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Topic: If you claim 0.999... is a rational number, then you must find
p/q such that 0.999... = p/q. 12/26/2017

Replies: 40   Last Post: Jan 4, 2018 7:37 AM

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 zelos.malum@gmail.com Posts: 1,176 Registered: 9/18/17
Re: If you claim 0.999... is a rational number, then you must find
p/q such that 0.999... = p/q. 12/26/2017

Posted: Dec 30, 2017 3:45 AM

Den fredag 29 december 2017 kl. 10:32:25 UTC+1 skrev WM:
> Am Freitag, 29. Dezember 2017 04:51:03 UTC+1 schrieb Zelos Malum:
> > >Yes, in case of irrationals. A sum of fractions will not yield an irrational value. The belief that infinitely many fractions could yield an irrational is a silly as the belief that infinitely many red points would yield a green line.
> >
> > Not at all, the former can be proved easily.

>
> It can be "proved" by refusing logics only. Every sober mind with minimum intellectual capacities knows and can prove that your claim is wrong. Simply assume logic:
>
> For all n in |N: 0.999...9_n =/= 1. This is true for every n. The infinite sequence does not contain more than every index. Therefore the claim that the infinite sequence would reach more than all finite sequences is as silly and interesting the as a chess competition of metally handicapped players.
>
> Same holds in case of irrational numbers.
>

> >
> > >One of the wrong "results" is the idea that Cantor proved the existence of transcendental numbers and the idea that the measure of the reals is 1 and the measure of the rationals is 0 in the unit interval.
> >
> > And how is that wrong? They are provably true.

>
> Again provable by refusing to apply logic only.
>
> There are transcendental numbers (better simply say transcendentals) as has been proved by Liouville. But that has nothing to do with Cantor's argument.
>
> There is no measure of rational points. For proof see "The meadow saffron dream" in https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p.332.
>
> Regards, WM

The proof for sums resulting in irrationals is fairly straight forward.

We know that Q is dense in R, pick an element x e R/Q, as Q is dense, that means there exists a sequence of elements in Q that converges to x. We can then make ssummants c_i by having c_i=x_i-Sum_{j<i} x_j

then sum_{j=1 -> oo} c_i = x

but because all c_i are finite sums of rational numbers (each x_i being rational as well from definition) that means c_i is rational, but x is not rational. Therefore we have that the infinite sum of rational numbers can be irrational.