliveiro said:
On Sat, 5 Apr 2025 10:10:44 +0300, Mikko wrote:
The proof is finite and complete.
It requires showing that there is no complete match among an infinity
of digits.
Cantor's proof and every proof with Cantor's diagonal method shows that.
On Sun, 6 Apr 2025 23:38:25 +0100, Richard Heathfield wrote:
On 06/04/2025 23:01, Lawrence D'Oliveiro wrote:
On Sun, 6 Apr 2025 07:53:06 +0100, Richard Heathfield wrote:
After infinitely many steps ...
I.e. never.
If you mean you can never know all the digits, hey, you're right.
No algorithm can derive the number. It's incomputable.
That’s not what “incomputable” means.
But "never" is a strong word.
Like anything in mathematics, you need proof before claiming something.
This is why we have proof-by-induction: it’s essentially the only way to make generalized statements about infinite sequences.
Instead of an
infinite number of propositions to be proved, it boils the whole lot down
to two:
P(1)
P(N) ⊢ P(N + 1)
I gave my proof-by-induction; it is up to you to try to tear it down. If
you can.
On Sun, 2025-04-06 at 13:35 +0300, Mikko wrote:
On 2025-04-06 07:15:51 +0000, wij said:
On Sun, 2025-04-06 at 06:43 +0000, Lawrence D'Oliveiro wrote:
On Sun, 6 Apr 2025 07:27:43 +0100, Richard Heathfield wrote:
On 06/04/2025 06:40, Lawrence D'Oliveiro wrote:
On Sat, 5 Apr 2025 09:07:22 +0100, Richard Heathfield wrote:
But to be computable, numbers must be computed in a finite number of >>>>>>> steps.
“Computable Number: A number which can be computed to any number of >>>>>> digits desired by a Turing machine.”
<https://mathworld.wolfram.com/ComputableNumber.html>
"The “computable” numbers may be described briefly as the real numbers
whose expressions as a decimal are calculable by finite means." - Alan >>>>> Turing.
And therefore, to be computable, numbers must be computed in a finite >>>>> number of steps.
I would say you are quoting Turing out of context. By your>> > >
(mis)interpretation of his words, even something like 1/3 is an>> > >
incomputable number, since its “expressions as a decimal are not>> > > >>>> calculable by finite means”.
Simply put, repeating decimals are irrational.
https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber2-en.txt/download
Repeating decimals are rational.
Prove it (be sure not to make mistakes shown in the link above)
On 06/04/2025 11:29, Mikko wrote:
On 2025-04-05 07:38:19 +0000, Lawrence D'Oliveiro said:
On Fri, 4 Apr 2025 09:16:17 +0100, Richard Heathfield wrote:
Since all elements (except your two openers) begin with a 3, none of
them start 12, and so after just two iterations we have already
constructed a number that's not in the infinite list.
Remember that the hypothesis of the Cantor “proof” is that the list is >>> already supposed to contain every computable number. The fact that the
contruction succeeds for your list examples does not mean it will succeed >>> with mine.
How can Cantor's construction fail to succeed on a list?
As I understand it, his argument can be summarised as follows:
1. Let C[inf][inf] be a list of all the digits of all the computable numbers.
2. Let D be the Cantor diagonal, eg via
for(n = 0; n <= inf; n++)
{
D[n] = (C[n][n] + 1) % 10;
}
3, Because we have computed D, it is a computable number, and therefore
it must have an entry in C[, so the construction of D must somehow be
in error.
The flaw, of course, is in overlooking that we required infinitely many
steps to derive D. for(n = 0; n <= inf; n++){whatever} is not an
algorithm, because by definition algorithms must have at most finitely
many steps.
On Sun, 6 Apr 2025 13:42:44 +0300, Mikko wrote:
On 2025-04-05 07:26:38 +0000, Lawrence D'Oliveiro said:
On Sat, 5 Apr 2025 10:10:44 +0300, Mikko wrote:
The proof is finite and complete.
It requires showing that there is no complete match among an infinity
of digits.
Cantor's proof and every proof with Cantor's diagonal method shows that.
Except for that example list I gave where I proved by induction that it
does not.
On 2025-04-06 10:52:50 +0000, Richard Heathfield said:
On 06/04/2025 11:29, Mikko wrote:
On 2025-04-05 07:38:19 +0000, Lawrence D'Oliveiro said:
On Fri, 4 Apr 2025 09:16:17 +0100, Richard Heathfield wrote:
Since all elements (except your two openers) begin with a 3,
none of
them start 12, and so after just two iterations we have already
constructed a number that's not in the infinite list.
Remember that the hypothesis of the Cantor “proof” is that
the list is
already supposed to contain every computable number. The fact
that the
contruction succeeds for your list examples does not mean it
will succeed
with mine.
How can Cantor's construction fail to succeed on a list?
As I understand it, his argument can be summarised as follows:
1. Let C[inf][inf] be a list of all the digits of all the
computable numbers.
2. Let D be the Cantor diagonal, eg via
for(n = 0; n <= inf; n++)
{
D[n] = (C[n][n] + 1) % 10;
}
3, Because we have computed D, it is a computable number, and
therefore it must have an entry in C[, so the construction of D
must somehow be in error.
The flaw, of course, is in overlooking that we required
infinitely many steps to derive D. for(n = 0; n <= inf;
n++){whatever} is not an algorithm, because by definition
algorithms must have at most finitely many steps.
The number D is not computable if the list C is not.
The counter-assumption was that there is a list of all
computable numbers but not that there is a computable list of
all computable numbers. As the number D is not computable its
absence in C is not a contradiction.
On Sun, 6 Apr 2025 13:42:44 +0300, Mikko wrote:
On 2025-04-05 07:26:38 +0000, Lawrence D'Oliveiro said:
On Sat, 5 Apr 2025 10:10:44 +0300, Mikko wrote:
The proof is finite and complete.
It requires showing that there is no complete match among an infinity
of digits.
Cantor's proof and every proof with Cantor's diagonal method shows that.
Except for that example list I gave where I proved by induction that it
does not.
On Sun, 6 Apr 2025 17:39:30 -0400, Richard Damon wrote:
... there doesn't need to be a master algorithm, that given k and n
gives the first n digits of the kth number, that can be shown to cover
the full set of [computable] numbers ...
The diagonal construction, given n, returns the nth digit of the nth
number in the set that is, by argument, claimed to contain the full set of numbers in the set under discussion (reals or computables).
On Mon, 2025-04-07 at 11:28 +0300, Mikko wrote:
On 2025-04-06 10:42:05 +0000, wij said:
On Sun, 2025-04-06 at 13:35 +0300, Mikko wrote:
On 2025-04-06 07:15:51 +0000, wij said:
On Sun, 2025-04-06 at 06:43 +0000, Lawrence D'Oliveiro wrote:
On Sun, 6 Apr 2025 07:27:43 +0100, Richard Heathfield wrote:
On 06/04/2025 06:40, Lawrence D'Oliveiro wrote:
On Sat, 5 Apr 2025 09:07:22 +0100, Richard Heathfield wrote:
But to be computable, numbers must be computed in a finite number of >>>>>>>>> steps.
“Computable Number: A number which can be computed to any number of >>>>>>>> digits desired by a Turing machine.”
<https://mathworld.wolfram.com/ComputableNumber.html>
"The “computable” numbers may be described briefly as the real numbers
whose expressions as a decimal are calculable by finite means." - Alan >>>>>>> Turing.
And therefore, to be computable, numbers must be computed in a finite >>>>>>> number of steps.
I would say you are quoting Turing out of context. By your>> > >
(mis)interpretation of his words, even something like 1/3 is an>> > > >>>>>> incomputable number, since its “expressions as a decimal are not>> > > >>>>>> calculable by finite means”.
Simply put, repeating decimals are irrational.
https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber2-en.txt/download
Repeating decimals are rational.
Prove it (be sure not to make mistakes shown in the link above)
See
https://math.stackexchange.com/questions/549254/why-is-a-repeating-decimal-a-rational-number
Still can't prove, except posting a copy from the internet?
On 04/04/2025 23:49, Lawrence D'Oliveiro wrote:
You should try and prove it formally, you'll see that it's you here
messing up definitions, even what a deductive system is, i.e. what
it means to prove things formally.
On Mon, 2025-04-07 at 11:50 +0100, Richard Heathfield wrote:<snip>
On 07/04/2025 11:31, wij wrote:
On Mon, 2025-04-07 at 11:28 +0300, Mikko wrote:
On 2025-04-06 10:42:05 +0000, wij said:
On Sun, 2025-04-06 at 13:35 +0300, Mikko wrote:
On 2025-04-06 07:15:51 +0000, wij said:
Simply put, repeating decimals are irrational.
https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber2-en.txt/download
Repeating decimals are rational.
Prove it (be sure not to make mistakes shown in the link above)
See
https://math.stackexchange.com/questions/549254/why-is-a-repeating-decimal-a-rational-number
Still can't prove, except posting a copy from the internet?
Posting a link to a proof /is/ proving it.
Let's work an example.
We take a repeating decimal such as r = 0.142857142857142857...
What's a million times that? Clearly it's 1000000r =
142857.142857142857142857...
Subtracting:
1000000r = 142857.142857142857142857...
r = 0.142857142857142857...
yields:
999999r = 142857
Dividing both sides by 142857:
7r = 1
Dividing both sides by 7:
r = 1/7
1 is an integer, 7 is an integer, so their ratio r is rational.
QED.
0.142857(142857)= 1/7
<=> 142857*(1000001000001...)/1000000... = 1/7 7*142857*(1000001000001...)/(5*2)^(6*n) = 1 7*3*3*3*11*13*37*(1000001000001...)....= (5*2)^(6*n)
QED.
Congratulation, you proved unique factorization theorem of positive integer is false
Read the file carefully:
You are quite wrong about an elementary proof and various of your misunderstandings have been pointed out to you ...
Richard Heathfield <rjh@cpax.org.uk> writes:
On 07/04/2025 08:33, Lawrence D'Oliveiro wrote:[...]
[...]That’s not what “incomputable” means.
Yeah, it is. We've already had this argument. Turing won: "The
"computable" numbers may be described briefly as the real numbers
whose expressions as a decimal are calculable by finite means."
That's a little too briefly.
Quoting Wikipedia:
In mathematics, computable numbers are the real numbers that can be
computed to within any desired precision by a finite, terminating
algorithm.
By dropping the "to within any desired precision"
your description
implies (unintentionally, I'm sure) that pi is not computable.
On Mon, 7 Apr 2025 02:10:23 -0600, Jeff Barnett wrote:
You are quite wrong about an elementary proof and various of your
misunderstandings have been pointed out to you ...
I gave a proof by induction why the Cantor construction fails on that flipped-integer list; no one has yet pointed out a flaw in that proof.
Somebody kept insisting that, even if my proposition is true for every element in the list,
it somehow goes false at the end. The end of an
infinite list! Yeah, right.
On Mon, 7 Apr 2025 02:10:23 -0600, Jeff Barnett wrote:
You are quite wrong about an elementary proof and various of your
misunderstandings have been pointed out to you ...
I gave a proof by induction why the Cantor construction fails on that flipped-integer list; no one has yet pointed out a flaw in that proof.
Somebody kept insisting that, even if my proposition is true for every element in the list, it somehow goes false at the end. The end of an
infinite list! Yeah, right.
Richard Heathfield <rjh@cpax.org.uk> writes:
On 07/04/2025 22:24, Keith Thompson wrote:
Richard Heathfield <rjh@cpax.org.uk> writes:
On 07/04/2025 08:33, Lawrence D'Oliveiro wrote:[...]
[...]That’s not what “incomputable” means.
Yeah, it is. We've already had this argument. Turing won: "The
"computable" numbers may be described briefly as the real numbers
whose expressions as a decimal are calculable by finite means."
That's a little too briefly.
I'll be sure to take it up with Turing next time I see him. What was
he thinking?
Quoting Wikipedia:
In mathematics, computable numbers are the real numbers that can
be computed to within any desired precision by a finite,
terminating algorithm.
By dropping the "to within any desired precision"
It wasn't there to drop.
your description implies (unintentionally, I'm sure) that pi is not
computable.
Not my description; Alan Turing's description. Those quotation marks
around the words weren't there just for fun.
Yes, that's what Turing wrote. I suggest that *if taken out of context*
it could be taken to imply that pi is not computable (at least that's
the implication I took from it).
I don't know the full context, but I
presume he clarified that an infinite number of steps might be required
in some cases.
I'm certain we're both in agreement that (a) all the digits of pi cannot
be computed by any algorithm or Turing machine in a finite number of
steps, but (b) any digit of pi can be computed in a finite number of
steps by some algorith or Turing machine.
Richard Heathfield <rjh@cpax.org.uk> writes:
On 07/04/2025 22:24, Keith Thompson wrote:
Richard Heathfield <rjh@cpax.org.uk> writes:
On 07/04/2025 08:33, Lawrence D'Oliveiro wrote:[...]
[...]That’s not what “incomputable” means.
Yeah, it is. We've already had this argument. Turing won: "The
"computable" numbers may be described briefly as the real numbers
whose expressions as a decimal are calculable by finite means."
That's a little too briefly.
I'll be sure to take it up with Turing next time I see him. What was
he thinking?
Quoting Wikipedia:
In mathematics, computable numbers are the real numbers that can
be computed to within any desired precision by a finite,
terminating algorithm.
By dropping the "to within any desired precision"
It wasn't there to drop.
your description implies (unintentionally, I'm sure) that pi is not
computable.
Not my description; Alan Turing's description. Those quotation marks
around the words weren't there just for fun.
Yes, that's what Turing wrote. I suggest that *if taken out of context*
it could be taken to imply that pi is not computable (at least that's
the implication I took from it). I don't know the full context, but I presume he clarified that an infinite number of steps might be required
in some cases.
Turing's paper is here:
https://www.cs.ox.ac.uk/activities/ieg/e-library/sources/tp2-ie.pdf
I'm certain we're both in agreement that (a) all the digits of pi cannot
be computed by any algorithm or Turing machine in a finite number of
steps, but (b) any digit of pi can be computed in a finite number of
steps by some algorith or Turing machine.
On Mon, 2025-04-07 at 11:28 +0300, Mikko wrote:
On 2025-04-06 10:42:05 +0000, wij said:
On Sun, 2025-04-06 at 13:35 +0300, Mikko wrote:https://math.stackexchange.com/questions/549254/why-is-a-repeating-decimal-a-rational-number
On 2025-04-06 07:15:51 +0000, wij said:
On Sun, 2025-04-06 at 06:43 +0000, Lawrence D'Oliveiro wrote:
On Sun, 6 Apr 2025 07:27:43 +0100, Richard Heathfield wrote:
On 06/04/2025 06:40, Lawrence D'Oliveiro wrote:
On Sat, 5 Apr 2025 09:07:22 +0100, Richard Heathfield wrote:
But to be computable, numbers must be computed in a finite number of >>>>>>>>> steps.
“Computable Number: A number which can be computed to any number of >>>>>>>> digits desired by a Turing machine.”
<https://mathworld.wolfram.com/ComputableNumber.html>
"The “computable” numbers may be described briefly as the real numbers
whose expressions as a decimal are calculable by finite means." - Alan >>>>>>> Turing.
And therefore, to be computable, numbers must be computed in a finite >>>>>>> number of steps.
I would say you are quoting Turing out of context. By your>> > >> > > > >>>>>> > (mis)interpretation of his words, even something like 1/3 is an>> > >>>>>> >> > > > > incomputable number, since its “expressions as a decimal are
calculable by finite means”.
