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Groups > sci.math > #603017 > unrolled thread

A simple argument. Acceptable?

Started byWM <askasker48@gmail.com>
First post2023-06-23 04:41 -0700
Last post2023-06-27 11:30 -0400
Articles 15 on this page of 75 — 7 participants

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  A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-23 04:41 -0700
    Re: A simple argument. Acceptable? Gus Gassmann <horand.gassmann@gmail.com> - 2023-06-23 05:09 -0700
    Re: A simple argument. Acceptable? Gus Gassmann <horand.gassmann@gmail.com> - 2023-06-23 05:12 -0700
      Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-23 09:58 -0700
    Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-23 07:10 -0700
    Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-23 07:12 -0700
      Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-23 07:25 -0700
      Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-23 09:56 -0700
      Re: A simple argument. Acceptable? Eram semper recta <thenewcalculus@gmail.com> - 2023-06-24 05:31 -0700
    Re: A simple argument. Acceptable? Jim Burns <james.g.burns@att.net> - 2023-06-23 11:54 -0400
      Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-23 09:59 -0700
        Re: A simple argument. Acceptable? Jim Burns <james.g.burns@att.net> - 2023-06-23 15:48 -0400
          Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-24 06:04 -0700
      Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-23 10:11 -0700
      Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-24 05:59 -0700
        Re: A simple argument. Acceptable? FromTheRafters <FTR@nomail.afraid.org> - 2023-06-24 10:07 -0400
        Re: A simple argument. Acceptable? Jim Burns <james.g.burns@att.net> - 2023-06-24 12:04 -0400
          Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-24 10:43 -0700
            Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-24 10:50 -0700
              Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-25 08:55 -0700
                Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-25 09:38 -0700
            Re: A simple argument. Acceptable? Jim Burns <james.g.burns@att.net> - 2023-06-25 10:16 -0400
              Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-25 09:04 -0700
                Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-25 09:51 -0700
          Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-25 08:59 -0700
            Re: A simple argument. Acceptable? Jim Burns <james.g.burns@att.net> - 2023-06-25 16:19 -0400
              Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-26 12:42 -0700
                Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 12:49 -0700
                Re: A simple argument. Acceptable? FromTheRafters <FTR@nomail.afraid.org> - 2023-06-26 16:10 -0400
                Re: A simple argument. Acceptable? Jim Burns <james.g.burns@att.net> - 2023-06-26 17:02 -0400
        Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-24 10:22 -0700
          Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-25 09:01 -0700
            Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-25 09:42 -0700
              Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-26 12:35 -0700
                Re: A simple argument. Acceptable? FromTheRafters <FTR@nomail.afraid.org> - 2023-06-26 16:15 -0400
            Re: A simple argument. Acceptable? Gus Gassmann <horand.gassmann@gmail.com> - 2023-06-25 13:24 -0700
              Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-26 12:43 -0700
                Re: A simple argument. Acceptable? Gus Gassmann <horand.gassmann@gmail.com> - 2023-06-26 18:57 -0700
                  Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 19:08 -0700
                    Re: A simple argument. Acceptable? Jim Burns <james.g.burns@att.net> - 2023-06-26 22:35 -0400
                  Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-28 10:34 -0700
                    Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-28 11:54 -0700
                      Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-29 05:45 -0700
                    Re: A simple argument. Acceptable? Gus Gassmann <horand.gassmann@gmail.com> - 2023-06-28 12:47 -0700
                      Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-29 05:45 -0700
                        Re: A simple argument. Acceptable? Gus Gassmann <horand.gassmann@gmail.com> - 2023-06-29 08:01 -0700
            Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-27 05:28 -0700
    Re: A simple argument. Acceptable? Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2023-06-23 11:51 -0700
      Re: A simple argument. Acceptable? Eram semper recta <thenewcalculus@gmail.com> - 2023-06-24 05:34 -0700
    Re: A simple argument. Acceptable? Eram semper recta <thenewcalculus@gmail.com> - 2023-06-24 05:29 -0700
      Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-24 06:08 -0700
        Re: A simple argument. Acceptable? FromTheRafters <FTR@nomail.afraid.org> - 2023-06-24 10:15 -0400
          Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-25 08:52 -0700
            Re: A simple argument. Acceptable? FromTheRafters <FTR@nomail.afraid.org> - 2023-06-25 12:06 -0400
              Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-25 09:56 -0700
                Re: A simple argument. Acceptable? FromTheRafters <FTR@nomail.afraid.org> - 2023-06-25 17:42 -0400
                Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-26 12:38 -0700
                  Re: A simple argument. Acceptable? FromTheRafters <FTR@nomail.afraid.org> - 2023-06-26 16:17 -0400
              Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-26 12:26 -0700
                Re: A simple argument. Acceptable? FromTheRafters <FTR@nomail.afraid.org> - 2023-06-26 16:25 -0400
            Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-25 09:35 -0700
              Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-26 12:30 -0700
        Re: A simple argument. Acceptable? Eram semper recta <thenewcalculus@gmail.com> - 2023-06-25 16:23 -0700
          Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-26 12:48 -0700
    Re: A simple argument. Acceptable? Eram semper recta <thenewcalculus@gmail.com> - 2023-06-26 05:19 -0700
      Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 08:12 -0700
        Re: A simple argument. Acceptable? Eram semper recta <thenewcalculus@gmail.com> - 2023-06-26 08:42 -0700
          Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 08:57 -0700
            Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-26 13:01 -0700
              Re: A simple argument. Acceptable? FromTheRafters <FTR@nomail.afraid.org> - 2023-06-26 16:27 -0400
              Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-27 02:55 -0700
      Re: A simple argument. Acceptable? WM <askasker48@gmail.com> - 2023-06-26 12:52 -0700
        Re: A simple argument. Acceptable? Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-27 03:06 -0700
          Re: A simple argument. Acceptable? Gus Gassmann <horand.gassmann@gmail.com> - 2023-06-27 04:20 -0700
          Re: A simple argument. Acceptable? FromTheRafters <FTR@nomail.afraid.org> - 2023-06-27 11:30 -0400

