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From: wij <wyniijj5@gmail.com>
Newsgroups: comp.theory
Subject: Re: Cantor Diagonal Proof
Date: Thu, 10 Apr 2025 00:45:35 +0800
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On Wed, 2025-04-09 at 17:32 +0100, Richard Heathfield wrote:
> On 09/04/2025 17:08, wij wrote:
> > 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 convincing
> > > > > > > > mathematics before I'll be ready to accept that 1/3 is irra=
tional.
> > > > > > >=20
> > > > > > > 	I don't think that's quite what Wij is claiming.=C2=A0 He th=
inks,
> > > > > > > rather, that 0.333... is different from 1/3.=C2=A0 No matter =
how far you
> > > > > > > pursue that sequence, you have a number that is slightly less=
 than
> > > > > > > 1/3.=C2=A0 In real analysis, the limit is 1/3 exactly.=C2=A0 =
In Wij-analysis,
> > > > > > > limits don't exist [as I understand it], because he doesn't a=
ccept
> > > > > > > that there are no infinitesimals.=C2=A0 It's like those who d=
ispute that
> > > > > > > 0.999... =3D=3D 1 [exactly], and when challenged to produce a=
 number
> > > > > > > between 0.999... and 1, produce 0.999...5.=C2=A0 They have a =
point, as
> > > > > > > the Archimedean axiom is not one of the things that gets ment=
ioned
> > > > > > > much at school or in many undergrad courses, and it seems lik=
e an
> > > > > > > arbitrary and unnecessary addition to the rules.=C2=A0 But we=
 have no good
> > > > > > > and widely-known notation for what can follow a "...", so the=
 Wijs of
> > > > > > > this world get mocked.=C2=A0 He doesn't help himself by refus=
ing to learn
> > > > > > > about the existing non-standard systems.
> > > > > >=20
> > > > > > Lots of excuses like POOH. You cannot hide the fact that you do=
n'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.
> > > > >=20
> > > > > I have read that document, about which I have a simple question.
> > > > >=20
> > > > > =C2=A0=C2=A0=C2=A0From Theorem 2 and Axiom 2, if x can be express=
ed 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.
> > > > >=20
> > > > > Let p =3D 1
> > > > > Let q =3D 3
> > > > >=20
> > > > > Is it or is it not your contention that p and q are "infinite"
> > > > > (non-natural) numbers?
> > > >=20
> > > > The audience of the file was originally intended to include 12 year=
s old kids.
> > > > Wordings in the file wont' be precise enough to meet rigorous requi=
rements.
> > > > The mentioned paragraph was revised (along with several others):
> > > >=20
> > > > Theorem 2: =E2=84=9A+=E2=84=9A=3D=E2=84=9A (the sum of a rational n=
umber and a rational number is still a
> > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 rational num=
ber), but it is only true for finite addition steps.
> > > > =C2=A0=C2=A0=C2=A0 Proof: Let Q'=3D{p/q| p,q=E2=88=88=E2=84=95, q=
=E2=89=A00 and p/q>0}, then Q'=E2=8A=82=E2=84=9A. Since the sum of any two
> > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 terms =
in Q' is greater than the individual terms, the sum q of the
> > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 infini=
te terms (q=3Dq=E2=82=81+q=E2=82=82+q=E2=82=83...) is not a fixed number.
> > > >=20
> > > > What I intended to mean is: 0.999...=3D 999.../1000... (in p/q form=
)
> > > > Since p,q will be infinitely long to denote/define 0.999..., p,q wo=
n't be
> > > > natural numbers. Thus, "=E2=84=9A+=E2=84=9A=3D=E2=84=9A" is conditi=
onally true (so false).
> > > >=20
> > > > But I still think your English is worse than olcott's (and mine).
> > >=20
> > > Charmed, I'm sure.
> > >=20
> > > > > Prediction: you will evade the question. Why not surprise me?
> > > > Ok, I evade more clarification.
> > >=20
> > > 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".
> > >=20
> > > And you're right. 1 and 3 are both integers. Natural numbers.
> > > Whole numbers. Finite numbers. Not infinite.
> > >=20
> > > 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!
> >=20
> > The correct equality is 1/3=3D 0.333... + nonzero_remainder.
>=20
> Keep on dividing the remainder, and what do you get? Oh look!=20
> More 3s.

???

> > If you use it to prove, that proof never finishes. Thus, invalid.
>=20
> And Achilles never catches the tortoise. Yeah, right.

[snit from the file]
....
Paradox: This example might help explain the concept of infinite =E2=88=9E:
  Let the sequence A(0)=3D0
  A(n)=3D (A(n-1)+1)/2, n=E2=88=88=E2=84=95.

  Question: What number of n can make A(n)=3D 1?
  Answer: Since 1/2+1/4+1/8...=3D 0.999...(also a kind of 0.999...) cannot =
be=20
          equal to 1. Therefore, there is no positive integer n (including=
=20
          infinity) that can make A(n)=3D1, that is, the value of n is not =
in the
          given meaning of the question. This answer also applies to the ba=
sic
          Zeno paradox, super tasks,... and other paradoxes. Usually this t=
ype
          of paradoxes also give additional speed and time information and
          force the question to be answered what the value of n is when the
          time is up.
---------- (a minor error found, I will fix it latter)

Stick to the problem. Such puzzle won't prove the "0.999... problem".

> --=20
> Richard Heathfield
> Email: rjh at cpax dot org dot uk
> "Usenet is a strange place" - dmr 29 July 1999
> Sig line 4 vacant - apply within
>=20