Simply put, repeating decimals are irrational.
https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber2-en.txt/download
Repeating decimals are rational.
Prove it (be sure not to make mistakes shown in the link above)
Still can't prove, except posting a copy from the internet?
On 07/04/2025 09:34, Mikko wrote:
On 2025-04-06 10:52:50 +0000, Richard Heathfield said:
On 06/04/2025 11:29, Mikko wrote:
On 2025-04-05 07:38:19 +0000, Lawrence D'Oliveiro said:
On Fri, 4 Apr 2025 09:16:17 +0100, Richard Heathfield wrote:
Since all elements (except your two openers) begin with a 3, none of >>>>>> them start 12, and so after just two iterations we have already
constructed a number that's not in the infinite list.
Remember that the hypothesis of the Cantor “proof” is that the list is
already supposed to contain every computable number. The fact that the >>>>> contruction succeeds for your list examples does not mean it will succeed >>>>> with mine.
How can Cantor's construction fail to succeed on a list?
As I understand it, his argument can be summarised as follows:
1. Let C[inf][inf] be a list of all the digits of all the computable numbers.
2. Let D be the Cantor diagonal, eg via
for(n = 0; n <= inf; n++)
{
D[n] = (C[n][n] + 1) % 10;
}
3, Because we have computed D, it is a computable number, and therefore
it must have an entry in C[, so the construction of D must somehow be
in error.
The flaw, of course, is in overlooking that we required infinitely many
steps to derive D. for(n = 0; n <= inf; n++){whatever} is not an
algorithm, because by definition algorithms must have at most finitely
many steps.
The number D is not computable if the list C is not.
Granted, but I would maintain that D would remain incomputable even if
you could wave a magic wand and compute C[].
The counter-assumption was that there is a list of all
computable numbers but not that there is a computable list of
all computable numbers. As the number D is not computable its absence
in C is not a contradiction.
Agreed. But I don't think we'll move Mr D'Oliveiro away from his claim
quite that easily, as he has evidently mistaken his slippery ground for bedrock.
On 2025-04-07 10:31:28 +0000, wij said:
On Mon, 2025-04-07 at 11:28 +0300, Mikko wrote:
On 2025-04-06 10:42:05 +0000, wij said:
On Sun, 2025-04-06 at 13:35 +0300, Mikko wrote:https://math.stackexchange.com/questions/549254/why-is-a-repeating-decimal-a-rational-number
On 2025-04-06 07:15:51 +0000, wij said:
Simply put, repeating decimals are irrational.
https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber2-en.txt/download
Repeating decimals are rational.
Prove it (be sure not to make mistakes shown in the link above)
Still can't prove, except posting a copy from the internet?
So you are only trolling?
My Gist, Cantor's diagonal argument (in Rocq): <https://gist.github.com/jp-diegidio/eb05f6265c38b35c85853514ed46ab47>
<< We go for the simplest construction, which
is not in terms of sets, just (type-theoretic)
functions plus the most basic functional
extensionality, i.e. eta-conversion. >>
In fact, learning some formal mathematics with some tool
that supports it I think makes one's informal mathematics
way sharper, especially for those who are [self-taught]
(whatever that means).
That said, I find my formalisation neither particularly
pretty nor particularly clear: corrections and suggestions
are welcome, especially in the direction of minimising
assumptions.
On Tue, 2025-04-08 at 09:29 +0100, Richard Heathfield wrote:
On 08/04/2025 08:31, Mikko wrote:
On 2025-04-07 10:31:28 +0000, wij said:
On Mon, 2025-04-07 at 11:28 +0300, Mikko wrote:
On 2025-04-06 10:42:05 +0000, wij said:
On Sun, 2025-04-06 at 13:35 +0300, Mikko wrote:https://math.stackexchange.com/questions/549254/why-is-a-repeating-decimal-a-rational-number
On 2025-04-06 07:15:51 +0000, wij said:
Simply put, repeating decimals are irrational.
https://sourceforge.net/projects/cscall/files/MisFiles/RealNumber2-en.txt/download
Repeating decimals are rational.
Prove it (be sure not to make mistakes shown in the link above)
Still can't prove, except posting a copy from the internet?
So you are only trolling?
Hasn't he just proved that by claiming that recurring decimals
are irrational? 1/3, 1/7, 1/11, 1/13, 1/17... all irrational!
Just remind you, your post showed that you don't understand 1. what the definition
is. 2. what a proof should be (the above statement is another evidence).
3. your understand is 'by belief'.
(d) Despite RichardH's comments, I think the jury is still just
about out on Wij.
Four fairly random comments:
(a) There have been several references in this thread to the fact
that
we can't store the whole of an infinite sequence and similar. Perhaps
worth noting that there is a fundamental difference between [pure] maths
and CS in this area. In maths, there are practical problems in
numerical analysis, in convergence, and the like, but worrying about
storage is /so/ 20thC. If you have a list of numbers, then you can let a[i,j] be the j-th decimal digit of the i-th number, and it just /is/. Similarly the number C whose n-th digit is an appropriate tweak of
a[n,n] just /is/, and it is easily seen to be different from any number
in your list. There is no question about waiting for the computation of
C to finish or of having to store it anywhere.
(b) Turing died in 1954, and published his best known paper in 1937,
ie
before WW2, before there were any electronic computers, before there was
any recognisable CS, and /long/ before there were any undergrad CS
courses.
He is, of course, a very important pioneer of the subject, but both
maths and CS have moved on a long way since 1937, and everything he
writes about computability needs to be treated with caution.
(c) I don't know whether I'm the only person to recognise the
convergence,
but both our resident cranks remind me of a Greek tragedy:
Petrence: Here is a spiffing new idea!
Chorus: It's neither new nor spiffing.
Lawer: Anyone who understands can see that I'm right.
Chorus: But the first mistake is [whatever].
Petrence: You don't understand, here is my proof.
Chorus: It's wrong because [whatever].
Lawer: No-one can point to a mistake.
Chorus: Yes, we can, and have!
Petrence: You don't understand, here is my proof.
Chorus: It's wrong because [whatever].
[Repeat ad inf or ad nauseam, take your pick]
I for one am happy to cut a newcomer with a problem or an idea some
slack, but not to be the third spear-carrier in the chorus. Clearly
some other knowledgeable posters feel the same way. Others ... less so.
(d) Despite RichardH's comments, I think the jury is still just about
out on Wij. His root problem is that he doesn't accept the Archimedean
[or Eudoxus] axiom [essentially that there are no infinitesimal real numbers]. Some of the weirder things he says are more-or-less correct
in some non-standard systems, but he refuses to learn about NS analysis
or about the surreals, and is [unsurprisingly] incapable of putting his theory onto a firm, axiomatic basis. Likewise with Peter's ideas about "geometric points".
It will, however, take me some extraordinarily convincing
mathematics before I'll be ready to accept that 1/3 is irrational.
It is correct to say that infinitesimals don't exist:
there is always a
positive real number smaller than any candidate positive infinitesimal -- this is basic logic.
On 08/04/2025 16:17, Richard Heathfield wrote:
It will, however, take me some extraordinarily convincing
mathematics before I'll be ready to accept that 1/3 is irrational.
I don't think that's quite what Wij is claiming. He thinks, rather, that 0.333... is different from 1/3. No matter how far you
pursue that sequence, you have a number that is slightly less than
1/3.
In real analysis, the limit is 1/3 exactly. In Wij-analysis,
limits don't exist [as I understand it], because he doesn't
accept that there are no infinitesimals. It's like those who
dispute that 0.999... == 1 [exactly], and when challenged to
produce a number between 0.999... and 1, produce 0.999...5.
They have a point, as the Archimedean axiom is not one of thehttps://en.wikipedia.org/wiki/Racing_flags#Chequered_flag
things that gets mentioned much at school or in many undergrad
courses, and it seems like an arbitrary and unnecessary
addition to the rules. But we have no good and widely-known
notation for what can follow a "..."
so the Wijs of this world get mocked.
On 08/04/2025 16:17, Richard Heathfield wrote:
It will, however, take me some extraordinarily convincing
mathematics before I'll be ready to accept that 1/3 is irrational.
I don't think that's quite what Wij is claiming. He thinks, rather, that 0.333... is different from 1/3. No matter how far you
pursue that sequence, you have a number that is slightly less than
1/3. In real analysis, the limit is 1/3 exactly. In Wij-analysis,
limits don't exist [as I understand it], because he doesn't accept
that there are no infinitesimals. It's like those who dispute that
0.999... == 1 [exactly], and when challenged to produce a number
between 0.999... and 1, produce 0.999...5. They have a point, as
the Archimedean axiom is not one of the things that gets mentioned
much at school or in many undergrad courses, and it seems like an
arbitrary and unnecessary addition to the rules. But we have no good
and widely-known notation for what can follow a "...", so the Wijs of
this world get mocked. He doesn't help himself by refusing to learn
about the existing non-standard systems.
I don't plan to dwell on this, but I'm tempted to wonder how
infinitesimals fare under division.
On 4/7/25 5:42 PM, Lawrence D'Oliveiro wrote:
Somebody kept insisting that, even if my proposition is true for every
element in the list, it somehow goes false at the end. The end of an
infinite list! Yeah, right.
And that is because while every element on the list has an algorithm to construct it, that list is infinite, so you can't just put them *ALL* in
to one finite algorithm to compute any one you need at the moment.
On 08/04/2025 18:00, Mr Flibble wrote:
It is correct to say that infinitesimals don't exist:
It's correct in standard analysis, because it's an axiom. It's
not correct in number systems that have infinitesimals [and therefore do
not have the Archimedean axiom].
there is always a
positive real number smaller than any candidate positive infinitesimal
--
this is basic logic.
Google for "non-standard analysis" and for "surreal" [other
search engines are available]. Be warned that, at least the last time I looked, Wiki is, as so often with maths topics, remarkably opaque for non-mathematicians seeking enlightenment; there are more user-friendly introductions out there. No, I'm not going to recommend one, it's a
matter of personal taste.
And an infinite listing of values doesn't need to be computable, even if every number in the list is computable.
On Mon, 7 Apr 2025 06:47:43 -0400, Richard Damon wrote:
And an infinite listing of values doesn't need to be computable, even if
every number in the list is computable.
Computability is a characteristic of particular numbers. It is a characteristic of all the numbers in the list, and of the number that the Cantor construction tries to construct from those numbers in the list.
The fact that you can’t apply that characteristic to the set as a whole is irrelevant, since the set itself is not a number.
On Mon, 7 Apr 2025 18:24:36 -0400, Richard Damon wrote:
On 4/7/25 5:42 PM, Lawrence D'Oliveiro wrote:
Somebody kept insisting that, even if my proposition is true for every
element in the list, it somehow goes false at the end. The end of an
infinite list! Yeah, right.
And that is because while every element on the list has an algorithm to
construct it, that list is infinite, so you can't just put them *ALL* in
to one finite algorithm to compute any one you need at the moment.
But that’s exactly how computable numbers work.
I don't have to google anything: if you can multiply an infinitesimal by
0.5 then it isn't an infinitesimal
-- it was twice as large as another
real and all real numbers can be multiplied by 0.5.
Richard Damon <richard@damon-family.org> writes:
[...]
And from what I see, the issue is that while each of the numbers in
the list could be defined as constructable, in that a algorithm exists
that given n, it will give at least n digits of that number, there
doesn't need to be a master algorithm, that given k and n gives the
first n digits of the kth number, that can be shown to cover the full
set of constructable numbers (but perhaps an countable infinite subset
of them). Without that master algorithm, the method of constructing
the diagonal isn't actually an implementable algorithm.
We can't just iterate through all possible machines, because not all
machines are halting, let alone meet the requirements for the
construction machines.
Thus, we don't have an actual algorithm that makes the diagonal number
constructable.
So it's the Halting Problem that makes it impossible to computationally generate a list of all computable numbers, even though there are only a countable infinity of them.
Cantor's proof applies correctly to the real numbers. Given a purported infinite list of all the real numbers, we can construct a real number
that's not in the list; therefore there is no such list.
The same proof does not apply to rational numbers. We can generate an infinite list of all the rational numbers, and the diagonal construction demonstrates the existence of a number that's not on the list, but any
such number is not rational, so there's no contradiction. Same thing
for algebraic numbers. The rational and algebraic numbers *can* be
placed into a one-to-one mapping with the integers.
There are only a countable infinity of computable numbers (right?),
but if I understand correctly the halting problem prevents us
from generating a list of them. Given a purported list of all the
computable numbers, diagonalization gives us a computable number
that's not in the list.
There is no list of real numbers because there are strictly more real
numbers than integers.
There is no list of computable numbers, but for a different reason.
There are *not* strictly more computable numbers than integers,
but the halting problem makes it impossible to construct a list
of them. (Generate an ordered list of all possible algorithms in
some notation. Eliminate the ones that don't generate a computable
number. But we can't perform the elimination step because it would
require solving the halting problem.)
It feels intuitively weird that there is a set of numbers that is
countably infinite but we can't generate an ordered list of them.
But sometimes math is like that.
I suspect I'm still missing something. For one thing, I'm not
sure whether "we can't computationally generate the list" and "the
list doesn't exist" are equivalent statements.
On 08/04/2025 21:19, Mr Flibble wrote:
I don't have to google anything: if you can multiply an infinitesimal by
0.5 then it isn't an infinitesimal
Are you under some strange impression that there can be only one infinitesimal? If f is infinitesimal and g == 0.5*f, is g not also infinitesimal? In the hyperreals and in the surreals, infinitesimals
have their own algebra [and of course can be combined with (standard)
reals in all the usual ways].
FTAOD, a positive infinitesimal is a number f s.t. there is no integer N s.t. N*f > 1. In the standard reals, there is no such number,
but that is /only/ by fiat of the Archimedean axiom, and other number
systems exist in which that axiom does not hold.
-- it was twice as large as another >> real and all real numbers can be multiplied by 0.5.
Twice as large as another /number/. Spot the difference.
Op 08.apr.2025 om 20:44 schreef Andy Walker:
On 08/04/2025 16:17, Richard Heathfield wrote:
It will, however, take me some extraordinarily convincing
mathematics before I'll be ready to accept that 1/3 is irrational.
I don't think that's quite what Wij is claiming. He thinks,
rather, that 0.333... is different from 1/3. No matter how far you
pursue that sequence, you have a number that is slightly less than
1/3. In real analysis, the limit is 1/3 exactly. In Wij-analysis,
limits don't exist [as I understand it], because he doesn't accept
that there are no infinitesimals. It's like those who dispute that
0.999... == 1 [exactly], and when challenged to produce a number
between 0.999... and 1, produce 0.999...5. They have a point, as
the Archimedean axiom is not one of the things that gets mentioned
much at school or in many undergrad courses, and it seems like an
arbitrary and unnecessary addition to the rules. But we have no good
and widely-known notation for what can follow a "...", so the Wijs of
this world get mocked. He doesn't help himself by refusing to learn
about the existing non-standard systems.
To me it seems that it comes down to the definition of real numbers.
One definition is in terms of limits of a series of numbers. Once one understands the definition of limits, it is clear that different series
can be used for the same real number. The real number 1, e.g., can be
defined by (amongst others) the following series:
0.9, 0.99, 0.999, 0.9999, ...
1, 1, 1, 1, ....
1.1, 1.01, 1.001, 1.0001, ...