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#603221

FromFritz Feldhase <franz.fritschee.ff@gmail.com>
Date2023-06-25 09:35 -0700
Message-ID<8cb60371-94e6-4936-be32-c6f2e9417358n@googlegroups.com>
In reply to#603214
On Sunday, June 25, 2023 at 5:52:58 PM UTC+2, WM wrote:

> If natnumbers are never mising, they are in the intersection. 

Indeed! But there are no such nutnumbers, Du hirloser Affe.

If n is a natnumber, then n is "missing" in the endsegment E(n+1). Hence it is NOT in the intersection of all endsegments.

Du bist wirklich selbst zum Scheißen zu blöde, Mann.

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#603365

FromWM <askasker48@gmail.com>
Date2023-06-26 12:30 -0700
Message-ID<11060fac-cb67-4a6b-a6ab-f665a55919a9n@googlegroups.com>
In reply to#603221
Fritz Feldhase schrieb am Sonntag, 25. Juni 2023 um 18:35:14 UTC+2:
> On Sunday, June 25, 2023 at 5:52:58 PM UTC+2, WM wrote: 
> 
> > If natnumbers are never mising, they are in the intersection. 
> 
> Indeed! But there are no such nutnumbers,

In infinite endegments there are infinitely many.
> 
> If n is a natnumber, then n is "missing" in the endsegment E(n+1). Hence it is NOT in the intersection of all endsegments. 

And it is not in the successor endsegments. What is in all infinite endsegments but missing in the intersecion?

