Re: Cantor Diagonal Proof

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 irrational.

>
I don't think that's quite what Wij is claiming.  He thinks,
rather, that 0.333... is different from 1/3.  No matter how far you
pursue that sequence, you have a number that is slightly less than
1/3.  In real analysis, the limit is 1/3 exactly.  In Wij-analysis,
limits don't exist [as I understand it], because he doesn't accept
that there are no infinitesimals.  It's like those who dispute that
0.999... == 1 [exactly], and when challenged to produce a number
between 0.999... and 1, produce 0.999...5.  They have a point, as
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.  But we have no good
and widely-known notation for what can follow a "...", so the Wijs of
this world get mocked.  He doesn't help himself by refusing to learn
about the existing non-standard systems.

>
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.

>
I have read that document, about which I have a simple question.
>
   From 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.
>
Let p = 1
Let q = 3
>
Is it or is it not your contention that p and q are "infinite"
(non-natural) numbers?

>
The audience of the file was originally intended to include 12 years old kids.
Wordings in the file wont' be precise enough to meet rigorous requirements.
The mentioned paragraph was revised (along with several others):
>
Theorem 2: ℚ+ℚ=ℚ (the sum of a rational number and a rational number is still a
          rational number), but it is only true for finite addition steps.
    Proof: Let Q'={p/q| p,q∈ℕ, q≠0 and p/q>0}, then Q'⊂ℚ. Since the sum of any two
           terms in Q' is greater than the individual terms, the sum q of the
           infinite terms (q=q₁+q₂+q₃...) is not a fixed number.
>
What I intended to mean is: 0.999...= 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, "ℚ+ℚ=ℚ" is conditionally true (so false).
>
But I still think your English is worse than olcott's (and mine).

>
Charmed, I'm sure.
>

Prediction: you will evade the question. Why not surprise me?

Ok, I evade more clarification.

>
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".
>
And you're right. 1 and 3 are both integers. Natural numbers.
Whole numbers. Finite numbers. Not infinite.
>
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!

 The correct equality is 1/3= 0.333... + nonzero_remainder.

Keep on dividing the remainder, and what do you get? Oh look! More 3s.

If you use it to prove, that proof never finishes. Thus, invalid.

And Achilles never catches the tortoise. Yeah, right.
--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
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