Two series X(n) and Y(n) indicate the same real number if for each small
ε one can find a number N so that for all n>N |X(n)-Y(n)| < ε.
(See https://en.wikipedia.org/wiki/Construction_of_the_real_numbers) Apparently the Wij-analysis is not about real numbers, but it is not
clear what alternative numbers are used.
Richard Damon <richard@damon-family.org> writes:
[...]
And from what I see, the issue is that while each of the numbers in
the list could be defined as constructable, in that a algorithm exists
that given n, it will give at least n digits of that number, there
doesn't need to be a master algorithm, that given k and n gives the
first n digits of the kth number, that can be shown to cover the full
set of constructable numbers (but perhaps an countable infinite subset
of them). Without that master algorithm, the method of constructing
the diagonal isn't actually an implementable algorithm.
We can't just iterate through all possible machines, because not all
machines are halting, let alone meet the requirements for the
construction machines.
Thus, we don't have an actual algorithm that makes the diagonal number
constructable.
So it's the Halting Problem that makes it impossible to computationally generate a list of all computable numbers, even though there are only a countable infinity of them.
Cantor's proof applies correctly to the real numbers. Given a purported infinite list of all the real numbers, we can construct a real number
that's not in the list; therefore there is no such list.
The same proof does not apply to rational numbers. We can generate an infinite list of all the rational numbers, and the diagonal construction demonstrates the existence of a number that's not on the list, but any
such number is not rational, so there's no contradiction. Same thing
for algebraic numbers. The rational and algebraic numbers *can* be
placed into a one-to-one mapping with the integers.
There are only a countable infinity of computable numbers (right?),
but if I understand correctly the halting problem prevents us
from generating a list of them.
Given a purported list of all the
computable numbers, diagonalization gives us a computable number
that's not in the list.
There is no list of real numbers because there are strictly more real
numbers than integers.
There is no list of computable numbers, but for a different reason.
There are *not* strictly more computable numbers than integers,
but the halting problem makes it impossible to construct a list
of them. (Generate an ordered list of all possible algorithms in
some notation. Eliminate the ones that don't generate a computable
number. But we can't perform the elimination step because it would
require solving the halting problem.)
It feels intuitively weird that there is a set of numbers that is
countably infinite but we can't generate an ordered list of them.
But sometimes math is like that.
I suspect I'm still missing something. For one thing, I'm not
sure whether "we can't computationally generate the list" and "the
list doesn't exist" are equivalent statements.
I suspect I'm still missing something. For one thing, I'm not
sure whether "we can't computationally generate the list" and "the
list doesn't exist" are equivalent statements.
On Tue, 2025-04-08 at 19:44 +0100, Andy Walker wrote:
On 08/04/2025 16:17, Richard Heathfield wrote:
It will, however, take me some extraordinarily convincing
mathematics before I'll be ready to accept that 1/3 is irrational.
I don't think that's quite what Wij is claiming. He thinks,
rather, that 0.333... is different from 1/3. No matter how far you
pursue that sequence, you have a number that is slightly less than
1/3. In real analysis, the limit is 1/3 exactly. In Wij-analysis,
limits don't exist [as I understand it], because he doesn't accept
that there are no infinitesimals. It's like those who dispute that
0.999... == 1 [exactly], and when challenged to produce a number
between 0.999... and 1, produce 0.999...5. They have a point, as
the Archimedean axiom is not one of the things that gets mentioned
much at school or in many undergrad courses, and it seems like an
arbitrary and unnecessary addition to the rules. But we have no good
and widely-known notation for what can follow a "...", so the Wijs of
this world get mocked. He doesn't help himself by refusing to learn
about the existing non-standard systems.
Lots of excuses like POOH. You cannot hide the fact that you don't have a valid proof in those kinds of argument.
If you propose a proof, be sure you checked against the file I provided.
I have no no time for garbage talk.
[Cross-posted to sci-logic. Please set follow-ups as needed.]
On 07/04/2025 17:16, Julio Di Egidio wrote:
My Gist, Cantor's diagonal argument (in Rocq):
<https://gist.github.com/jp-diegidio/eb05f6265c38b35c85853514ed46ab47>
<< We go for the simplest construction, which
is not in terms of sets, just (type-theoretic)
functions plus the most basic functional
extensionality, i.e. eta-conversion. >>
In fact, learning some formal mathematics with some tool
that supports it I think makes one's informal mathematics
way sharper, especially for those who are [self-taught]
(whatever that means).
That said, I find my formalisation neither particularly
pretty nor particularly clear: corrections and suggestions
are welcome, especially in the direction of minimising
assumptions.
In particular, I think I am botching this:
```
(* given hypothesis [f = g] *)
assert (En : forall n, f n = g n)
by congruence. (** indisc. of ident. *)
```
That is not indiscernibility of identicals,
which rather looks like `forall P, P f = P g`,
but it's not eta-expansion either. What is it?
Mr Flibble <flibble@red-dwarf.jmc.corp> writes:
[...]
I don't have to google anything: if you can multiply an infinitesimal
by 0.5 then it isn't an infinitesimal -- it was twice as large as
another real and all real numbers can be multiplied by 0.5.
Infinitesimals are not real numbers.
There are number systems that include infinitesimals in addition to all
the real numbers, and that are internally consistent. The rules can
vary from one such number system to another, so there's probably no one answer to what happens when you multiply an infinitesimal by 0.5 in such
a system.
If you're interested in learning more, search for "surreal numbers" or "hyperreal numbers". If you're not, don't.
On 4/8/25 6:59 PM, Andy Walker wrote:
On 08/04/2025 21:19, Mr Flibble wrote:And then you get to the real weird fact, that when you get to
I don't have to google anything: if you can multiply an infinitesimal
by 0.5 then it isn't an infinitesimal
Are you under some strange impression that there can be only
one
infinitesimal? If f is infinitesimal and g == 0.5*f, is g not also
infinitesimal? In the hyperreals and in the surreals, infinitesimals
have their own algebra [and of course can be combined with (standard)
reals in all the usual ways].
FTAOD, a positive infinitesimal is a number f s.t. there is no
integer N s.t. N*f > 1. In the standard reals, there is no such
number,
but that is /only/ by fiat of the Archimedean axiom, and other number
systems exist in which that axiom does not hold.
-- it was twice as large as
another
real and all real numbers can be multiplied by 0.5.
Twice as large as another /number/. Spot the difference.
real-scaled infintesimals (where you can talk about 0.5 * the base infinitesimal) you also get the fact that there is a strange gap between
the "smallest" real-scaled-infinitisimal and zero, allowing us to create
a second-level infinitesimal, and then a third, and so on.
Consistant definitions of Mathematics exist that handles all of that,
but some things get a bit strange, and you need to take care of things
that aren't an issue with "normal" Real Numbers.
On Tue, 08 Apr 2025 15:46:54 -0700, Keith Thompson wrote:
If you're interested in learning more, search for "surreal numbers" or
"hyperreal numbers". If you're not, don't.
Surreal numbers are bullshit as they don't actually exist, logically (as I have show). Bullshit can be internally consistent with itself.
/Flibble
Mr Flibble <flibble@red-dwarf.jmc.corp> wrote:
On Tue, 08 Apr 2025 15:46:54 -0700, Keith Thompson wrote:
[ .... ]
If you're interested in learning more, search for "surreal numbers" or
"hyperreal numbers". If you're not, don't.
Surreal numbers are bullshit as they don't actually exist, logically
(as I have show). Bullshit can be internally consistent with itself.
What exactly do you mean by a mathematical entity "not existing"? What
is your test which partitions such entities into "existing" and "non-existing"?
/Flibble
On Wed, 2025-04-09 at 13:48 +0100, Richard Heathfield wrote:
On 09/04/2025 13:25, wij wrote:
On Tue, 2025-04-08 at 19:44 +0100, Andy Walker wrote:
On 08/04/2025 16:17, Richard Heathfield wrote:
It will, however, take me some extraordinarily convincing
mathematics before I'll be ready to accept that 1/3 is irrational.
I don't think that's quite what Wij is claiming. He thinks,
rather, that 0.333... is different from 1/3. No matter how far you
pursue that sequence, you have a number that is slightly less than
1/3. In real analysis, the limit is 1/3 exactly. In Wij-analysis,
limits don't exist [as I understand it], because he doesn't accept
that there are no infinitesimals. It's like those who dispute that
0.999... == 1 [exactly], and when challenged to produce a number
between 0.999... and 1, produce 0.999...5. They have a point, as
the Archimedean axiom is not one of the things that gets mentioned
much at school or in many undergrad courses, and it seems like an
arbitrary and unnecessary addition to the rules. But we have no good >>>> and widely-known notation for what can follow a "...", so the Wijs of
this world get mocked. He doesn't help himself by refusing to learn
about the existing non-standard systems.
Lots of excuses like POOH. You cannot hide the fact that you don't have a >>> valid proof in those kinds of argument.
If you propose a proof, be sure you checked against the file I provided. >>> I have no no time for garbage talk.
I have read that document, about which I have a simple question.
From Theorem 2 and Axiom 2, if x can be expressed in the form of
p/q, then p and q will be infinite numbers (non-natural numbers).
Therefore, x is not a rational number. And since a non-rational
number is an irrational number, the proposition is proved.
Let p = 1
Let q = 3
Is it or is it not your contention that p and q are "infinite"
(non-natural) numbers?
The audience of the file was originally intended to include 12 years old kids.
Wordings in the file wont' be precise enough to meet rigorous requirements. The mentioned paragraph was revised (along with several others):
Theorem 2: ℚ+ℚ=ℚ (the sum of a rational number and a rational number is still a
rational number), but it is only true for finite addition steps.
Proof: Let Q'={p/q| p,q∈ℕ, q≠0 and p/q>0}, then Q'⊂ℚ. Since the sum of any two
terms in Q' is greater than the individual terms, the sum q of the
infinite terms (q=q₁+q₂+q₃...) is not a fixed number.
What I intended to mean is: 0.999...= 999.../1000... (in p/q form)
Since p,q will be infinitely long to denote/define 0.999..., p,q won't be natural numbers. Thus, "ℚ+ℚ=ℚ" is conditionally true (so false).
But I still think your English is worse than olcott's (and mine).
Prediction: you will evade the question. Why not surprise me?Ok, I evade more clarification.
Mr Flibble <flibble@red-dwarf.jmc.corp> wrote:
On Wed, 09 Apr 2025 14:11:54 +0000, Alan Mackenzie wrote:
Mr Flibble <flibble@red-dwarf.jmc.corp> wrote:
On Tue, 08 Apr 2025 15:46:54 -0700, Keith Thompson wrote:
[ .... ]
If you're interested in learning more, search for "surreal numbers"
or "hyperreal numbers". If you're not, don't.
Surreal numbers are bullshit as they don't actually exist, logically
(as I have show). Bullshit can be internally consistent with itself.
What exactly do you mean by a mathematical entity "not existing"?
What is your test which partitions such entities into "existing" and
"non-existing"?
/Flibble
Simple: things that make no logical sense don't exist: ....
Surreal numbers do make logical sense. They form an ordered field which
has the real numbers as a subfield.
.... logically a real number always has a number smaller than it ....
Every stricly positive surreal number has a number smaller than it, too.
.... so trying to put a "surreal" infinitesimal on the same number line
as a "real" makes no logical sense: in fact I will go as far to say
that it is a category error.
The surreal number line is not the real number line, so trying to put a surreal on the latter indeed makes no sense. It might even constitute a category error, as you suggest.
That, however, has no bearing on the existence of surreal numbers. They don't create inconsistencies, hence do exist, and have been studied intensively.
/Flibble
On Wed, 09 Apr 2025 14:11:54 +0000, Alan Mackenzie wrote:
Mr Flibble <flibble@red-dwarf.jmc.corp> wrote:
On Tue, 08 Apr 2025 15:46:54 -0700, Keith Thompson wrote:
[ .... ]
If you're interested in learning more, search for "surreal numbers" or >>>> "hyperreal numbers". If you're not, don't.
Surreal numbers are bullshit as they don't actually exist, logically
(as I have show). Bullshit can be internally consistent with itself.
What exactly do you mean by a mathematical entity "not existing"? What
is your test which partitions such entities into "existing" and
"non-existing"?
/Flibble
Simple: things that make no logical sense don't exist: ....
.... logically a real number always has a number smaller than it ....
.... so trying to put a "surreal" infinitesimal on the same number line
as a "real" makes no logical sense: in fact I will go as far to say
that it is a category error.
/Flibble
On Wed, 2025-04-09 at 15:48 +0100, Richard Heathfield wrote:
On 09/04/2025 15:31, wij wrote:
On Wed, 2025-04-09 at 13:48 +0100, Richard Heathfield wrote:
On 09/04/2025 13:25, wij wrote:
On Tue, 2025-04-08 at 19:44 +0100, Andy Walker wrote:
On 08/04/2025 16:17, Richard Heathfield wrote:
It will, however, take me some extraordinarily convincingI don't think that's quite what Wij is claiming. He thinks,
mathematics before I'll be ready to accept that 1/3 is irrational. >>>>>>
rather, that 0.333... is different from 1/3. No matter how far you >>>>>> pursue that sequence, you have a number that is slightly less than >>>>>> 1/3. In real analysis, the limit is 1/3 exactly. In Wij-analysis, >>>>>> limits don't exist [as I understand it], because he doesn't accept >>>>>> that there are no infinitesimals. It's like those who dispute that >>>>>> 0.999... == 1 [exactly], and when challenged to produce a number
between 0.999... and 1, produce 0.999...5. They have a point, as >>>>>> the Archimedean axiom is not one of the things that gets mentioned >>>>>> much at school or in many undergrad courses, and it seems like an
arbitrary and unnecessary addition to the rules. But we have no good >>>>>> and widely-known notation for what can follow a "...", so the Wijs of >>>>>> this world get mocked. He doesn't help himself by refusing to learn >>>>>> about the existing non-standard systems.
Lots of excuses like POOH. You cannot hide the fact that you don't have a >>>>> valid proof in those kinds of argument.
If you propose a proof, be sure you checked against the file I provided. >>>>> I have no no time for garbage talk.
I have read that document, about which I have a simple question.
From Theorem 2 and Axiom 2, if x can be expressed in the form of
p/q, then p and q will be infinite numbers (non-natural numbers).
Therefore, x is not a rational number. And since a non-rational
number is an irrational number, the proposition is proved.
Let p = 1
Let q = 3
Is it or is it not your contention that p and q are "infinite"
(non-natural) numbers?
The audience of the file was originally intended to include 12 years old kids.
Wordings in the file wont' be precise enough to meet rigorous requirements. >>> The mentioned paragraph was revised (along with several others):
Theorem 2: ℚ+ℚ=ℚ (the sum of a rational number and a rational number is still a
rational number), but it is only true for finite addition steps.
Proof: Let Q'={p/q| p,q∈ℕ, q≠0 and p/q>0}, then Q'⊂ℚ. Since the sum of any two
terms in Q' is greater than the individual terms, the sum q of the
infinite terms (q=q₁+q₂+q₃...) is not a fixed number.
What I intended to mean is: 0.999...= 999.../1000... (in p/q form)
Since p,q will be infinitely long to denote/define 0.999..., p,q won't be >>> natural numbers. Thus, "ℚ+ℚ=ℚ" is conditionally true (so false).
But I still think your English is worse than olcott's (and mine).
Charmed, I'm sure.
Prediction: you will evade the question. Why not surprise me?Ok, I evade more clarification.
I deduce from what you intended to mean (and that's very classy
English, so well done you) that you didn't intend to mean that 1
and 3 are "infinite".