Regards, WM

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#603280

FromEram semper recta <thenewcalculus@gmail.com>
Date2023-06-25 16:23 -0700
Message-ID<316d231e-b31b-4019-beac-b5258529b9ebn@googlegroups.com>
In reply to#603130
On Saturday, 24 June 2023 at 09:08:58 UTC-4, WM wrote:
> Eram semper recta schrieb am Samstag, 24. Juni 2023 um 14:29:13 UTC+2: 
> > On Friday, 23 June 2023 at 07:41:05 UTC-4, WM wrote: 
> > > We assume that all endsegments E(n) = {n, n+1, n+2, ...} of natural numbers are visible and infinite and have an empty intersection. Then there exists at least one visible endsegment E(k) containing at least one natural number j ≥ k that is not in at least one of its predecessors. 
> > > 
> > > Acceptable?
> > Yes, your claim is 100% correct.
> Thank you. It is relieving to see that intelligent persons can understand. I am sure my students will become educated as well. 

You should post these questions and your students' responses here. That would be interesting to see.

> 
> Regards, WM

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#603374

FromWM <askasker48@gmail.com>
Date2023-06-26 12:48 -0700
Message-ID<53d42c03-2585-4b5b-873f-3c5132d95761n@googlegroups.com>
In reply to#603280
Eram semper recta schrieb am Montag, 26. Juni 2023 um 01:23:35 UTC+2:
> On Saturday, 24 June 2023 at 09:08:58 UTC-4, WM wrote: 
> > Eram semper recta schrieb am Samstag, 24. Juni 2023 um 14:29:13 UTC+2: 
> > > On Friday, 23 June 2023 at 07:41:05 UTC-4, WM wrote: 
> > > > We assume that all endsegments E(n) = {n, n+1, n+2, ...} of natural numbers are visible and infinite and have an empty intersection. Then there exists at least one visible endsegment E(k) containing at least one natural number j ≥ k that is not in at least one of its predecessors. 
> > > > 
> > > > Acceptable? 
> > > Yes, your claim is 100% correct. 
> > Thank you. It is relieving to see that intelligent persons can understand. I am sure my students will become educated as well.
> You should post these questions and your students' responses here. That would be interesting to see. 

I will do so. But my lesson will begin only in October (because in summer I do not teach) and the students will learn all this only by the end of the year.

On the other hand it is clear: Unless all natnumbers have gone, the intersection is not empty. And as long as all endsegments are infinite, infinitely many natnumbers have not gone.

Regards, WM

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#603329

FromEram semper recta <thenewcalculus@gmail.com>
Date2023-06-26 05:19 -0700
Message-ID<159272f7-cb81-4320-a534-9f81e4bc3c68n@googlegroups.com>
In reply to#603017
On Friday, 23 June 2023 at 07:41:05 UTC-4, WM wrote:
> We assume that all endsegments E(n) = {n, n+1, n+2, ...} of natural numbers are visible and infinite and have an empty intersection. Then there exists at least one visible endsegment E(k) containing at least one natural number j ≥ k that is not in at least one of its predecessors. 
> 
> Acceptable? 
> 
> Regards, WM

I shared your question with ChatGPT.

WM:
I have a question about set theory.

Let's assume that all end segments E(n) = {n, n+1, n+2, ...} of natural numbers are visible and infinite and have an empty intersection. Then there exists at least one visible end segment E(k) containing at least one natural number j ≥ k that is not in at least one of its predecessors.

Is this conclusion correct?  Please keep your answers directed and avoid adding any extraneous or non-relevant information.

ChatGPT:
Yes, the conclusion is correct. If all end segments E(n) of natural numbers are visible, infinite, and have an empty intersection, then for any given natural number k, there exists an end segment E(k) that contains at least one natural number j ≥ k that is not present in any of its predecessors.

This can be inferred from the fact that each end segment E(k) contains all natural numbers greater than or equal to k. Therefore, there will always be numbers in E(k) that are not present in any of its predecessors E(n) where n < k.

-----------------------------------------------------------------------------------------------------

This is the opinion of ChatGPT which is not necessarily reliable, but on first inspection, the natural conclusion is that your initial claim is correct.

