Validity of a proof of the existence of $\sqrt2$

Question

In my searchings for proofs I found this page

Prove the existence of the square root of $2$.

In one of the comments there someone had posted this proof link:

http://www.public.iastate.edu/~roettger/201/sqrt2.pdf

How is this proof even valid? I think it has a "broken twist" at the end of it.

Could you write in your text which argument you do not believe? We do not want to search somewhere in a pdf-file. And you have a valid proof in the first link, right? Did you compare it? — Dietrich Burde, Dec 19 '17 at 21:17
Please check that the edits I suggested still reflect what you intended $\ddot\smile$ — gen-ℤ ready to perish, Dec 19 '17 at 21:29
Working solely within the rational numbers is fine and $\sqrt{2}$ is not a rational number and in that context would not exist as a rational number. If you wish to talk about the real number $\sqrt{2}$, then you will first need to understand how a real number is defined in the first place. Depending on how you defined the set of real numbers, the existence of $\sqrt{2}$ as a real number should follow from the definitions and/or the least upper bound property. — JMoravitz, Dec 19 '17 at 21:35
This question is a duplicated of https://math.stackexchange.com/questions/1415235/prove-the-existence-of-the-square-root-of-2? Excuse me for the user, I have a block on my account and i can not formulate any answer. — , Dec 19 '17 at 21:35

score 2 · Answer 1 · answered Dec 19 '17 at 21:24

2

In the proof, $\alpha$ is defined as the least upper bound of the set $A$ of real numbers $x\in\mathbb{R}$ such that $x^2<2$.

It is then shown that $\alpha^2<2$ is not true, and that $\alpha^2>2$ is also not true. By the law of trichotomy (I believe this is the "broken twist" at the end to which you refer), we must have one of

$$\alpha^2>2\ \mathrm{or}\ \alpha^2<2\ \mathrm{or}\ \alpha^2=2.$$

As neither of the first two occur, the third must.

answered Dec 19 '17 at 21:24

Carl Schildkraut

32,931

The "twist" is that I do not see how $x^2<2<y^2$ says anything about the second contradiction. – יהודה שמחה ולדמן Dec 19 '17 at 21:30
whare $x,y$ in this comment? – Andres Mejia Dec 19 '17 at 21:34
In that second link. – יהודה שמחה ולדמן Dec 19 '17 at 21:49
@יהודהשמחהולדמן It says, in the link, that $x^2<2<y^2$, which means $x<y$, for all $x\in A$, so $y$ is an upper bound of $A$. However, $y<\alpha$, so $\alpha$ is not the least upper bound, a contradiction. – Carl Schildkraut Dec 19 '17 at 21:50
So why couldn't he just say that $b<y<α\ \implies\ y^2<α^2$, then from the AGM inequality $2<y^2<α^2$ ? – יהודה שמחה ולדמן Dec 19 '17 at 22:03
@יהודהשמחהולדמן That's essentially what's happening. – Carl Schildkraut Dec 19 '17 at 22:06
Wait, was he trying to prove just $2<y^2<α^2$ or $x<y<α$ ? – יהודה שמחה ולדמן Dec 19 '17 at 22:18
@יהודהשמחהולדמן Once we know $2<y^2<\alpha^2$, we have $x^2<2<y^2<\alpha^2 \implies x^2<y^2<\alpha^2$. Since all numbers involved are positive and obviously squares of real numbers, we have $x<y<\alpha$. – Carl Schildkraut Dec 20 '17 at 00:53
Got it.
The biggest problem with the proof's end is that the writer, Instead of using a few symbols, used "words".
– יהודה שמחה ולדמן Dec 21 '17 at 01:53

Validity of a proof of the existence of $\sqrt2$

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