1) Let every unit interval after a natural number on the real axis be coloured white with exception of the powers of 2 which are colouredAre the intervals closed or not?
black. Is it possible to shift the black intervals so that the whole
real axis becomes black?
2) Let every unit interval after a natural number on the real axis be coloured as above with exception of the intervals after the odd primewith the exception of = but instead
numbers which are coloured red. Is it possible to shift the red
intervals so that the whole real axis becomes red?
What colour has the real axis after you have solved both tasks?If you have "solved" them, I suppose it is black if you do the
Am Wed, 20 Nov 2024 18:02:40 +0100 schrieb WM:
1) Let every unit interval after a natural number on the real axis beAre the intervals closed or not?
coloured white with exception of the powers of 2 which are coloured
black. Is it possible to shift the black intervals so that the whole
real axis becomes black?
2) Let every unit interval after a natural number on the real axis bewith the exception of = but instead
coloured as above with exception of the intervals after the odd prime
numbers which are coloured red. Is it possible to shift the red
intervals so that the whole real axis becomes red?
What colour has the real axis after you have solved both tasks?If you have "solved" them, I suppose it is black if you do the
second one first.
On 11/20/2024 9:02 AM, WM wrote:
What colour has the real axis after you have solved both tasks?
I can make a graph on it to see it visually.
On 20.11.2024 19:06, joes wrote:You forget that you can push open intervals closer together.
Am Wed, 20 Nov 2024 18:02:40 +0100 schrieb WM:Irrelevant, but assume closed.
1) Let every unit interval after a natural number on the real axis beAre the intervals closed or not?
coloured white with exception of the powers of 2 which are coloured
black. Is it possible to shift the black intervals so that the whole
real axis becomes black?
Badly written.Well understood.2) Let every unit interval after a natural number on the real axis bewith the exception of = but instead
coloured as above with exception of the intervals after the odd prime
numbers which are coloured red. Is it possible to shift the red
intervals so that the whole real axis becomes red?
Formally: lim n->oo n/(2^n) = 0No. The density of coloured intervals within the first n intervals is a sequence converging to zero. Every positive eps is undercut. The limitWhat colour has the real axis after you have solved both tasks?If you have "solved" them, I suppose it is black if you do the second
one first.
cannot be 1.
Am Wed, 20 Nov 2024 19:32:38 +0100 schrieb WM:
The density of coloured intervals within the first n intervals is a
sequence converging to zero. Every positive eps is undercut. The limit
cannot be 1.
Formally: lim n->oo n/(2^n) = 0
On 21.11.2024 01:18, joes wrote:I don't believe in falsehoods. How do you derive the above? Both the denominator and numerator diverge. The expression oo/oo is undefined.
Am Wed, 20 Nov 2024 19:32:38 +0100 schrieb WM:But you don't believe it?
The density of coloured intervals within the first n intervals is aFormally: lim n->oo n/(2^n) = 0
sequence converging to zero. Every positive eps is undercut. The limit
cannot be 1.
Am Thu, 21 Nov 2024 12:54:22 +0100 schrieb WM:
On 21.11.2024 01:18, joes wrote:I don't believe in falsehoods. How do you derive the above? Both the denominator and numerator diverge. The expression oo/oo is undefined.
Am Wed, 20 Nov 2024 19:32:38 +0100 schrieb WM:But you don't believe it?
The density of coloured intervals within the first n intervals is aFormally: lim n->oo n/(2^n) = 0
sequence converging to zero. Every positive eps is undercut. The limit >>>> cannot be 1.
On 21.11.2024 22:03, joes wrote:Right. What even is oo^3.
Am Thu, 21 Nov 2024 12:54:22 +0100 schrieb WM:like lim n->oo n/n^3?
On 21.11.2024 01:18, joes wrote:I don't believe in falsehoods. How do you derive the above? Both the
Am Wed, 20 Nov 2024 19:32:38 +0100 schrieb WM:But you don't believe it?
The density of coloured intervals within the first n intervals is aFormally: lim n->oo n/(2^n) = 0
sequence converging to zero. Every positive eps is undercut. The
limit cannot be 1.
denominator and numerator diverge. The expression oo/oo is undefined.
Op 21/11/2024 om 22:03 schreef joes:
Am Thu, 21 Nov 2024 12:54:22 +0100 schrieb WM:
On 21.11.2024 01:18, joes wrote:I don't believe in falsehoods. How do you derive the above? Both the
Am Wed, 20 Nov 2024 19:32:38 +0100 schrieb WM:But you don't believe it?
The density of coloured intervals within the first n intervals is aFormally: lim n->oo n/(2^n) = 0
sequence converging to zero. Every positive eps is undercut. The limit >>>>> cannot be 1.
denominator and numerator diverge. The expression oo/oo is undefined.
lim n->oo n/(2^n) = lim n->oo 1/(2^(n-(ln(n)/ln(2)))) = 0
lim n->oo n/(n^3) = lim n->oo 1/n^2 = 0
https://www.desmos.com/calculator/ulookjqjcq
https://www.wolframalpha.com/input?i=lim+n+to+infinity+1%2F%282%5E%28n-%28ln%28n%29%2Fln%282%29%29%29%29
Am Thu, 21 Nov 2024 12:54:22 +0100 schrieb WM:
On 21.11.2024 01:18, joes wrote:I don't believe in falsehoods. How do you derive the above? Both the denominator and numerator diverge. The expression oo/oo is undefined.
Am Wed, 20 Nov 2024 19:32:38 +0100 schrieb WM:But you don't believe it?
The density of coloured intervals within the first n intervals is aFormally: lim n->oo n/(2^n) = 0
sequence converging to zero. Every positive eps is undercut. The limit >>>> cannot be 1.
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