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From: wij <wyniijj5@gmail.com>
Newsgroups: comp.theory
Subject: Re: Cantor Diagonal Proof
Date: Wed, 09 Apr 2025 22:31:09 +0800
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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.
> > >=20
> > > 	I don't think that's quite what Wij is claiming.=C2=A0 He thinks,
> > > 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-a=
nalysis,
> > > limits don't exist [as I understand it], because he doesn't accept
> > > that there are no infinitesimals.=C2=A0 It's like those who dispute t=
hat
> > > 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, a=
s
> > > 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.=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 refusing to l=
earn
> > > about the existing non-standard systems.
> >=20
> > 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.
>=20
> I have read that document, about which I have a simple question.
>=20
> =C2=A0From Theorem 2 and Axiom 2, if x can be expressed in the form of=
=20
> p/q, then p and q will be infinite numbers (non-natural numbers).=20
> 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"=20
> (non-natural) numbers?

The audience of the file was originally intended to include 12 years old ki=
ds.
Wordings in the file wont' be precise enough to meet rigorous requirements.=
=20
The mentioned paragraph was revised (along with several others):

Theorem 2: =E2=84=9A+=E2=84=9A=3D=E2=84=9A (the sum of a rational number an=
d a rational number is still a
        rational number), but it is only true for finite addition steps.
  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
         terms in Q' is greater than the individual terms, the sum q of the
         infinite terms (q=3Dq=E2=82=81+q=E2=82=82+q=E2=82=83...) is not a =
fixed number.

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 won't be
natural numbers. Thus, "=E2=84=9A+=E2=84=9A=3D=E2=84=9A" is conditionally t=
rue (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.