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From: wij <wyniijj5@gmail.com>
Newsgroups: comp.theory
Subject: Re: Cantor Diagonal Proof
Date: Thu, 10 Apr 2025 00:08:56 +0800
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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 irration=
al.
> > > > >=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 tha=
n
> > > > > 1/3.=C2=A0 In real analysis, the limit is 1/3 exactly.=C2=A0 In W=
ij-analysis,
> > > > > limits don't exist [as I understand it], because he doesn't accep=
t
> > > > > that there are no infinitesimals.=C2=A0 It's like those who dispu=
te that
> > > > > 0.999... =3D=3D 1 [exactly], and when challenged to produce a num=
ber
> > > > > between 0.999... and 1, produce 0.999...5.=C2=A0 They have a poin=
t, as
> > > > > the Archimedean axiom is not one of the things that gets mentione=
d
> > > > > much at school or in many undergrad courses, and it seems like an
> > > > > arbitrary and unnecessary addition to the rules.=C2=A0 But we hav=
e no good
> > > > > and widely-known notation for what can follow a "...", so the Wij=
s of
> > > > > this world get mocked.=C2=A0 He doesn't help himself by refusing =
to learn
> > > > > 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 prov=
ided.
> > > > 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=A0From Theorem 2 and Axiom 2, if x can be expressed 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 years ol=
d kids.
> > Wordings in the file wont' be precise enough to meet rigorous requireme=
nts.
> > 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 numbe=
r and a rational number is still a
> > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 rational number), but =
it is only true for finite addition steps.
> > =C2=A0=C2=A0 Proof: Let Q'=3D{p/q| p,q=E2=88=88=E2=84=95, q=E2=89=A00 a=
nd 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 terms in Q' is g=
reater 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 infinite 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 won't =
be
> > natural numbers. Thus, "=E2=84=9A+=E2=84=9A=3D=E2=84=9A" is conditional=
ly 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=20
> English, so well done you) that you didn't intend to mean that 1=20
> and 3 are "infinite".
>=20
> And you're right. 1 and 3 are both integers. Natural numbers.=20
> Whole numbers. Finite numbers. Not infinite.
>=20
> Let us calculate the ratio of these two integers, 1/3. Oh look,=20
> it's 0.3r. So 0.3r is the ratio of two integers (i.e. rational)=20
> after all. Quelle surprise!

The correct equality is 1/3=3D 0.333... + nonzero_remainder.
If you use it to prove, that proof never finishes. Thus, invalid.