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#603337

FromFritz Feldhase <franz.fritschee.ff@gmail.com>
Date2023-06-26 08:12 -0700
Message-ID<24f738f6-6331-4d62-87b0-4dfe0da8a27dn@googlegroups.com>
In reply to#603329
On Monday, June 26, 2023 at 2:19:40 PM UTC+2, Eram semper recta wrote:

> ChatGPT: 
>
> Yes, the conclusion is correct. [...]
> 
> This can be inferred from the fact that each end segment E(k) contains all natural numbers greater than or equal to k. Therefore, there will always be numbers in E(k) that are not present in any of its predecessors E(n) where n < k. 
> 
> ----------------------------------------------------------------------------------------------------- 
> 
> This is the opinion of ChatGPT which is not necessarily reliable, but on first inspection, the natural conclusion is that your initial claim is correct.

If you had a brain, you would see that ChatGPT's ARGUMENT (i.e. its claim "This can <bla>") is wrong.

It claims: "This can be inferred from the fact that each end segment E(k) contains all natural numbers greater than or equal to k."

Actually, it cannot be infered from that "fact". In fact, the CONTRARY can be inferred from that "fact"!

Hint: "The fact" (due to ChatGPT) is: E(k) := {m e IN : m >= k}. And this definition is indeed fine (i.e. the usual one or at least equivalent with common definitions).

Now ChatGPT claims:

"Therefore, there will always be numbers in E(k) that are not present in any of its predecessors E(n) where n < k."

But this [that there will always be numbers in E(k) that are not present in any of its predecessors E(n) where n < k] is OBVIOUSLY NOT the case.

Hint: x in E(k) implies that x in IN and _x > k_ (by definition of E(.), see above). Now, if n < k, then clearly _x > n_ too (since x > k > n). Hence (again by definition of E(.)) x in E(n). Hence for each and every x: if x in E(k), then x in E(n). In other words, E(k) c E(n), where n < k. ==> ALL numbers in E(k) "are present" in any of its predecessors E(n) where n < k.

Not that hard, is it?

This just proves that WM, JG and ChatGPT are complete math idiots.

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#603340

FromEram semper recta <thenewcalculus@gmail.com>
Date2023-06-26 08:42 -0700
Message-ID<2c98b292-ac73-4091-87c4-d9753513a231n@googlegroups.com>
In reply to#603337
On Monday, 26 June 2023 at 11:12:38 UTC-4, Fritz Feldhase wrote:
> On Monday, June 26, 2023 at 2:19:40 PM UTC+2, Eram semper recta wrote: 
> 
> > ChatGPT: 
> >
> > Yes, the conclusion is correct. [...]
> > 
> > This can be inferred from the fact that each end segment E(k) contains all natural numbers greater than or equal to k. Therefore, there will always be numbers in E(k) that are not present in any of its predecessors E(n) where n < k. 
> > 
> > ----------------------------------------------------------------------------------------------------- 
> > 
> > This is the opinion of ChatGPT which is not necessarily reliable, but on first inspection, the natural conclusion is that your initial claim is correct.
> If you had a brain, you would see that ChatGPT's ARGUMENT (i.e. its claim "This can <bla>") is wrong. 
> 
> It claims: "This can be inferred from the fact that each end segment E(k) contains all natural numbers greater than or equal to k." 
> 
> Actually, it cannot be infered from that "fact". In fact, the CONTRARY can be inferred from that "fact"! 

You're wrong, moron.  It can be inferred and it is inferred.