And you're right. 1 and 3 are both integers. Natural numbers.
Whole numbers. Finite numbers. Not infinite.
Let us calculate the ratio of these two integers, 1/3. Oh look,
it's 0.3r. So 0.3r is the ratio of two integers (i.e. rational)
after all. Quelle surprise!
The correct equality is 1/3= 0.333... + nonzero_remainder.
If you use it to prove, that proof never finishes. Thus, invalid.
On Wed, 09 Apr 2025 16:17:37 +0000, Alan Mackenzie wrote:
Mr Flibble <flibble@red-dwarf.jmc.corp> wrote:
On Wed, 09 Apr 2025 14:11:54 +0000, Alan Mackenzie wrote:
Mr Flibble <flibble@red-dwarf.jmc.corp> wrote:
On Tue, 08 Apr 2025 15:46:54 -0700, Keith Thompson wrote:
[ .... ]
If you're interested in learning more, search for "surreal numbers" >>>>>> or "hyperreal numbers". If you're not, don't.
Surreal numbers are bullshit as they don't actually exist, logically >>>>> (as I have show). Bullshit can be internally consistent with itself.
What exactly do you mean by a mathematical entity "not existing"?
What is your test which partitions such entities into "existing" and
"non-existing"?
/Flibble
Simple: things that make no logical sense don't exist: ....
Surreal numbers do make logical sense. They form an ordered field which
has the real numbers as a subfield.
.... logically a real number always has a number smaller than it ....
Every stricly positive surreal number has a number smaller than it, too.
.... so trying to put a "surreal" infinitesimal on the same number line
as a "real" makes no logical sense: in fact I will go as far to say
that it is a category error.
The surreal number line is not the real number line, so trying to put a
surreal on the latter indeed makes no sense. It might even constitute a
category error, as you suggest.
That, however, has no bearing on the existence of surreal numbers. They
don't create inconsistencies, hence do exist, and have been studied
intensively.
/Flibble
The category error I identified runs contrary to your claim that the reals are a sub-field of the surreals ....
.... as that would suggest that reals and surreals can exist on the
same number line as a real is-a surreal which is logically unsound for
the reason I have already given.
/Flibble
/Flibble
Mr Flibble <flibble@red-dwarf.jmc.corp> wrote:
On Wed, 09 Apr 2025 16:17:37 +0000, Alan Mackenzie wrote:
Mr Flibble <flibble@red-dwarf.jmc.corp> wrote:
On Wed, 09 Apr 2025 14:11:54 +0000, Alan Mackenzie wrote:
Mr Flibble <flibble@red-dwarf.jmc.corp> wrote:
On Tue, 08 Apr 2025 15:46:54 -0700, Keith Thompson wrote:
[ .... ]
If you're interested in learning more, search for "surreal
numbers"
or "hyperreal numbers". If you're not, don't.
Surreal numbers are bullshit as they don't actually exist,
logically (as I have show). Bullshit can be internally consistent >>>>>> with itself.
What exactly do you mean by a mathematical entity "not existing"?
What is your test which partitions such entities into "existing" and >>>>> "non-existing"?
/Flibble
Simple: things that make no logical sense don't exist: ....
Surreal numbers do make logical sense. They form an ordered field
which has the real numbers as a subfield.
.... logically a real number always has a number smaller than it ....
Every stricly positive surreal number has a number smaller than it,
too.
.... so trying to put a "surreal" infinitesimal on the same number
line as a "real" makes no logical sense: in fact I will go as far to
say that it is a category error.
The surreal number line is not the real number line, so trying to put
a surreal on the latter indeed makes no sense. It might even
constitute a category error, as you suggest.
That, however, has no bearing on the existence of surreal numbers.
They don't create inconsistencies, hence do exist, and have been
studied intensively.
/Flibble
The category error I identified runs contrary to your claim that the
reals are a sub-field of the surreals ....
It's not my claim. It's an established mathematical fact.
.... as that would suggest that reals and surreals can exist on the
same number line as a real is-a surreal which is logically unsound for
the reason I have already given.
You're mistaken. You appear not to understand the implications and
meaning of the word "is" in that last sentence. In this subthread, you haven't given any reason for your assertion. And even if you had, you'd still be mistaken.
/Flibble
/Flibble
Stick to the problem. Such puzzle won't prove the "0.999... problem".
Richard Damon <richard@damon-family.org> writes:
[...]
And from what I see, the issue is that while each of the numbers in
the list could be defined as constructable, in that a algorithm exists
that given n, it will give at least n digits of that number, there
doesn't need to be a master algorithm, that given k and n gives the
first n digits of the kth number, that can be shown to cover the full
set of constructable numbers (but perhaps an countable infinite subset
of them). Without that master algorithm, the method of constructing
the diagonal isn't actually an implementable algorithm.
We can't just iterate through all possible machines, because not all
machines are halting, let alone meet the requirements for the
construction machines.
Thus, we don't have an actual algorithm that makes the diagonal number
constructable.
So it's the Halting Problem that makes it impossible to computationally generate a list of all computable numbers, even though there are only a countable infinity of them.
Cantor's proof applies correctly to the real numbers. Given a purported infinite list of all the real numbers, we can construct a real number
that's not in the list; therefore there is no such list.
The same proof does not apply to rational numbers. We can generate an infinite list of all the rational numbers, and the diagonal construction demonstrates the existence of a number that's not on the list, but any
such number is not rational, so there's no contradiction. Same thing
for algebraic numbers. The rational and algebraic numbers *can* be
placed into a one-to-one mapping with the integers.
There are only a countable infinity of computable numbers (right?),
but if I understand correctly the halting problem prevents us
from generating a list of them. Given a purported list of all the
computable numbers, diagonalization gives us a computable number
that's not in the list.
There is no list of real numbers because there are strictly more real
numbers than integers.
There is no list of computable numbers, but for a different reason.
There are *not* strictly more computable numbers than integers,
but the halting problem makes it impossible to construct a list
of them. (Generate an ordered list of all possible algorithms in
some notation. Eliminate the ones that don't generate a computable
number. But we can't perform the elimination step because it would
require solving the halting problem.)
It feels intuitively weird that there is a set of numbers that is
countably infinite but we can't generate an ordered list of them.
But sometimes math is like that.
I suspect I'm still missing something. For one thing, I'm not
sure whether "we can't computationally generate the list" and "the
list doesn't exist" are equivalent statements.
On Wed, 2025-04-09 at 18:10 +0100, Richard Heathfield wrote:
On 09/04/2025 17:45, wij wrote:
<snip>
Stick to the problem. Such puzzle won't prove the "0.999... problem".
You know what? You're right. 1/3 is irrational, 1 is infinite, 3
is unnatural, and Achilles never did catch the tortoise. I don't
know what I was thinking, claiming that rationals are rational!
Clearly the arithmetic I learned at school was deeply flawed, and
I was naif to think it could be trusted. How happy I am, now that
I have learned the truth from some faceless bloke on the
Internet. Well done you, eh? 1/3 is irrational; who knew?
1/3 is rational "by definition".
0.333... (repeating decimal) is irrational.
Achilles puzzle gives us two models. One can catch the tortoise, one
cannot. Since we believe Achilles can, so it make? people to believe
0.99.... will finish (false from the other model. Also the term
'repeating' implies (define) never terminate)
On Wed, 09 Apr 2025 16:49:39 +0000, Alan Mackenzie wrote:
Mr Flibble <flibble@red-dwarf.jmc.corp> wrote:
On Wed, 09 Apr 2025 16:17:37 +0000, Alan Mackenzie wrote:
Mr Flibble <flibble@red-dwarf.jmc.corp> wrote:
On Wed, 09 Apr 2025 14:11:54 +0000, Alan Mackenzie wrote:
Mr Flibble <flibble@red-dwarf.jmc.corp> wrote:
On Tue, 08 Apr 2025 15:46:54 -0700, Keith Thompson wrote:
[ .... ]
If you're interested in learning more, search for "surreal
numbers"
or "hyperreal numbers". If you're not, don't.
Surreal numbers are bullshit as they don't actually exist,
logically (as I have show). Bullshit can be internally consistent >>>>>>> with itself.
What exactly do you mean by a mathematical entity "not existing"?
What is your test which partitions such entities into "existing" and >>>>>> "non-existing"?
/Flibble
Simple: things that make no logical sense don't exist: ....
Surreal numbers do make logical sense. They form an ordered field
which has the real numbers as a subfield.
.... logically a real number always has a number smaller than it ....
Every stricly positive surreal number has a number smaller than it,
too.
.... so trying to put a "surreal" infinitesimal on the same number
line as a "real" makes no logical sense: in fact I will go as far to >>>>> say that it is a category error.
The surreal number line is not the real number line, so trying to put
a surreal on the latter indeed makes no sense. It might even
constitute a category error, as you suggest.
That, however, has no bearing on the existence of surreal numbers.
They don't create inconsistencies, hence do exist, and have been
studied intensively.
/Flibble
The category error I identified runs contrary to your claim that the
reals are a sub-field of the surreals ....
It's not my claim. It's an established mathematical fact.
.... as that would suggest that reals and surreals can exist on the
same number line as a real is-a surreal which is logically unsound for
the reason I have already given.
You're mistaken. You appear not to understand the implications and
meaning of the word "is" in that last sentence. In this subthread, you
haven't given any reason for your assertion. And even if you had, you'd
still be mistaken.
/Flibble
/Flibble
No, it is you who is mistaken: my arguments are logically sound and yours
are not.
/Flibble
Mr Flibble <flibble@red-dwarf.jmc.corp> wrote:
On Wed, 09 Apr 2025 16:49:39 +0000, Alan Mackenzie wrote:
Mr Flibble <flibble@red-dwarf.jmc.corp> wrote:
On Wed, 09 Apr 2025 16:17:37 +0000, Alan Mackenzie wrote:
Mr Flibble <flibble@red-dwarf.jmc.corp> wrote:
On Wed, 09 Apr 2025 14:11:54 +0000, Alan Mackenzie wrote:
Mr Flibble <flibble@red-dwarf.jmc.corp> wrote:
On Tue, 08 Apr 2025 15:46:54 -0700, Keith Thompson wrote:
[ .... ]
If you're interested in learning more, search for "surreal
numbers"
or "hyperreal numbers". If you're not, don't.
Surreal numbers are bullshit as they don't actually exist,
logically (as I have show). Bullshit can be internally
consistent with itself.
What exactly do you mean by a mathematical entity "not existing"? >>>>>>> What is your test which partitions such entities into "existing" >>>>>>> and "non-existing"?
/Flibble
Simple: things that make no logical sense don't exist: ....
Surreal numbers do make logical sense. They form an ordered field
which has the real numbers as a subfield.
.... logically a real number always has a number smaller than it
....
Every stricly positive surreal number has a number smaller than it,
too.
.... so trying to put a "surreal" infinitesimal on the same number >>>>>> line as a "real" makes no logical sense: in fact I will go as far
to say that it is a category error.
The surreal number line is not the real number line, so trying to
put a surreal on the latter indeed makes no sense. It might even
constitute a category error, as you suggest.
That, however, has no bearing on the existence of surreal numbers.
They don't create inconsistencies, hence do exist, and have been
studied intensively.
/Flibble
The category error I identified runs contrary to your claim that the
reals are a sub-field of the surreals ....
It's not my claim. It's an established mathematical fact.
.... as that would suggest that reals and surreals can exist on the
same number line as a real is-a surreal which is logically unsound
for the reason I have already given.
You're mistaken. You appear not to understand the implications and
meaning of the word "is" in that last sentence. In this subthread,
you haven't given any reason for your assertion. And even if you had,
you'd still be mistaken.
/Flibble
/Flibble
No, it is you who is mistaken: my arguments are logically sound and
yours are not.
Wrong. You are ignorant of the maths involved. You seem to be a crank
and a troll. Do you even understand what an ordered field is? I
strongly suspect not.
I haven't been debating with you, I've been trying to educate you,
trying to get you to see that there's a lot more worth learning which is currently outside of your rather limited perspective.
If you don't want to learn, it's not my loss.
On 07/04/2025 08:33, Lawrence D'Oliveiro wrote:
On Sun, 6 Apr 2025 23:38:25 +0100, Richard Heathfield wrote:
On 06/04/2025 23:01, Lawrence D'Oliveiro wrote:
On Sun, 6 Apr 2025 07:53:06 +0100, Richard Heathfield wrote:
After infinitely many steps ...
I.e. never.
If you mean you can never know all the digits, hey, you're right.
No algorithm can derive the number. It's incomputable.
That’s not what “incomputable” means.
Yeah, it is. We've already had this argument. Turing won: "The
"computable" numbers may be described briefly as the real numbers whose expressions as a decimal are calculable by finite means."
Like anything in mathematics, if you're going to overturn a
long-established proof you're going to need a better argument than that
you don't understand what it proves.
The paper clearly talks about the process continuing indefinitely.
On 4/8/25 5:02 PM, Lawrence D'Oliveiro wrote:
On Mon, 7 Apr 2025 18:24:36 -0400, Richard Damon wrote:
And that is because while every element on the list has an algorithm
to construct it, that list is infinite, so you can't just put them
*ALL* in to one finite algorithm to compute any one you need at the
moment.
But that’s exactly how computable numbers work.
No, *A* computable number has a finite algorithm that computes it.
Finite in having finite instructions in its algorithm and finite states
to process.
The problem is your "master" algorithm need the algorithms of *ALL* the computable numbers within it, which is an infinite number of algorithms,
and thus isn't itself a finite algorithm.
On Tue, 8 Apr 2025 18:57:17 -0400, Richard Damon wrote:
On 4/8/25 5:02 PM, Lawrence D'Oliveiro wrote:
On Mon, 7 Apr 2025 18:24:36 -0400, Richard Damon wrote:
And that is because while every element on the list has an algorithm
to construct it, that list is infinite, so you can't just put them
*ALL* in to one finite algorithm to compute any one you need at the
moment.
But that’s exactly how computable numbers work.
No, *A* computable number has a finite algorithm that computes it.
Finite in having finite instructions in its algorithm and finite states
to process.
The problem is your "master" algorithm need the algorithms of *ALL* the
computable numbers within it, which is an infinite number of algorithms,
and thus isn't itself a finite algorithm.
That is the problem with the Cantor construction, not with my disproof of
it. My disproof of it only needs a finite number of list elements at any point in the proof. The Cantor construction needs them all.
Even there, proving that a computable list of all computable numbers
does not exist.
The normal maths way of putting all this is that the set of computable numbers is countable (can be "listed" / there exists a one-to-one
mapping from N to the computable numbers), and when the diagonal
argument is applied it results (as above) in a non-computable number
that's (obviously) not in the list.
The same proof does not apply to rational numbers. We can generate an infinite list of all the rational numbers, and the diagonal construction demonstrates the existence of a number that's not on the list ...
There is no list of computable numbers, but for a different reason.
There are *not* strictly more computable numbers than integers,
but the halting problem makes it impossible to construct a list of them.
(Generate an ordered list of all possible algorithms in some notation. Eliminate the ones that don't generate a computable number. But we
can't perform the elimination step because it would require solving the halting problem.)
On Mon, 7 Apr 2025 20:48:27 -0400, Richard Damon wrote:
The paper clearly talks about the process continuing indefinitely.
Note the key point about any computation of a computable number is that
the answer *converges* to the exact result in the limit. As you compute
more and more digits, the discrepancy between your approximation and the correct answer can be made as close to zero as you like, just as long as
you don’t ask for it to be zero.
