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  • collective and individual removal

    From WM@21:1/5 to All on Mon Apr 28 15:15:48 2025
    We can remove collectively all terms from the harmonic series. Nothing
    remains.

    If we restrict the removal to individually definable terms, then an
    infinity remains.

    Is this remainder caused by the impossibility to define infinitely many
    terms individually? No! It is sufficient to define one term, namely the
    last one definable.

    This is an irrefutable proof of the existence of dark numbers.

    Regards, WM

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  • From Alan Mackenzie@21:1/5 to wolfgang.mueckenheim@tha.de on Mon Apr 28 13:50:46 2025
    WM <wolfgang.mueckenheim@tha.de> wrote:
    We can remove collectively all terms from the harmonic series. Nothing remains.

    If we restrict the removal to individually definable terms, then an
    infinity remains.

    Is this remainder caused by the impossibility to define infinitely many
    terms individually? No! It is sufficient to define one term, namely the
    last one definable.

    This is an irrefutable proof of the existence of dark numbers.

    <Yawn>

    Regards, WM

    --
    Alan Mackenzie (Nuremberg, Germany).

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  • From WM@21:1/5 to FromTheRafters on Mon Apr 28 21:54:31 2025
    On 28.04.2025 16:04, FromTheRafters wrote:
    WM explained on 4/28/2025 :
    We can remove collectively all terms from the harmonic series. Nothing
    remains.

    If we restrict the removal to individually definable terms, then an
    infinity remains.

    Is this remainder caused by the impossibility to define infinitely
    many terms individually? No! It is sufficient to define one term,
    namely the last one definable.

    But there is no last one definable

    That is true. They are a potentially infinite sequence.

    since they all are.

    That is wrong because you cannot remove all natural numbers by removing
    only definable numbers

    Regards, WM

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  • From WM@21:1/5 to Alan Mackenzie on Mon Apr 28 21:55:40 2025
    On 28.04.2025 15:50, Alan Mackenzie wrote:
    WM <wolfgang.mueckenheim@tha.de> wrote:
    We can remove collectively all terms from the harmonic series. Nothing
    remains.

    If we restrict the removal to individually definable terms, then an
    infinity remains.

    Is this remainder caused by the impossibility to define infinitely many
    terms individually? No! It is sufficient to define one term, namely the
    last one definable.

    This is an irrefutable proof of the existence of dark numbers.

    <Yawn>

    Some people never get the clue.

    Regards, WM

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  • From Python@21:1/5 to All on Mon Apr 28 21:28:19 2025
    Le 28/04/2025 à 21:55, WM a écrit :
    On 28.04.2025 15:50, Alan Mackenzie wrote:
    WM <wolfgang.mueckenheim@tha.de> wrote:
    We can remove collectively all terms from the harmonic series. Nothing
    remains.

    If we restrict the removal to individually definable terms, then an
    infinity remains.

    Is this remainder caused by the impossibility to define infinitely many
    terms individually? No! It is sufficient to define one term, namely the
    last one definable.

    This is an irrefutable proof of the existence of dark numbers.

    <Yawn>

    Some people never get the clue.

    When you are the only one considering that you have a clue (no, you cannot count students that you are abusing at Hochschule Augsburg by using your illegitimate authority), then you should consider that you may have no
    clue at all.

    As a matter of fact, you have no clue. You are just an senile crank
    abusing students and posting nonsense on Usenet.

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  • From WM@21:1/5 to FromTheRafters on Tue Apr 29 14:14:46 2025
    On 29.04.2025 01:30, FromTheRafters wrote:
    WM formulated the question :

    That is wrong because you cannot remove all natural numbers by
    removing only definable numbers

    Sure you can.

    I cannot. Show it if you can.

    Regards, WM

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  • From WM@21:1/5 to FromTheRafters on Wed Apr 30 14:43:46 2025
    On 29.04.2025 15:34, FromTheRafters wrote:
    on 4/29/2025, WM supposed :
    On 29.04.2025 01:30, FromTheRafters wrote:
    WM formulated the question :

    That is wrong because you cannot remove all natural numbers by
    removing only definable numbers

    Sure you can.

    I cannot. Show it if you can.

    Each and every non-initial natural number is *DEFINED* as being one more
    than the previously defined one.