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#603342

FromFritz Feldhase <franz.fritschee.ff@gmail.com>
Date2023-06-26 08:57 -0700
Message-ID<d3a6f937-6ebe-4e9d-a00e-9ff30d9abd9en@googlegroups.com>
In reply to#603340
On Monday, June 26, 2023 at 5:42:16 PM UTC+2, Eram semper recta wrote:
> On Monday, 26 June 2023 at 11:12:38 UTC-4, Fritz Feldhase wrote: 
> > On Monday, June 26, 2023 at 2:19:40 PM UTC+2, Eram semper recta wrote: 
> > >
> > > ChatGPT: 
> > > 
> > > Yes, the conclusion is correct. [...] 
> > > 
> > > This can be inferred from the fact that each end segment E(k) contains all natural numbers greater than or equal to k. Therefore, there will always be numbers in E(k) that are not present in any of its predecessors E(n) where n < k. 
> > > 
> > > ----------------------------------------------------------------------------------------------------- 
> > > 
> > > This is the opinion of ChatGPT which is not necessarily reliable, but on first inspection, the natural conclusion is that your initial claim is correct. 
> > >
> > If you had a brain, you would see that ChatGPT's ARGUMENT (i.e. its claim "This can <bla>") is wrong. 
> > 
> > It claims: "This can be inferred from the fact that each end segment E(k) contains all natural numbers greater than or equal to k." 
> > 
> > Actually, it cannot be infered from that "fact". In fact, the CONTRARY can be inferred from that "fact"!
> >
> You're wrong, moron. It can be inferred and it is inferred.

Nope.

Hint: "The fact" (due to ChatGPT) is: E(k) := {m e IN : m >= k}. And this definition is indeed fine (i.e. the usual one or at least equivalent with common definitions).

Now ChatGPT claims:

"Therefore, there will always be numbers in E(k) that are not present in any of its predecessors E(n) where n < k."

But this [that there will always be numbers in E(k) that are not present in any of its predecessors E(n) where n < k] is OBVIOUSLY NOT the case.

Hint: x in E(k) implies that x in IN and _x > k_ (by definition of E(.), see above). Now, if n < k, then clearly _x > n_ too (since x > k > n). Hence (again by definition of E(.)) x in E(n). Hence for each and every x: if x in E(k), then x in E(n). In other words, E(k) c E(n), where n < k. ==> ALL numbers in E(k) "are present" in any of its predecessors E(n) where n < k.

Not that hard, is it?

This just proves that WM, JG and ChatGPT are complete math idiots.

----------------------------------------------------------------------------------------------------- 

Summary: The definition E(j) := {m e IN : m >= j} (where j e IN) implies E(k) ⊆ E(n) for all k,n e IN where k > n. But ChatGPT and JG are too dumb to get that.

Hint: E(1) ⊇ E(2) ⊇ E(3) ⊇ ...

Since: E(1) = {1, 2, 3, ...}, E(2) = {2, 3, 4, ...}, E(3) = {3, 4, 5, ...}, ...

Hence there will be NO "numbers in E(k) that are not present in any of its predecessors E(n) where n < k."

Therefore, ChatGPT's "argument" is WRONG.

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#603380

FromWM <askasker48@gmail.com>
Date2023-06-26 13:01 -0700
Message-ID<262b0a63-df2a-4df6-9191-d3b1110b5b47n@googlegroups.com>
In reply to#603342
Fritz Feldhase schrieb am Montag, 26. Juni 2023 um 17:57:28 UTC+2:

> Therefore, ChatGPT's "argument" is WRONG.

It is wrong that all infinite endsegments have an empty intersection. The argument "n is not in E(n+1)" removes all natnumbers from the sequence of endsegments as well as from their intersection.

Regards, WM

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#603394

FromFromTheRafters <FTR@nomail.afraid.org>
Date2023-06-26 16:27 -0400
Message-ID<u7csbo$12jl9$1@dont-email.me>
In reply to#603380
WM formulated the question :
> Fritz Feldhase schrieb am Montag, 26. Juni 2023 um 17:57:28 UTC+2:
>
>> Therefore, ChatGPT's "argument" is WRONG.
>
> It is wrong that all infinite endsegments have an empty intersection. The 
> argument "n is not in E(n+1)" removes all natnumbers from the sequence of 
> endsegments as well as from their intersection.

That's just crazy talk!