The Cantor construction does not converge.
On Tue, 8 Apr 2025 18:57:17 -0400, Richard Damon wrote:
On 4/8/25 5:02 PM, Lawrence D'Oliveiro wrote:
On Mon, 7 Apr 2025 18:24:36 -0400, Richard Damon wrote:
And that is because while every element on the list has an algorithm
to construct it, that list is infinite, so you can't just put them
*ALL* in to one finite algorithm to compute any one you need at the
moment.
But that’s exactly how computable numbers work.
No, *A* computable number has a finite algorithm that computes it.
Finite in having finite instructions in its algorithm and finite states
to process.
The problem is your "master" algorithm need the algorithms of *ALL* the
computable numbers within it, which is an infinite number of algorithms,
and thus isn't itself a finite algorithm.
That is the problem with the Cantor construction, not with my disproof of
it. My disproof of it only needs a finite number of list elements at any point in the proof. The Cantor construction needs them all.
On Mon, 7 Apr 2025 09:21:25 +0100, Richard Heathfield wrote:
On 07/04/2025 08:33, Lawrence D'Oliveiro wrote:
On Sun, 6 Apr 2025 23:38:25 +0100, Richard Heathfield wrote:
On 06/04/2025 23:01, Lawrence D'Oliveiro wrote:
On Sun, 6 Apr 2025 07:53:06 +0100, Richard Heathfield wrote:
After infinitely many steps ...
I.e. never.
If you mean you can never know all the digits, hey, you're right.
No algorithm can derive the number. It's incomputable.
That’s not what “incomputable” means.
Yeah, it is. We've already had this argument. Turing won: "The
"computable" numbers may be described briefly as the real numbers whose
expressions as a decimal are calculable by finite means."
You didn’t even read to the end of his first paragraph: “According to my definition, a number is computable if its decimal can be written down by a machine”.
Second paragraph: “The include, for instance, the real parts of all algebraic numbers, the real parts of the zeros of the Bessel functions,
the numbers π, e, etc”.
Tell me your interpretation of how the digits of π “as a decimal are calculable by finite means”.
Like anything in mathematics, if you're going to overturn a
long-established proof you're going to need a better argument than that
you don't understand what it proves.
Go on, then, show the flaw in my proof by induction. Because if proof-by- induction is not a “long-established proof”, then tell me what is.
On Mon, 7 Apr 2025 11:40:58 +0300, Mikko wrote:
Even there, proving that a computable list of all computable numbers
does not exist.
But the ability to construct an infinite list of reals or computables is a basic assumption of the Cantor construction.
On Mon, 7 Apr 2025 11:40:58 +0300, Mikko wrote:
Even there, proving that a computable list of all computable numbers
does not exist.
But the ability to construct an infinite list of reals or computables is a basic assumption of the Cantor construction. If you’re saying it’s not possible to construct such a list, then that’s the end of Cantor’s proof.
On Mon, 7 Apr 2025 20:48:27 -0400, Richard Damon wrote:
The paper clearly talks about the process continuing indefinitely.
Note the key point about any computation of a computable number is that
the answer *converges* to the exact result in the limit. As you compute
more and more digits, the discrepancy between your approximation and the correct answer can be made as close to zero as you like, just as long as
you don’t ask for it to be zero.
The Cantor construction does not converge.
On Wed, 9 Apr 2025 18:49:54 +0100, Mike Terry wrote:
The normal maths way of putting all this is that the set of computable
numbers is countable (can be "listed" / there exists a one-to-one
mapping from N to the computable numbers), and when the diagonal
argument is applied it results (as above) in a non-computable number
that's (obviously) not in the list.
Assume the list consists of algorithms for all computable numbers which
are guaranteed to terminate, ordered according to some Gödel numbering.
Apply the Cantor construction; that algorithm is also guaranteed to terminate. Therefore it must have a Gödel number, and be located at a
finite place in the list -- call it Nₙ.
So what happens when you ask the Cantor construction to compute digit Nₙ
of its number? It gets stuck in an endless loop. That means it is not guaranteed to terminate. Therefore it cannot occur in the list.
But if you take it out of the list, then it *will* terminate, because all
the rest of the elements in the list do so. Put it in, it doesn’t belong: take it out, it does belong.
So, by reductio ad absurdum, the assumption that the Cantor construction
for such a list even *exists* is false.
On 4/9/25 8:41 PM, Lawrence D'Oliveiro wrote:
On Tue, 8 Apr 2025 18:57:17 -0400, Richard Damon wrote:
The problem is your "master" algorithm need the algorithms of *ALL*
the computable numbers within it, which is an infinite number of
algorithms, and thus isn't itself a finite algorithm.
That is the problem with the Cantor construction, not with my disproof
of it. My disproof of it only needs a finite number of list elements at
any point in the proof. The Cantor construction needs them all.
But a finite list can't get you to the needed arbitrary precision
needed.
Not also, Cantor only claimed that the numbers weren't countable, he
didn't talk about computable. That is a later work looking at the proof.
The problem with the "Cantor Construction" is that one of the steps,
"Compute to the Nth digit of the Nth number" isn't computable with a
finite algorithm.
On 2025-04-10 00:50:10 +0000, Lawrence D'Oliveiro said:
On Mon, 7 Apr 2025 20:48:27 -0400, Richard Damon wrote:
The paper clearly talks about the process continuing indefinitely.
Note the key point about any computation of a computable number is that
the answer *converges* to the exact result in the limit. As you compute
more and more digits, the discrepancy between your approximation and
the correct answer can be made as close to zero as you like, just as
long as you don’t ask for it to be zero.
The Cantor construction does not converge.
If it is a computable number it does converge.
On 4/8/25 5:05 PM, Lawrence D'Oliveiro wrote:
On Mon, 7 Apr 2025 06:47:43 -0400, Richard Damon wrote:
And an infinite listing of values doesn't need to be computable, even
if every number in the list is computable.
Computability is a characteristic of particular numbers. It is a
characteristic of all the numbers in the list, and of the number that
the Cantor construction tries to construct from those numbers in the
list.
The fact that you can’t apply that characteristic to the set as a whole
is irrelevant, since the set itself is not a number.
Right, so the DIAGONAL number, which you claim to be computable, needs a finite algorithm to do so.
The algorithm described is NOT FINITE, as it includes the infinite
number of algorithms to compute all the other numbers.
On 10/04/2025 02:06, Lawrence D'Oliveiro wrote:
Assume the list consists of algorithms for all computable numbers which
are guaranteed to terminate, ordered according to some Gödel numbering.
Please clarify what the above means: is it
a) a list of [all algorithms for { computable numbers which are
guaranteed to terminate } ], ordered according to some Gödel numbering.
or
b) a list of [all algorithms for computable numbers] (which are
guaranteed to terminate), ordered according to some Gödel numbering.
If (a) what do you mean by a "computable number that is guaranteed to terminate"?
If (b), the "which are guaranteed to terminate" is just a clarification
since the computable number algorithm is indeed specified as terminating after producing the requested digit. (no problem)
Also, I'm taking it that you consider an "algorithm for a computable
number" to be an algorithm (let's say a TM, to be definite) that takes a number n as input, and outputs the n'th digit of the computable number
and then terminates. Right?
I'll add further comments below when this is cleared up.
Possibly you are just confused because the Cantor diagonal
argument is not a "computation". It's a definition of a particular
number, which is subsequently shown to be missing from the given list.
The missing number in general might or might not be computable.
On Tue, 8 Apr 2025 18:59:02 -0400, Richard Damon wrote:
On 4/8/25 5:05 PM, Lawrence D'Oliveiro wrote:
On Mon, 7 Apr 2025 06:47:43 -0400, Richard Damon wrote:
And an infinite listing of values doesn't need to be computable, even
if every number in the list is computable.
Computability is a characteristic of particular numbers. It is a
characteristic of all the numbers in the list, and of the number that
the Cantor construction tries to construct from those numbers in the
list.
The fact that you can’t apply that characteristic to the set as a whole >>> is irrelevant, since the set itself is not a number.
Right, so the DIAGONAL number, which you claim to be computable, needs a
finite algorithm to do so.
The algorithm described is NOT FINITE, as it includes the infinite
number of algorithms to compute all the other numbers.
Potentially, *any* computable number could need an infinite amount of
storage (as well as an infinite number of computation steps) to compute an infinite number of digits. But the storage needed is always finite for a finite number of digits. That is true of the Cantor construction as well.
Remember, if you’re saying that the list cannot be constructed in the
first place, then you’re destroying the Cantor proof as well. I am
assuming it can be constructed, because the Cantor proof assumes it can be constructed. You can’t knock out my proof and keep the Cantor one by
saying the list cannot be constructed.
On Wed, 9 Apr 2025 22:00:15 -0400, Richard Damon wrote:
The problem with the "Cantor Construction" is that one of the steps,
"Compute to the Nth digit of the Nth number" isn't computable with a
finite algorithm.
But N is always finite, therefore the construction is always finite.
On Thu, 10 Apr 2025 10:37:49 +0300, Mikko wrote:
On 2025-04-10 00:50:10 +0000, Lawrence D'Oliveiro said:
On Mon, 7 Apr 2025 20:48:27 -0400, Richard Damon wrote:
The paper clearly talks about the process continuing indefinitely.
Note the key point about any computation of a computable number is that
the answer *converges* to the exact result in the limit. As you compute
more and more digits, the discrepancy between your approximation and
the correct answer can be made as close to zero as you like, just as
long as you don’t ask for it to be zero.
The Cantor construction does not converge.
If it is a computable number it does converge.
That’s a key point of my proof: if it converges, then the number is
already in the list. The only way it can come up with a number not in the list is by never converging.
On Wed, 9 Apr 2025 22:00:11 -0400, Richard Damon wrote:
On 4/9/25 8:41 PM, Lawrence D'Oliveiro wrote:
On Tue, 8 Apr 2025 18:57:17 -0400, Richard Damon wrote:
The problem is your "master" algorithm need the algorithms of *ALL*
the computable numbers within it, which is an infinite number of
algorithms, and thus isn't itself a finite algorithm.
That is the problem with the Cantor construction, not with my disproof
of it. My disproof of it only needs a finite number of list elements at
any point in the proof. The Cantor construction needs them all.
But a finite list can't get you to the needed arbitrary precision
needed.
I was going to say, sure it can, because the size of the list is a
function of the precision you ask for.
But that’s irrelevant, because if you’re saying an infinite list cannot be
constructed, then that blows the Cantor construction out of the water,
since an infinite list is precisely what it assumes.
Not also, Cantor only claimed that the numbers weren't countable, he
didn't talk about computable. That is a later work looking at the proof.
True. But interesting things happen when you try to apply his construction
to other kinds of number lists.
On Thu, 10 Apr 2025 17:11:43 +0100, Mike Terry wrote:
On 10/04/2025 02:06, Lawrence D'Oliveiro wrote:
Assume the list consists of algorithms for all computable numbers which
are guaranteed to terminate, ordered according to some Gödel numbering.
Please clarify what the above means: is it
a) a list of [all algorithms for { computable numbers which are
guaranteed to terminate } ], ordered according to some Gödel numbering.
or
b) a list of [all algorithms for computable numbers] (which are
guaranteed to terminate), ordered according to some Gödel numbering.
If (a) what do you mean by a "computable number that is guaranteed to
terminate"?
I didn’t come up with that phrase, you did.
If (b), the "which are guaranteed to terminate" is just a clarification
since the computable number algorithm is indeed specified as terminating
after producing the requested digit. (no problem)
Also, I'm taking it that you consider an "algorithm for a computable
number" to be an algorithm (let's say a TM, to be definite) that takes a
number n as input, and outputs the n'th digit of the computable number
and then terminates. Right?
I'll add further comments below when this is cleared up.
I should have thought that was obvious.
Apply the Cantor construction; that algorithm is also guaranteed to
terminate.
Therefore it must have a Gödel number, and be located at a
finite place in the list -- call it Nₙ.
So what happens when you ask the Cantor construction to compute digit Nₙ
of its number? It gets stuck in an endless loop. That means it is not
guaranteed to terminate. Therefore it cannot occur in the list.
But if you take it out of the list, then it *will* terminate, because all
the rest of the elements in the list do so. Put it in, it doesn’t belong:
take it out, it does belong.
So, by reductio ad absurdum, the assumption that the Cantor construction
for such a list even *exists* is false.
Possibly you are just confused because the Cantor diagonal
argument is not a "computation". It's a definition of a particular
number, which is subsequently shown to be missing from the given list.
The missing number in general might or might not be computable.
Are you saying the Cantor construction is not an algorithm?
Remember, if you’re saying that the list cannot be constructed in the
first place, then you’re destroying the Cantor proof as well.
That’s a key point of my proof: if it converges, then the number is
already in the list.
On 4/10/25 8:28 PM, Lawrence D'Oliveiro wrote:
On Wed, 9 Apr 2025 22:00:11 -0400, Richard Damon wrote:
But a finite list can't get you to the needed arbitrary precision
needed.
I was going to say, sure it can, because the size of the list is a
function of the precision you ask for.
But the function needs to be prepared to handle ANY precision, and thus
needs to be infinite.
But you need to remember that he wasn't "constructing" it in the manner
you are assuming, it isn't being constructed by a finite function, as
that wasn't the domain he was talking about.
On 4/10/25 8:30 PM, Lawrence D'Oliveiro wrote:
On Wed, 9 Apr 2025 22:00:15 -0400, Richard Damon wrote:
The problem with the "Cantor Construction" is that one of the steps,
"Compute to the Nth digit of the Nth number" isn't computable with a
finite algorithm.
But N is always finite, therefore the construction is always finite.
N my be finite, but is unbounded, and thus the construction is unbounded
in size, and thus not finite.
Your problem is you assume you can compute the nth value from the value
of n, but that requires you master algorithm include an infinite number
of algorithms in itself to choose from to build that number.
On Mon, 7 Apr 2025 06:51:02 -0400, Richard Damon wrote:
Your problem is you assume you can compute the nth value from the value
of n, but that requires you master algorithm include an infinite number
of algorithms in itself to choose from to build that number.
But the Cantor construction assumes you can construct that list.
So if you
object to the assumption of the existence of such a list, then you knock
down Cantor’s proof as well.
On Thu, 10 Apr 2025 10:37:49 +0300, Mikko wrote:
On 2025-04-10 00:50:10 +0000, Lawrence D'Oliveiro said:
On Mon, 7 Apr 2025 20:48:27 -0400, Richard Damon wrote:
The paper clearly talks about the process continuing indefinitely.
Note the key point about any computation of a computable number is that
the answer *converges* to the exact result in the limit. As you compute
more and more digits, the discrepancy between your approximation and
the correct answer can be made as close to zero as you like, just as
long as you don’t ask for it to be zero.
The Cantor construction does not converge.
If it is a computable number it does converge.
That’s a key point of my proof: if it converges, then the number is
already in the list.
On Thu, 2025-04-10 at 17:23 -0700, Keith Thompson wrote:
wij <wyniijj5@gmail.com> writes:
[...]
"lim(x->c) f(x)=L" means the limit of f approaching c is L, not
f(c)=L 'eventually'. f at c is not defined (handled) in limit.
Correct.
lim 0.333...=1/3 ... The *limit* is 1/3, not 0.333...=1/3
0.3+0.33+0.333+... ... The sequence converges to 1/3
Σ(n=1,∞) 3/10^n ... The sum converges to 1/3 (or you can use lim)
The limit as the number of 3s increases without bound *is exactly what
we mean* by the notation "0.333...". Once you understand that, it's
obvious that 0.333... is exactly equal to 1/3, and that 0.333... is a
rational number.