    This chain fails. Otherwise you could remove all numbers by removing
    only defined ones. But that is impossible.

    Your set of
    definable natural numbers is the same set as the natural numbers.

    |ℕ \ ℕ| = 0.
    |ℕ \ ℕ_def| = ℵo.

    Regards, WM

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  • From joes@21:1/5 to All on Wed Apr 30 13:15:27 2025
    Am Wed, 30 Apr 2025 14:43:46 +0200 schrieb WM:
    On 29.04.2025 15:34, FromTheRafters wrote:
    on 4/29/2025, WM supposed :
    On 29.04.2025 01:30, FromTheRafters wrote:
    WM formulated the question :

    That is wrong because you cannot remove all natural numbers by
    removing only definable numbers
    Each and every non-initial natural number is *DEFINED* as being one
    more than the previously defined one.
    This chain fails. Otherwise you could remove all numbers by removing
    only defined ones. But that is impossible.
    It could only fail if one didn't have a successor.

    Your set of definable natural numbers is the same set as the natural
    numbers.
    |ℕ \ ℕ| = 0.
    |ℕ \ ℕ_def| = ℵo.
    That's just, like, your opinion, man.

    --
    Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
    It is not guaranteed that n+1 exists for every n.

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  • From WM@21:1/5 to FromTheRafters on Wed Apr 30 22:32:00 2025
    On 30.04.2025 16:13, FromTheRafters wrote:
    on 4/30/2025, WM supposed :
    On 29.04.2025 15:34, FromTheRafters wrote:
    on 4/29/2025, WM supposed :
    On 29.04.2025 01:30, FromTheRafters wrote:
    WM formulated the question :

    That is wrong because you cannot remove all natural numbers by
    removing only definable numbers

    Sure you can.

    I cannot. Show it if you can.

    Each and every non-initial natural number is *DEFINED* as being one
    more than the previously defined one.

    This chain fails. Otherwise you could remove all numbers by removing
    only defined ones. But that is impossible.

    You are begging the question.

    No. Try it. Fail.

    Regards, WM

    https://en.wikipedia.org/wiki/Begging_the_question

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  • From WM@21:1/5 to joes on Wed Apr 30 22:34:26 2025
    On 30.04.2025 15:15, joes wrote:
    Am Wed, 30 Apr 2025 14:43:46 +0200 schrieb WM:
    On 29.04.2025 15:34, FromTheRafters wrote:
    on 4/29/2025, WM supposed :
    On 29.04.2025 01:30, FromTheRafters wrote:
    WM formulated the question :

    That is wrong because you cannot remove all natural numbers by
    removing only definable numbers
    Each and every non-initial natural number is *DEFINED* as being one
    more than the previously defined one.
    This chain fails. Otherwise you could remove all numbers by removing
    only defined ones. But that is impossible.
    It could only fail if one didn't have a successor.#

    It fails. It is impossible to remove all natural numbers by defining
    each one.

    Regards, WM

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  • From joes@21:1/5 to All on Thu May 1 15:11:15 2025
    Am Wed, 30 Apr 2025 22:34:26 +0200 schrieb WM:
    On 30.04.2025 15:15, joes wrote:
    Am Wed, 30 Apr 2025 14:43:46 +0200 schrieb WM:
    On 29.04.2025 15:34, FromTheRafters wrote:
    on 4/29/2025, WM supposed :
    On 29.04.2025 01:30, FromTheRafters wrote:
    WM formulated the question :

    That is wrong because you cannot remove all natural numbers by
    removing only definable numbers
    Each and every non-initial natural number is *DEFINED* as being one
    more than the previously defined one.
    This chain fails. Otherwise you could remove all numbers by removing
    only defined ones. But that is impossible.
    It could only fail if one didn't have a successor.#
    It fails. It is impossible to remove all natural numbers by defining
    each one.
    No. It is possible, every one has a successor, there are none that don't,
    those are all naturals, numbers that don't aren't naturals.

    --
    Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
    It is not guaranteed that n+1 exists for every n.

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  • From WM@21:1/5 to joes on Thu May 1 21:41:42 2025
    On 01.05.2025 17:11, joes wrote:
    Am Wed, 30 Apr 2025 22:34:26 +0200 schrieb WM:

    It is impossible to remove all natural numbers by defining
    each one.

    No. It is possible,

    Show it.

    Regards, WM

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