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#603461

FromFritz Feldhase <franz.fritschee.ff@gmail.com>
Date2023-06-27 02:55 -0700
Message-ID<72e3d895-3025-4380-bfd9-1041ec8f5e21n@googlegroups.com>
In reply to#603380
On Monday, June 26, 2023 at 10:01:55 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Montag, 26. Juni 2023 um 17:57:28 UTC+2: 
> >
> > Therefore, ChatGPT's "argument" is WRONG.
> >
> It is wrong that all infinite endsegments have an empty intersection.

Nope. It's a trivial mathematical fact.

ChatGTP: "it is true that the intersection of all infinite endsegments defined as E(n) = {n, n+1, n+2, n+3, ...} is the empty set."

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#603376

FromWM <askasker48@gmail.com>
Date2023-06-26 12:52 -0700
Message-ID<bdafcd85-a9c6-4b5f-a6de-ae936a1f73efn@googlegroups.com>
In reply to#603329
Eram semper recta schrieb am Montag, 26. Juni 2023 um 14:19:40 UTC+2:
> On Friday, 23 June 2023 at 07:41:05 UTC-4, WM wrote: 
> > We assume that all endsegments E(n) = {n, n+1, n+2, ...} of natural numbers are visible and infinite and have an empty intersection. Then there exists at least one visible endsegment E(k) containing at least one natural number j ≥ k that is not in at least one of its predecessors. 
> > 
> > Acceptable? 
> >
> > Regards, WM 
> 
> I shared your question with ChatGPT. 
> 
> WM: 
> I have a question about set theory. 
> 
> Let's assume that all end segments E(n) = {n, n+1, n+2, ...} of natural numbers are visible and infinite and have an empty intersection. Then there exists at least one visible end segment E(k) containing at least one natural number j ≥ k that is not in at least one of its predecessors. 
> 
> Is this conclusion correct? Please keep your answers directed and avoid adding any extraneous or non-relevant information. 
> 
> ChatGPT: 
> Yes, the conclusion is correct. If all end segments E(n) of natural numbers are visible, infinite, and have an empty intersection, then for any given natural number k, there exists an end segment E(k) that contains at least one natural number j ≥ k that is not present in any of its predecessors. 
> 
> This can be inferred from the fact that each end segment E(k) contains all natural numbers greater than or equal to k. Therefore, there will always be numbers in E(k) that are not present in any of its predecessors E(n) where n < k. 
> 
> ----------------------------------------------------------------------------------------------------- 
> 
> This is the opinion of ChatGPT which is not necessarily reliable, but on first inspection, the natural conclusion is that your initial claim is correct.

Thank you. Another piece of circumstantial evidence is the denial of all matheolgians to say what remains in the infinite endsegments when all natural numbers n have been lost in E(n+1).

Regards, WM

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#603462

FromFritz Feldhase <franz.fritschee.ff@gmail.com>
Date2023-06-27 03:06 -0700
Message-ID<75f5a993-4879-458a-9575-3e9d0415f7c4n@googlegroups.com>
In reply to#603376
On Monday, June 26, 2023 at 9:52:07 PM UTC+2, WM wrote:

> Thank you.

3 questions, 3 answers:


(1) Question: "An endsegement is defined as E(n) = {n, n+1, n+2, n+3, ...}. Is it true that the intersection of all these endsegments is the empty set."

ChatGTP: "Yes, you are correct. The intersection of all the endsegments defined as E(n) = {n, n+1, n+2, n+3, ...} is indeed the empty set.

If we consider [all] of these endsegments, we can see that they have no common elements. Each endsegment starts at a different number, and as we move along the sequence, the numbers only increase. Therefore, there is no number that belongs to all of the endsegments simultaneously.

Since the intersection of sets refers to the elements that are common to all sets being considered, and in this case, there are no common elements, the intersection is empty."

ChatGPT is completely right here!


(2) Question: "An endsegement is defined as E(n) = {n, n+1, n+2, n+3, ...}. Is it true that the intersection of all infinite endsegments is the empty set."