You agree "f at c is not defined (handled) in limit", yet, on the other hand ASSERTING 0.333... is 'exactly' 1/3 from limit? Are you nut?
.... is entirely correct.The limit as the number of 3s increases without bound *is exactly what
we mean* by the notation "0.333...".
As usual, you need to prove what you say. Or you are just showing yourself another olcott, just blink belief, nothing else.
On Fri, 2025-04-11 at 04:21 -0700, Keith Thompson wrote:
wij <wyniijj5@gmail.com> writes:
On Thu, 2025-04-10 at 17:23 -0700, Keith Thompson wrote:
wij <wyniijj5@gmail.com> writes:
[...]
"lim(x->c) f(x)=L" means the limit of f approaching c is L, not
f(c)=L 'eventually'.
f at c is not defined (handled) in limit.
Correct.
lim 0.333...=1/3 ... The *limit* is 1/3, not 0.333...=1/3
0.3+0.33+0.333+... ... The sequence converges to 1/3 Σ(n=1,∞)
3/10^n ... The sum converges to 1/3 (or you can use lim)
The limit as the number of 3s increases without bound *is exactly
what we mean* by the notation "0.333...". Once you understand
that, it's obvious that 0.333... is exactly equal to 1/3, and that
0.333... is a rational number.
You agree "f at c is not defined (handled) in limit", yet, on the
other hand ASSERTING 0.333... is 'exactly' 1/3 from limit? Are you
nut?
As usual, you need to prove what you say. Or you are just showing
yourself another olcott, just blink belief, nothing else.
Keep the insults to yourself. Last warning.
I still think 'nut' is a common word, at least a terse word for people
saying one thing and doing the other (or a liar more appropriate?)
My assertion is simply about what the "..." notation means.
Do you agree that the limit of 0.3, 0.33, 0.333, as the number of 3s
increases without bound, is exactly 1/3? (You said so above.)
Increases without bound -> yes is exactly 1/3 -> no such logic
What exactly do you think the notation "0.333..." means? I and many
others use that notation to mean the limit, which you agree is exactly
1/3.
Is this a lie? I have always consistently claiming "repeating decimals
are irrational".
On Fri, 2025-04-11 at 12:04 +0000, Mr Flibble wrote:
On Fri, 11 Apr 2025 19:52:20 +0800, wij wrote:
On Fri, 2025-04-11 at 04:21 -0700, Keith Thompson wrote:
wij <wyniijj5@gmail.com> writes:
On Thu, 2025-04-10 at 17:23 -0700, Keith Thompson wrote:
wij <wyniijj5@gmail.com> writes:
[...]
"lim(x->c) f(x)=L" means the limit of f approaching c is L,
not f(c)=L 'eventually'.
f at c is not defined (handled) in limit.
Correct.
lim 0.333...=1/3 ... The *limit* is 1/3, not 0.333...=1/3The limit as the number of 3s increases without bound *is
0.3+0.33+0.333+... ... The sequence converges to 1/3
Σ(n=1,∞)
3/10^n ... The sum converges to 1/3 (or you can use lim) >> > > > >
exactly what we mean* by the notation "0.333...". Once you
understand that, it's obvious that 0.333... is exactly equal to
1/3, and that 0.333... is a rational number.
You agree "f at c is not defined (handled) in limit", yet, on the
other hand ASSERTING 0.333... is 'exactly' 1/3 from limit? Are
you nut?
As usual, you need to prove what you say. Or you are just showing
yourself another olcott, just blink belief, nothing else.
Keep the insults to yourself. Last warning.
I still think 'nut' is a common word, at least a terse word for
people saying one thing and doing the other (or a liar more
appropriate?)
My assertion is simply about what the "..." notation means.
Do you agree that the limit of 0.3, 0.33, 0.333, as the number of
3s increases without bound, is exactly 1/3? (You said so above.)
Increases without bound -> yes is exactly 1/3 -> no such logic
What exactly do you think the notation "0.333..." means? I and
many others use that notation to mean the limit, which you agree is
exactly 1/3.
Is this a lie? I have always consistently claiming "repeating
decimals are irrational".
The decimals only repeat in certain bases:
Agree
it is wrong to think that any part of mathematics relies on a specific
base such as base 10.
Did I claimed what you said "any part of mathematics relies on a
specific base"?
On Fri, 2025-04-11 at 09:50 +0000, Alan Mackenzie wrote:
wij <wyniijj5@gmail.com> wrote:
On Thu, 2025-04-10 at 17:23 -0700, Keith Thompson wrote:
wij <wyniijj5@gmail.com> writes:
[...]
"lim(x->c) f(x)=L" means the limit of f approaching c is L, not
f(c)=L 'eventually'. f at c is not defined (handled) in limit.
Correct.
lim 0.333...=1/3 ... The *limit* is 1/3, not 0.333...=1/3
0.3+0.33+0.333+... ... The sequence converges to 1/3
Σ(n=1,∞) 3/10^n ... The sum converges to 1/3 (or you can use lim)
The limit as the number of 3s increases without bound *is exactly what >>>> we mean* by the notation "0.333...". Once you understand that, it's
obvious that 0.333... is exactly equal to 1/3, and that 0.333... is a
rational number.
You agree "f at c is not defined (handled) in limit", yet, on the other hand
ASSERTING 0.333... is 'exactly' 1/3 from limit? Are you nut?
No, Keith Thompson is simply correct, here. It is you who are nuts,
making unfounded claims about mathematics without having studied it.
The sentence ....
.... is entirely correct.The limit as the number of 3s increases without bound *is exactly what >>>> we mean* by the notation "0.333...".
As usual, you need to prove what you say. Or you are just showing yourself >>> another olcott, just blink belief, nothing else.
No, one doesn't need continually to prove standard mathematical
definitions and results. One could spend the whole day, every day, doing >> nothing else.
It is _you_ who needs to prove your remarkable assertions. You can't, of >> course, because they're false. What you could do, of course, is to show
a bit of respect for those who have studied and learnt mathematics.
I am not interesting to blind beliefs.
As I may guess from your posts, your knowledge is essentially 'what people say'
without knowing the meaning of words.
You may say it is 'standard', 'mainstream'...,etc. But whatever it is, simply no logical proof.
Remind you, the so called 'standard', 'mainstream' is on the side of logical proof.
They may evolve/change from errors. It is not a static thing and not the source of fact.
To save garbage talks, provide your logical proof (as usual, I believe NONE).
On Fri, 2025-04-11 at 12:04 +0000, Mr Flibble wrote:
On Fri, 11 Apr 2025 19:52:20 +0800, wij wrote:
On Fri, 2025-04-11 at 04:21 -0700, Keith Thompson wrote:
wij <wyniijj5@gmail.com> writes:
On Thu, 2025-04-10 at 17:23 -0700, Keith Thompson wrote:
wij <wyniijj5@gmail.com> writes:
[...]
"lim(x->c) f(x)=L" means the limit of f approaching c is L, not
f(c)=L 'eventually'.
f at c is not defined (handled) in limit.
Correct.
lim 0.333...=1/3 ... The *limit* is 1/3, not 0.333...=1/3
0.3+0.33+0.333+... ... The sequence converges to 1/3 Σ(n=1,∞) >>>>>>> 3/10^n ... The sum converges to 1/3 (or you can use lim)
The limit as the number of 3s increases without bound *is exactly
what we mean* by the notation "0.333...". Once you understand
that, it's obvious that 0.333... is exactly equal to 1/3, and that >>>>>> 0.333... is a rational number.
You agree "f at c is not defined (handled) in limit", yet, on the
other hand ASSERTING 0.333... is 'exactly' 1/3 from limit? Are you
nut?
As usual, you need to prove what you say. Or you are just showing
yourself another olcott, just blink belief, nothing else.
Keep the insults to yourself. Last warning.
I still think 'nut' is a common word, at least a terse word for people
saying one thing and doing the other (or a liar more appropriate?)
My assertion is simply about what the "..." notation means.
Do you agree that the limit of 0.3, 0.33, 0.333, as the number of 3s
increases without bound, is exactly 1/3? (You said so above.)
Increases without bound -> yes is exactly 1/3 -> no such logic
What exactly do you think the notation "0.333..." means? I and many
others use that notation to mean the limit, which you agree is exactly >>>> 1/3.
Is this a lie? I have always consistently claiming "repeating decimals
are irrational".
The decimals only repeat in certain bases:
Agree
On Mon, 7 Apr 2025 06:51:02 -0400, Richard Damon wrote:
Your problem is you assume you can compute the nth value from the value
of n, but that requires you master algorithm include an infinite number
of algorithms in itself to choose from to build that number.
But the Cantor construction assumes you can construct that list. So if you object to the assumption of the existence of such a list, then you knock
down Cantor’s proof as well.
On Thu, 10 Apr 2025 21:29:51 -0400, Richard Damon wrote:
On 4/10/25 8:30 PM, Lawrence D'Oliveiro wrote:
On Wed, 9 Apr 2025 22:00:15 -0400, Richard Damon wrote:
The problem with the "Cantor Construction" is that one of the steps,
"Compute to the Nth digit of the Nth number" isn't computable with a
finite algorithm.
But N is always finite, therefore the construction is always finite.
N my be finite, but is unbounded, and thus the construction is unbounded
in size, and thus not finite.
Why is N allowed to be finite but unbounded, but not the construction?
On Thu, 10 Apr 2025 21:21:18 -0400, Richard Damon wrote:
On 4/10/25 8:28 PM, Lawrence D'Oliveiro wrote:
On Wed, 9 Apr 2025 22:00:11 -0400, Richard Damon wrote:
But a finite list can't get you to the needed arbitrary precision
needed.
I was going to say, sure it can, because the size of the list is a
function of the precision you ask for.
But the function needs to be prepared to handle ANY precision, and thus
needs to be infinite.
That’s true of computable numbers in general, so unless you’re objecting to the very existence of the concept, it’s still irrelevant.
But you need to remember that he wasn't "constructing" it in the manner
you are assuming, it isn't being constructed by a finite function, as
that wasn't the domain he was talking about.
Given the example list I gave elsewhere, there is a fundamental conflict between a proof by induction (a well-established technique) and a proof by his construction (which has to be seen as something novel). I would say
that points to a logical weakness in his construction.
others may even imply I claim 1/3 is irrational
On Fri, 2025-04-11 at 09:07 -0400, Richard Damon wrote:
On 4/11/25 7:32 AM, wij wrote:
On Fri, 2025-04-11 at 09:50 +0000, Alan Mackenzie wrote:
wij <wyniijj5@gmail.com> wrote:
On Thu, 2025-04-10 at 17:23 -0700, Keith Thompson wrote:
wij <wyniijj5@gmail.com> writes:
[...]
"lim(x->c) f(x)=L" means the limit of f approaching c is L, not
f(c)=L 'eventually'. f at c is not defined (handled) in limit.
Correct.
lim 0.333...=1/3 ... The *limit* is 1/3, not 0.333...=1/3
0.3+0.33+0.333+... ... The sequence converges to 1/3
Σ(n=1,∞) 3/10^n ... The sum converges to 1/3 (or you can use lim)
The limit as the number of 3s increases without bound *is exactly what >>>>>> we mean* by the notation "0.333...". Once you understand that, it's >>>>>> obvious that 0.333... is exactly equal to 1/3, and that 0.333... is a >>>>>> rational number.
You agree "f at c is not defined (handled) in limit", yet, on the other hand
ASSERTING 0.333... is 'exactly' 1/3 from limit? Are you nut?
No, Keith Thompson is simply correct, here. It is you who are nuts,
making unfounded claims about mathematics without having studied it.
The sentence ....
.... is entirely correct.The limit as the number of 3s increases without bound *is exactly what >>>>>> we mean* by the notation "0.333...".
As usual, you need to prove what you say. Or you are just showing yourself
another olcott, just blink belief, nothing else.
No, one doesn't need continually to prove standard mathematical
definitions and results. One could spend the whole day, every day, doing >>>> nothing else.
It is _you_ who needs to prove your remarkable assertions. You can't, of >>>> course, because they're false. What you could do, of course, is to show >>>> a bit of respect for those who have studied and learnt mathematics.
I am not interesting to blind beliefs.
As I may guess from your posts, your knowledge is essentially 'what people say'
without knowing the meaning of words.
You may say it is 'standard', 'mainstream'...,etc. But whatever it is, simply
no logical proof.
Remind you, the so called 'standard', 'mainstream' is on the side of logical proof.
They may evolve/change from errors. It is not a static thing and not the source of fact.
To save garbage talks, provide your logical proof (as usual, I believe NONE).
Remmeber, the claim is that 0.33333... is 1/3 in the limit, i.e. that
for any possible epsilon, no matter how small, but still positive, there
is a point in the sequence of generatation of 0.3333... that all points
after that will be closer to the limit then epsilon.
We can compute that point, and thus show the limit is that value.
We do that by taking the log base 10 of epsilon, taking its floor (the
largest integer that is less than or equal to the value), negate it, and
use that many 3's (but at least one if we start with a big epsilon).
For instance, an epsilon of 0.001 has a log base 10 of -3, so we say
that all number of the pattern with at least 3 3's are that close.
we can show the example as 1/3 - 0.333 will be 0.0003333... which is
less than 0.0004 which is less than 0.001, and adding more 3s to the
number just makes us closer.
Firstly, we are now talking about limit, nothing to do with "repeating decimals
are irrational".
Your statement above is sloppy, cannot be verified or refuted. It contains too
many concepts to be defined. So, you just jump to the conclusion (or assertion)
you like. So, no valid proof is taken.
limit only says the *limit* is 1/3, all others are your wishes (or most people,
'standard', 'mainstream',... whatever you like, it does not matter).
1. No one disagree that the sequence 0.3,0,33..... 0.3333 can go on forever.
2. No one disagree that we can choose an arbitrary epsolon/delta whatever,
to make the error arbitrarily close to the limit (i.e L or 1/3).
So, don't make implications that I disagrees with these basics. (your are just
slight, others may even imply I claim 1/3 is irrational. Smear as proof?).
In logic language and point of view, the premise (i.e. the sequence 0.3, 0.33...)
does not contain 1/3 (also whatever epsilon/delta you like), therefore,
no possibility a valid logical proof can yield the conclusion 1/3 (QED).
Thus we have the limit, and thus the proof by the definition.
It seems your problem is you don't actually believe in the concept of
limit as a rigerous mathematical process, which just means you aren't
following the defintions.
On Fri, 2025-04-11 at 12:42 -0400, Richard Damon wrote:
On 4/11/25 11:43 AM, wij wrote:
On Fri, 2025-04-11 at 09:07 -0400, Richard Damon wrote:
On 4/11/25 7:32 AM, wij wrote:
On Fri, 2025-04-11 at 09:50 +0000, Alan Mackenzie wrote:Remmeber, the claim is that 0.33333... is 1/3 in the limit, i.e.
wij <wyniijj5@gmail.com> wrote:
On Thu, 2025-04-10 at 17:23 -0700, Keith Thompson wrote:
wij <wyniijj5@gmail.com> writes:
[...]
"lim(x->c) f(x)=L" means the limit of f approaching c is
L, not f(c)=L 'eventually'. f at c is not defined
(handled) in limit.