ChatGTP: "Yes, it is true that the intersection of all infinite endsegments defined as E(n) = {n, n+1, n+2, n+3, ...} is the empty set."

If we consider [all] infinite endsegments, they will have no common elements. Each endsegment starts at a different number, and as we move along the sequence, the numbers only increase. Thus, there is no number that belongs to all of the endsegments simultaneously.

Since the intersection of sets refers to the elements that are common to all sets being considered, and in this case, there are no common elements, the intersection is empty."

Once more ChatGPT is completely right here! (Actually, the answer agrees with the answer to question 1., since all endsegments are infinite. Siee question 3.)


(3) Question: "An endsegement is defined as E(n) = {n, n+1, n+2, n+3, ...}. Is it true that all endsegments are infinite "

ChatGTP: "Yes, all endsegments defined as E(n) = {n, n+1, n+2, n+3, ...} are indeed infinite.

By definition, an endsegment extends infinitely in one direction. In this case, each endsegment starts at the number n and continues indefinitely, incrementing by 1 with each subsequent element. Therefore, the set of elements in each endsegment is infinite.

For example, if we take E(1), it would be {1, 2, 3, ...}, which represents the set of all positive integers, extending infinitely in the positive direction. Similarly, E(2) would be {2, 3, 4, ...}, E(3) would be {3, 4, 5, ...}, and so on. In each case, the endsegment continues infinitely without [an] endpoint."

Again, ChatGPT is completely right here!


You are welcome!

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#603463

FromGus Gassmann <horand.gassmann@gmail.com>
Date2023-06-27 04:20 -0700
Message-ID<7fef4eaa-4172-4f15-b419-bab102b02f7dn@googlegroups.com>
In reply to#603462
On Tuesday, 27 June 2023 at 07:06:07 UTC-3, Fritz Feldhase wrote:
> On Monday, June 26, 2023 at 9:52:07 PM UTC+2, WM wrote: 
> 
> > Thank you. 
> 
> 3 questions, 3 answers: 
> 
> 
> (1) Question: "An endsegement is defined as E(n) = {n, n+1, n+2, n+3, ...}. Is it true that the intersection of all these endsegments is the empty set." 
> 
> ChatGTP: "Yes, you are correct. The intersection of all the endsegments defined as E(n) = {n, n+1, n+2, n+3, ...} is indeed the empty set. 
> 
> If we consider [all] of these endsegments, we can see that they have no common elements. Each endsegment starts at a different number, and as we move along the sequence, the numbers only increase. Therefore, there is no number that belongs to all of the endsegments simultaneously. 
> 
> Since the intersection of sets refers to the elements that are common to all sets being considered, and in this case, there are no common elements, the intersection is empty." 
> 
> ChatGPT is completely right here! 
> 
> 
> (2) Question: "An endsegement is defined as E(n) = {n, n+1, n+2, n+3, ...}. Is it true that the intersection of all infinite endsegments is the empty set." 
> 
> ChatGTP: "Yes, it is true that the intersection of all infinite endsegments defined as E(n) = {n, n+1, n+2, n+3, ...} is the empty set." 
> 
> If we consider [all] infinite endsegments, they will have no common elements. Each endsegment starts at a different number, and as we move along the sequence, the numbers only increase. Thus, there is no number that belongs to all of the endsegments simultaneously. 
> 
> Since the intersection of sets refers to the elements that are common to all sets being considered, and in this case, there are no common elements, the intersection is empty." 
> 
> Once more ChatGPT is completely right here! (Actually, the answer agrees with the answer to question 1., since all endsegments are infinite. Siee question 3.) 
> 
> 
> (3) Question: "An endsegement is defined as E(n) = {n, n+1, n+2, n+3, ...}. Is it true that all endsegments are infinite " 
> 
> ChatGTP: "Yes, all endsegments defined as E(n) = {n, n+1, n+2, n+3, ...} are indeed infinite. 
> 
> By definition, an endsegment extends infinitely in one direction. In this case, each endsegment starts at the number n and continues indefinitely, incrementing by 1 with each subsequent element. Therefore, the set of elements in each endsegment is infinite. 
> 
> For example, if we take E(1), it would be {1, 2, 3, ...}, which represents the set of all positive integers, extending infinitely in the positive direction. Similarly, E(2) would be {2, 3, 4, ...}, E(3) would be {3, 4, 5, ...}, and so on. In each case, the endsegment continues infinitely without [an] endpoint." 
> 
> Again, ChatGPT is completely right here! 

Can't be!!! ChatGTP learns from absorbing vast amounts of material on the internet. Of course this includes overwhelming amounts of stuff written by mathematicians, and only a small handful of authors such as WM, JG, AP who contradict. Evil mathematicians have gotten to ChatGTP!

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#603479

FromFromTheRafters <FTR@nomail.afraid.org>
Date2023-06-27 11:30 -0400
Message-ID<u7evb2$1e811$1@dont-email.me>
In reply to#603462
Fritz Feldhase formulated on Tuesday :
> On Monday, June 26, 2023 at 9:52:07 PM UTC+2, WM wrote:
>
>> Thank you.
>
> 3 questions, 3 answers:
>
>
> (1) Question: "An endsegement is defined as E(n) = {n, n+1, n+2, n+3, ...}. 
> Is it true that the intersection of all these endsegments is the empty set."
>
> ChatGTP: "Yes, you are correct. The intersection of all the endsegments 
> defined as E(n) = {n, n+1, n+2, n+3, ...} is indeed the empty set.
>
> If we consider [all] of these endsegments, we can see that they have no 
> common elements. Each endsegment starts at a different number, and as we move 
> along the sequence, the numbers only increase. Therefore, there is no number 
> that belongs to all of the endsegments simultaneously.
>
> Since the intersection of sets refers to the elements that are common to all 
> sets being considered, and in this case, there are no common elements, the 
> intersection is empty."
>
> ChatGPT is completely right here!
>
>
> (2) Question: "An endsegement is defined as E(n) = {n, n+1, n+2, n+3, ...}. 
> Is it true that the intersection of all infinite endsegments is the empty 
> set."
>
> ChatGTP: "Yes, it is true that the intersection of all infinite endsegments 
> defined as E(n) = {n, n+1, n+2, n+3, ...} is the empty set."
>
> If we consider [all] infinite endsegments, they will have no common elements. 
> Each endsegment starts at a different number, and as we move along the 
> sequence, the numbers only increase. Thus, there is no number that belongs to 
> all of the endsegments simultaneously.
>
> Since the intersection of sets refers to the elements that are common to all 
> sets being considered, and in this case, there are no common elements, the 
> intersection is empty."
>
> Once more ChatGPT is completely right here! (Actually, the answer agrees with 
> the answer to question 1., since all endsegments are infinite. Siee question 
> 3.)
>
>
> (3) Question: "An endsegement is defined as E(n) = {n, n+1, n+2, n+3, ...}. 
> Is it true that all endsegments are infinite "
>
> ChatGTP: "Yes, all endsegments defined as E(n) = {n, n+1, n+2, n+3, ...} are 
> indeed infinite.
>
> By definition, an endsegment extends infinitely in one direction. In this 
> case, each endsegment starts at the number n and continues indefinitely, 
> incrementing by 1 with each subsequent element. Therefore, the set of 
> elements in each endsegment is infinite.
>
> For example, if we take E(1), it would be {1, 2, 3, ...}, which represents 
> the set of all positive integers, extending infinitely in the positive 
> direction. Similarly, E(2) would be {2, 3, 4, ...}, E(3) would be {3, 4, 5, 
> ...}, and so on. In each case, the endsegment continues infinitely without 
> [an] endpoint."
>
> Again, ChatGPT is completely right here!
>
>
> You are welcome!

I would rather believe you, than it.

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