Correct.
lim 0.333...=1/3 ... The *limit* is 1/3, not
0.333...=1/3 0.3+0.33+0.333+... ... The sequence
converges to 1/3 Σ(n=1,∞) 3/10^n ... The sum
converges to 1/3 (or you can use lim)
The limit as the number of 3s increases without bound *is
exactly what we mean* by the notation "0.333...". Once you
understand that, it's obvious that 0.333... is exactly
equal to 1/3, and that 0.333... is a rational number.
You agree "f at c is not defined (handled) in limit", yet, on
the other hand ASSERTING 0.333... is 'exactly' 1/3 from
limit? Are you nut?
No, Keith Thompson is simply correct, here. It is you who are
nuts, making unfounded claims about mathematics without having
studied it.
The sentence ....
.... is entirely correct.The limit as the number of 3s increases without bound *is
exactly what we mean* by the notation "0.333...".
As usual, you need to prove what you say. Or you are just
showing yourself another olcott, just blink belief, nothing
else.
No, one doesn't need continually to prove standard mathematical
definitions and results. One could spend the whole day, every
day, doing nothing else.
It is _you_ who needs to prove your remarkable assertions. You
can't, of course, because they're false. What you could do, of
course, is to show a bit of respect for those who have studied
and learnt mathematics.
I am not interesting to blind beliefs.
As I may guess from your posts, your knowledge is essentially
'what people say'
without knowing the meaning of words.
You may say it is 'standard', 'mainstream'...,etc. But whatever
it is, simply no logical proof.
Remind you, the so called 'standard', 'mainstream' is on the side
of logical proof. They may evolve/change from errors. It is not a
static thing and not the source of fact.
To save garbage talks, provide your logical proof (as usual, I
believe NONE).
that for any possible epsilon, no matter how small, but still
positive, there is a point in the sequence of generatation of
0.3333... that all points after that will be closer to the limit
then epsilon.
We can compute that point, and thus show the limit is that value.
We do that by taking the log base 10 of epsilon, taking its floor
(the largest integer that is less than or equal to the value),
negate it, and use that many 3's (but at least one if we start with
a big epsilon).
For instance, an epsilon of 0.001 has a log base 10 of -3, so we
say that all number of the pattern with at least 3 3's are that
close.
we can show the example as 1/3 - 0.333 will be 0.0003333... which
is less than 0.0004 which is less than 0.001, and adding more 3s to
the number just makes us closer.
Firstly, we are now talking about limit, nothing to do with
"repeating decimals are irrational".
But the limit is what DEFINES what a repeating decimal represents, at
least within the normal Real Number System.
Your statement above is sloppy, cannot be verified or refuted. It
contains too many concepts to be defined. So, you just jump to the
conclusion (or assertion) you like. So, no valid proof is taken.
Of course it can be verified and thus refuted if it was wrong.
What concepts used were not defined in normal mathematics?
limit only says the *limit* is 1/3, all others are your wishes (or
most people, 'standard', 'mainstream',... whatever you like, it does
not matter).
1. No one disagree that the sequence 0.3,0,33..... 0.3333 can go on
forever.
Right, and the question is, what value does that sequence, when taken
"to the end" become. Since we can't actually do the infinite operation
to the end, we define it "in the limit".
Of course, in some hyper-mathematics which uses trans-finite values,
like the infintesimals, we might be able to come up with other
definitions, but then we are not working with what are called the "Real
Numbers", but some Hyper-Real number system, which you claim not to be
doing,
2. No one disagree that we can choose an arbitrary epsolon/delta
whatever,
to make the error arbitrarily close to the limit (i.e L or
1/3).
Which means that the limit of the sequence, which is the definition of
what that notation means, is defined and found.
So, don't make implications that I disagrees with these basics. (your
are just slight, others may even imply I claim 1/3 is irrational.
Smear as proof?).
But if the value of 0.333... but the definition of the representation
*IS* 1/3, and you claim that 0.333... is irrational, you are claiming
that 1/3 is irrational. That or your logic doesn't support the axiom
of equivalence (if A = B and B = C, thus A = C)
In logic language and point of view, the premise (i.e. the sequence
0.3, 0.33...)
does not contain 1/3 (also whatever epsilon/delta you like),
therefore,
no possibility a valid logical proof can yield the conclusion 1/3
(QED).
But no one says that the series contains the limit, just that the
"value" of the series is that limit.
OK, I will make one exception for you.
"lim(x->c) f(x)=L" means the limit of f approaching c is L, not f(c)=L.
f at c is not defined (handled) in limit.
Do you agree?
It seems that you problem is that you want to try to say you are usingDon't be that fast. You will have problem to eat your own words.
mathematics at least compatible with the "standard" mathematics, but
you want to try to define somethings differently, but can't find the
axioms you need, so you try to just define the results, which isn't how
mathematics works.
If you want to be working in a Hyper-Real number system, you need to
admit that, and then you can likely find a lot of formulations that get
what you want to do, they just admit they are not working in the "Real
Number System" as standardly defined.
Thus we have the limit, and thus the proof by the definition.
It seems your problem is you don't actually believe in the concept
of limit as a rigerous mathematical process, which just means you
aren't following the defintions.
On Fri, 2025-04-11 at 12:42 -0400, Richard Damon wrote:
On 4/11/25 11:43 AM, wij wrote:
On Fri, 2025-04-11 at 09:07 -0400, Richard Damon wrote:
On 4/11/25 7:32 AM, wij wrote:
On Fri, 2025-04-11 at 09:50 +0000, Alan Mackenzie wrote:
wij <wyniijj5@gmail.com> wrote:I am not interesting to blind beliefs.
On Thu, 2025-04-10 at 17:23 -0700, Keith Thompson wrote:
wij <wyniijj5@gmail.com> writes:
[...]
"lim(x->c) f(x)=L" means the limit of f approaching c is L, not >>>>>>>>> f(c)=L 'eventually'. f at c is not defined (handled) in limit. >>>>>>Correct.
lim 0.333...=1/3 ... The *limit* is 1/3, not 0.333...=1/3 >>>>>>>>> 0.3+0.33+0.333+... ... The sequence converges to 1/3
Σ(n=1,∞) 3/10^n ... The sum converges to 1/3 (or you can use lim)
The limit as the number of 3s increases without bound *is exactly what >>>>>>>> we mean* by the notation "0.333...". Once you understand that, it's >>>>>>>> obvious that 0.333... is exactly equal to 1/3, and that 0.333... is a >>>>>>>> rational number.
You agree "f at c is not defined (handled) in limit", yet, on the other hand
ASSERTING 0.333... is 'exactly' 1/3 from limit? Are you nut?
No, Keith Thompson is simply correct, here. It is you who are nuts, >>>>>> making unfounded claims about mathematics without having studied it. >>>>>>
The sentence ....
.... is entirely correct.The limit as the number of 3s increases without bound *is exactly what >>>>>>>> we mean* by the notation "0.333...".
As usual, you need to prove what you say. Or you are just showing yourself
another olcott, just blink belief, nothing else.
No, one doesn't need continually to prove standard mathematical
definitions and results. One could spend the whole day, every day, doing
nothing else.
It is _you_ who needs to prove your remarkable assertions. You can't, of
course, because they're false. What you could do, of course, is to show
a bit of respect for those who have studied and learnt mathematics. >>>>>
As I may guess from your posts, your knowledge is essentially 'what people say'
without knowing the meaning of words.
You may say it is 'standard', 'mainstream'...,etc. But whatever it is, simply
no logical proof.
Remind you, the so called 'standard', 'mainstream' is on the side of logical proof.
They may evolve/change from errors. It is not a static thing and not the source of fact.
To save garbage talks, provide your logical proof (as usual, I believe NONE).
Remmeber, the claim is that 0.33333... is 1/3 in the limit, i.e. that
for any possible epsilon, no matter how small, but still positive, there >>>> is a point in the sequence of generatation of 0.3333... that all points >>>> after that will be closer to the limit then epsilon.
We can compute that point, and thus show the limit is that value.
We do that by taking the log base 10 of epsilon, taking its floor (the >>>> largest integer that is less than or equal to the value), negate it, and >>>> use that many 3's (but at least one if we start with a big epsilon).
For instance, an epsilon of 0.001 has a log base 10 of -3, so we say
that all number of the pattern with at least 3 3's are that close.
we can show the example as 1/3 - 0.333 will be 0.0003333... which is
less than 0.0004 which is less than 0.001, and adding more 3s to the
number just makes us closer.
Firstly, we are now talking about limit, nothing to do with "repeating decimals
are irrational".
But the limit is what DEFINES what a repeating decimal represents, at
least within the normal Real Number System.
Your statement above is sloppy, cannot be verified or refuted. It contains too
many concepts to be defined. So, you just jump to the conclusion (or assertion)
you like. So, no valid proof is taken.
Of course it can be verified and thus refuted if it was wrong.
What concepts used were not defined in normal mathematics?
limit only says the *limit* is 1/3, all others are your wishes (or most people,
'standard', 'mainstream',... whatever you like, it does not matter).
1. No one disagree that the sequence 0.3,0,33..... 0.3333 can go on forever.
Right, and the question is, what value does that sequence, when taken
"to the end" become. Since we can't actually do the infinite operation
to the end, we define it "in the limit".
Of course, in some hyper-mathematics which uses trans-finite values,
like the infintesimals, we might be able to come up with other
definitions, but then we are not working with what are called the "Real
Numbers", but some Hyper-Real number system, which you claim not to be
doing,
2. No one disagree that we can choose an arbitrary epsolon/delta whatever, >>> to make the error arbitrarily close to the limit (i.e L or 1/3).
Which means that the limit of the sequence, which is the definition of
what that notation means, is defined and found.
So, don't make implications that I disagrees with these basics. (your are justBut if the value of 0.333... but the definition of the representation
slight, others may even imply I claim 1/3 is irrational. Smear as proof?). >>
*IS* 1/3, and you claim that 0.333... is irrational, you are claiming
that 1/3 is irrational. That or your logic doesn't support the axiom of
equivalence (if A = B and B = C, thus A = C)
In logic language and point of view, the premise (i.e. the sequence 0.3, 0.33...)
does not contain 1/3 (also whatever epsilon/delta you like), therefore,
no possibility a valid logical proof can yield the conclusion 1/3 (QED).
But no one says that the series contains the limit, just that the
"value" of the series is that limit.
OK, I will make one exception for you.
"lim(x->c) f(x)=L" means the limit of f approaching c is L, not f(c)=L.
f at c is not defined (handled) in limit.
Do you agree?
It seems that you problem is that you want to try to say you are using
mathematics at least compatible with the "standard" mathematics, but you
want to try to define somethings differently, but can't find the axioms
you need, so you try to just define the results, which isn't how
mathematics works.
If you want to be working in a Hyper-Real number system, you need to
admit that, and then you can likely find a lot of formulations that get
what you want to do, they just admit they are not working in the "Real
Number System" as standardly defined.
Don't be that fast. You will have problem to eat your own words.
Thus we have the limit, and thus the proof by the definition.
It seems your problem is you don't actually believe in the concept of
limit as a rigerous mathematical process, which just means you aren't >>>> following the defintions.
On 4/11/25 3:27 AM, Lawrence D'Oliveiro wrote:
On Thu, 10 Apr 2025 21:29:51 -0400, Richard Damon wrote:
On 4/10/25 8:30 PM, Lawrence D'Oliveiro wrote:
On Wed, 9 Apr 2025 22:00:15 -0400, Richard Damon wrote:
The problem with the "Cantor Construction" is that one of the steps, >>>>> "Compute to the Nth digit of the Nth number" isn't computable with a >>>>> finite algorithm.
But N is always finite, therefore the construction is always finite.
N my be finite, but is unbounded, and thus the construction is
unbounded in size, and thus not finite.
Why is N allowed to be finite but unbounded, but not the construction?
Because N is a variable, but the construction is supposed to be a
constant.
On Fri, 11 Apr 2025 09:15:38 -0400, Richard Damon wrote:
On 4/11/25 3:27 AM, Lawrence D'Oliveiro wrote:
On Thu, 10 Apr 2025 21:29:51 -0400, Richard Damon wrote:
On 4/10/25 8:30 PM, Lawrence D'Oliveiro wrote:
On Wed, 9 Apr 2025 22:00:15 -0400, Richard Damon wrote:
The problem with the "Cantor Construction" is that one of the steps, >>>>>> "Compute to the Nth digit of the Nth number" isn't computable with a >>>>>> finite algorithm.
But N is always finite, therefore the construction is always finite.
N my be finite, but is unbounded, and thus the construction is
unbounded in size, and thus not finite.
Why is N allowed to be finite but unbounded, but not the construction?
Because N is a variable, but the construction is supposed to be a
constant.
Do you know how loops work in algorithms? A finite amount of code can be executed an unbounded number of times. Without this capability, we would
have no “halting problem”. The construction can be conveyed with a finite amount of information, which makes it finite.
(See also “primitive recursive”.)
On Fri, 2025-04-11 at 04:21 -0700, Keith Thompson wrote:
Keep the insults to yourself. Last warning.
I still think 'nut' is a common word,
at least a terse word for people saying
one thing and doing the other (or a liar more appropriate?)
Is this a lie? I have always consistently claiming "repeating decimals are irrational".
Would, say "0.333<∞>" be clearer? Could you agree that that refers to
the limit and gives a result that's exactly 1/3?
0.333... approaches 1/3 --> no problem.
0.333... equals exactly to 1/3 --> no way (I have provided proofs and you don't).
If so, why do you
object to "..." but not to "<∞>" as a symbol for the limit? (Note that >> "..." is easier to type, unless you happen to have an ∞ key your
keyboard.)
Who say I object the use of "..."?
As said, it is 'the limit', not exactly equal (as explained)
As usual, you still only have irrelevant garbage talk, no valid logic proof.
If so, I can choose to stop responding.
Yes, but since you need the algorithms to compute ALL the numbers in
your code, you can't put them all in.
The problem with your "induction" is you assumed the existance of a computation that doesn't exist, a computation that given the nth digit
of the nth computable number for any value of n.
On Fri, 11 Apr 2025 21:41:48 -0400, Richard Damon wrote:
Yes, but since you need the algorithms to compute ALL the numbers in
your code, you can't put them all in.
But the Cantor construction relies on constructing precisely such a list.
If you can’t put together such a list, then you can’t perform the Cantor construction.
Lawrence D'Oliveiro <ldo@nz.invalid> writes:
On Fri, 11 Apr 2025 21:41:48 -0400, Richard Damon wrote:
Yes, but since you need the algorithms to compute ALL the numbers in
your code, you can't put them all in.
But the Cantor construction relies on constructing precisely such a list.
If you can’t put together such a list, then you can’t perform the Cantor >> construction.
The Cantor construction *assumes* the existence of such a list,
demonstrates that that assumption leads to a contradiction, and
concludes that no such list can exist.
On Fri, 11 Apr 2025 09:24:57 -0400, Richard Damon wrote:
The problem with your "induction" is you assumed the existance of a
computation that doesn't exist, a computation that given the nth digit
of the nth computable number for any value of n.
If you don’t have that, then how can you have the Cantor construction?
Here’s another way to look at the Cantor construction.
It is possible to construct a list of numbers, ostensibly from ℝ, where it is provable, by induction, that the Cantor construction cannot produce a number not in the list. This shows that you cannot prove, a priori, the validity of the Cantor construction.
But then, you cannot prove, a priori, the validity of proof by induction, either. Instead, it has to be added as an explicit axiom when constructing the integers ℤ.
In the same way, you can add “the Cantor construction works” as an axiom when constructing the set of reals ℝ. Otherwise, ℝ is just the set of computable numbers.
The Cantor construction *assumes* the existence of such a list ...
On Sun, 13 Apr 2025 15:00:34 -0700, Keith Thompson wrote:
The Cantor construction *assumes* the existence of such a list ...
Clearly, then, there must not be any objections to the creation of such a list, other than the Cantor construction itself.
Also, if you like mathematical bullshit then you probably won't like the
fact that negative zero doesn't exist: IEEE 754 has a defect because of
this fact.
On Sun, 13 Apr 2025 15:00:34 -0700, Keith Thompson wrote:
The Cantor construction *assumes* the existence of such a list ...
Clearly, then, there must not be any objections to the creation of such a list, other than the Cantor construction itself.
On Thu, 10 Apr 2025 10:37:49 +0300, Mikko wrote:
On 2025-04-10 00:50:10 +0000, Lawrence D'Oliveiro said:
On Mon, 7 Apr 2025 20:48:27 -0400, Richard Damon wrote:
The paper clearly talks about the process continuing indefinitely.
Note the key point about any computation of a computable number is that
the answer *converges* to the exact result in the limit. As you compute
more and more digits, the discrepancy between your approximation and
the correct answer can be made as close to zero as you like, just as
long as you don’t ask for it to be zero.
The Cantor construction does not converge.
If it is a computable number it does converge.
That’s a key point of my proof: if it converges, then the number is
already in the list. The only way it can come up with a number not in the list is by never converging.
On Wed, 09 Apr 2025 14:07:28 GMT, Mr Flibble wrote:
Also, if you like mathematical bullshit then you probably won't like
the fact that negative zero doesn't exist: IEEE 754 has a defect
because of this fact.
What is that “defect”, exactly?
Lawrence D'Oliveiro <ldo@nz.invalid> writes:
On Sun, 13 Apr 2025 15:00:34 -0700, Keith Thompson wrote:
The Cantor construction *assumes* the existence of such a list ...
Clearly, then, there must not be any objections to the creation of such
a list, other than the Cantor construction itself.
Why would you assume that?
On 4/11/25 3:28 AM, Lawrence D'Oliveiro wrote:
On Mon, 7 Apr 2025 06:51:02 -0400, Richard Damon wrote:
Your problem is you assume you can compute the nth value from the
value of n, but that requires you master algorithm include an infinite
number of algorithms in itself to choose from to build that number.
But the Cantor construction assumes you can construct that list. So if
you object to the assumption of the existence of such a list, then you
knock down Cantor’s proof as well.
But Cantors arguement wasn't about Computable Numbers ...
On 4/13/25 5:30 PM, Lawrence D'Oliveiro wrote:
Here’s another way to look at the Cantor construction.
It is possible to construct a list of numbers, ostensibly from ℝ, where
it is provable, by induction, that the Cantor construction cannot
produce a number not in the list. This shows that you cannot prove, a
priori, the validity of the Cantor construction.
Why not?
But he didn't use "induction" to make his proof.
He used basic inspection.
That is merely an application of simple logic.
But we don't need an axiom to show it works.
On Fri, 11 Apr 2025 09:35:15 -0400, Richard Damon wrote:
On 4/11/25 3:28 AM, Lawrence D'Oliveiro wrote:
On Mon, 7 Apr 2025 06:51:02 -0400, Richard Damon wrote:
Your problem is you assume you can compute the nth value from the
value of n, but that requires you master algorithm include an infinite >>>> number of algorithms in itself to choose from to build that number.
But the Cantor construction assumes you can construct that list. So if
you object to the assumption of the existence of such a list, then you
knock down Cantor’s proof as well.
But Cantors arguement wasn't about Computable Numbers ...
Doesn’t matter. If such a list can be assumed for the purposes of one proof, it can be assumed for the purposes of another. You can’t argue by saying it can only be used for purposes that you agree with.
On Sun, 13 Apr 2025 19:57:14 -0400, Richard Damon wrote:
On 4/13/25 5:30 PM, Lawrence D'Oliveiro wrote:
Here’s another way to look at the Cantor construction.
It is possible to construct a list of numbers, ostensibly from ℝ, where >>> it is provable, by induction, that the Cantor construction cannot
produce a number not in the list. This shows that you cannot prove, a
priori, the validity of the Cantor construction.
Why not?
See my construction of the flipped-integer list.
But he didn't use "induction" to make his proof.
Precisely.
He used basic inspection.
How do you conduct an infinity of “basic inspections”?
That is merely an application of simple logic.
“We hold these truths to be self-evident”, is it?
But we don't need an axiom to show it works.
Yes we do. Because otherwise we can use induction to show it doesn’t.
Here’s my counterexample list: write out the whole numbers
(non-negative integers) from 0 in increasing order, and flip the
digits of each one so that the digit from the 10⁰ place goes to the
10¯¹ place, 10¹ to 10¯² etc:
0.0000000000000...
0.1000000000000...
0.2000000000000...
0.3000000000000...
...
0.9000000000000...
0.0100000000000...
0.1100000000000...
0.2100000000000...
0.3100000000000...
...
0.9100000000000...
0.0200000000000...
...
Notice an interesting property of the list:
So even in a list which we already know does not contain every
possible real number, the Cantor construction fails to find one of the missing ones.
On Fri, 11 Apr 2025 09:35:15 -0400, Richard Damon wrote:
On 4/11/25 3:28 AM, Lawrence D'Oliveiro wrote:
On Mon, 7 Apr 2025 06:51:02 -0400, Richard Damon wrote:
Your problem is you assume you can compute the nth value from the
value of n, but that requires you master algorithm include an infinite >>>> number of algorithms in itself to choose from to build that number.
But the Cantor construction assumes you can construct that list. So if
you object to the assumption of the existence of such a list, then you
knock down Cantor’s proof as well.
But Cantors arguement wasn't about Computable Numbers ...
Doesn’t matter. If such a list can be assumed for the purposes of one proof, it can be assumed for the purposes of another. You can’t argue by saying it can only be used for purposes that you agree with.
On Sun, 13 Apr 2025 17:15:35 -0700, Keith Thompson wrote:
Lawrence D'Oliveiro <ldo@nz.invalid> writes:
On Sun, 13 Apr 2025 15:00:34 -0700, Keith Thompson wrote:
The Cantor construction *assumes* the existence of such a list ...
Clearly, then, there must not be any objections to the creation of such
a list, other than the Cantor construction itself.
Why would you assume that?
Because I don’t see any objections to Cantor making that assumption.
Here’s my counterexample list: write out the whole numbers
(non-negative integers) from 0 in increasing order, and flip the
digits of each one so that the digit from the 10⁰ place goes to the
10¯¹ place, 10¹ to 10¯² etc:
0.0000000000000...
0.1000000000000...
0.2000000000000...
0.3000000000000...
...
0.9000000000000...
0.0100000000000...
0.1100000000000...
0.2100000000000...
0.3100000000000...
...
0.9100000000000...
0.0200000000000...
...
Notice an interesting property of the list:
* If you look at the first digit after the decimal point, then in
every run of 10 consecutive list entries, you will find every
possible value of that digit.
* If you look at the first two digits after the decimal point, then in
every run of 100 consecutive list entries, you will find every
possible combination of values of those two digits.
...
* If you look at the first N digits after the decimal point, then in
every run of 10**N consecutive list entries, you will find every
possible combination of values of those N digits.
(Combinatorial explosion? Of course it’s a combinatorial explosion.
But there’s plenty of room for a combinatorial explosion or two, or
many, in an infinite list!)
This is a priori not a *complete* list of all the reals (the original
point of the Cantor construction). You’d think it would make things
easier for the Cantor construction, but it doesn’t.
Step 1 of the Cantor construction: choose a digit in the 10¯¹ place different from that of the first item in the list. There are 9
possibilities we could pick. But all the 10 possibilities for that
first digit occur in the following 10 numbers, so our pick will
definitely match one of them.
Step 2: choose the next digit, in the 10¯² place, different from that
of the second item in the list. There are, again, 9 possibilities we
could pick. But all the 100 combinations of possibilities for the
first two digits occur in the following 100 numbers, so our picks so
far will definitely match one of them.
And so on: at step N, we pick a digit in the Nth decimal place, to be different from that of the Nth number in the list. But all the 10**N possibilities for the digits we have picked so far occur in the
following 10**N numbers, so the number we have constructed so far will provably match one of them.
Note this is a proof by induction: if our choices at step N match some existing entry in the list, then so will the addition of our next
choice at step N + 1. Since our first choice already matches some
existing entry in the list, it follows that, however many digits we
choose, the result will always match some existing entry in the list.
So even in a list which we already know does not contain every
possible real number, the Cantor construction fails to find one of the missing ones.
QED.
Here’s my counterexample list: write out the whole numbers
(non-negative integers) from 0 in increasing order, and flip the
digits of each one so that the digit from the 10⁰ place goes to the
10¯¹ place, 10¹ to 10¯² etc:
0.0000000000000...
0.1000000000000...
0.2000000000000...
0.3000000000000...
...
0.9000000000000...
0.0100000000000...
0.1100000000000...
0.2100000000000...
0.3100000000000...
...
0.9100000000000...
0.0200000000000...
...
Notice an interesting property of the list:
* If you look at the first digit after the decimal point, then in
every run of 10 consecutive list entries, you will find every
possible value of that digit.
* If you look at the first two digits after the decimal point, then in
every run of 100 consecutive list entries, you will find every
possible combination of values of those two digits.
...
* If you look at the first N digits after the decimal point, then in
every run of 10**N consecutive list entries, you will find every
possible combination of values of those N digits.
(Combinatorial explosion? Of course it’s a combinatorial explosion.
But there’s plenty of room for a combinatorial explosion or two, or
many, in an infinite list!)
This is a priori not a *complete* list of all the reals (the original
point of the Cantor construction). You’d think it would make things
easier for the Cantor construction, but it doesn’t.
Step 1 of the Cantor construction: choose a digit in the 10¯¹ place different from that of the first item in the list. There are 9
possibilities we could pick. But all the 10 possibilities for that
first digit occur in the following 10 numbers, so our pick will
definitely match one of them.
Step 2: choose the next digit, in the 10¯² place, different from that
of the second item in the list. There are, again, 9 possibilities we
could pick. But all the 100 combinations of possibilities for the
first two digits occur in the following 100 numbers, so our picks so
far will definitely match one of them.
And so on: at step N, we pick a digit in the Nth decimal place, to be different from that of the Nth number in the list. But all the 10**N possibilities for the digits we have picked so far occur in the
following 10**N numbers, so the number we have constructed so far will provably match one of them.
Note this is a proof by induction: if our choices at step N match some existing entry in the list, then so will the addition of our next
choice at step N + 1. Since our first choice already matches some
existing entry in the list, it follows that, however many digits we
choose, the result will always match some existing entry in the list.
So even in a list which we already know does not contain every
possible real number, the Cantor construction fails to find one of the missing ones.
It is not a proof by induction as it makes no use of an induction
hypothesis PHI(n), and does not have any essential induction step PHI(n)
PHI(n+1).
Nonsense! RH's example using your very list constructs the
anti-diagonal 0.111111... which is NOT IN YOUR LIST.
You misunderstand what it means for a number (such as the Cantor anti-diagonal D) to "be in the list" or not.
On Tue, 15 Apr 2025 19:44:11 +0100, Mike Terry wrote:
It is not a proof by induction as it makes no use of an induction
hypothesis PHI(n), and does not have any essential induction step PHI(n)
PHI(n+1).
It does. It shows that, if the first N digits match, then so does the
(N+1)th digit. Given that it matches the first digit, those are your two requirements for proof by induction.
Nonsense! RH's example using your very list constructs the
anti-diagonal 0.111111... which is NOT IN YOUR LIST.
But at every stage, by induction, it matches an element in the list. You
only get a number that’s not in the list when you get to the end of the construction. But it’s an infinite construction!
Your fallacy is in assuming that the human reader will extrapolate the obvious pattern in that digit sequence to construct, in their mind, a
number that is not in the list. That’s not how logic works.
You misunderstand what it means for a number (such as the Cantor
anti-diagonal D) to "be in the list" or not.
Or maybe you misunderstand that there is no inherent logical validity to
the Cantor construction, just as there is no inherent logical validity to proof by induction; induction had to be added as one of the axioms in the construction of the integers, so that we could reason with it. But given
that it is there, you cannot prove induction false; because if you do, you have proven that there is a logical contradiction in the system of
integers.
Let’s start again, with the assumption that we have a list mapping all the reals 1:1 to the positive integers. So given any real, we can assign it a position N ∈ ℤ ≥ 1.
So now we apply the Cantor construction, to try to come up with a number
not in the list. But a consequence of the starting assumption is that the number being constructed must be somewhere in the list, and therefore the Cantor construction must map to some positive integer Nₙ.
So the question is: what is digit Nₙ of this number?
The answer is, it must be different from digit Nₙ of itself!
So you see, the assumption that you *can* perform the Cantor construction
on a list of the reals leads to a contradiction. Therefore the
construction cannot be performed. QED.
What we have here is duelling assumptions: either the list can be constructed, or (according to the Cantor construction) it cannot. There is
no “self-evident” reason to say one argument is valid while the other is not.
Therefore I suggest that the Cantor construction is similarly an axiom,
that has to be added before you can construct the reals. Without it, the ℝ you construct consists solely of computable numbers.
On Tue, 15 Apr 2025 19:44:11 +0100, Mike Terry wrote:
It is not a proof by induction as it makes no use of an induction
hypothesis PHI(n), and does not have any essential induction step PHI(n)
PHI(n+1).
It does. It shows that, if the first N digits match, then so does the
(N+1)th digit. Given that it matches the first digit, those are your two requirements for proof by induction.
Nonsense! RH's example using your very list constructs the
anti-diagonal 0.111111... which is NOT IN YOUR LIST.
But at every stage, by induction, it matches an element in the list. You
only get a number that’s not in the list when you get to the end of the construction. But it’s an infinite construction!
Let’s start again, with the assumption that we have a list mapping all the reals 1:1 to the positive integers. So given any real, we can assign it a position N ∈ ℤ ≥ 1.
So now we apply the Cantor construction, to try to come up with a number
not in the list. But a consequence of the starting assumption is that the number being constructed must be somewhere in the list, and therefore the Cantor construction must map to some positive integer Nₙ.
So the question is: what is digit Nₙ of this number?
The answer is, it must be different from digit Nₙ of itself!
So you see, the assumption that you *can* perform the Cantor construction
on a list of the reals leads to a contradiction. Therefore the
construction cannot be performed. QED.
What we have here is duelling assumptions: either the list can be constructed, or (according to the Cantor construction) it cannot. There is
no “self-evident” reason to say one argument is valid while the other is not.
Therefore I suggest that the Cantor construction is similarly an axiom,
that has to be added before you can construct the reals. Without it, the ℝ you construct consists solely of computable numbers.
On Tue, 15 Apr 2025 19:44:11 +0100, Mike Terry wrote:
It is not a proof by induction as it makes no use of an induction
hypothesis PHI(n), and does not have any essential induction step PHI(n)
PHI(n+1).
It does. It shows that, if the first N digits match, then so does the
(N+1)th digit. Given that it matches the first digit, those are your two requirements for proof by induction.
Nonsense! RH's example using your very list constructs the
anti-diagonal 0.111111... which is NOT IN YOUR LIST.
But at every stage, by induction, it matches an element in the list. You
only get a number that’s not in the list when you get to the end of the construction. But it’s an infinite construction!
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