# Irrational proof

Let $$n \in \mathbb{N}$$ be odd. Prove that there is no $$r \in \mathbb{Q}$$ such that $$r^2=2n$$

In my book an odd natural number $$n$$ has to be defined as $$n=2k-1$$ for $$k \in \mathbb{N}$$ This is because $$0$$ is not included in the natural numbers.

Proof With Flaws Proof#1: Assume $$r$$ rational,Let $$r=\frac{m}{n}$$ where $$\frac{m}{n} \in \mathbb{Q}$$ then $$r^2=\frac{m^2}{n^2}=2(2k-1)$$

$$\implies$$ $$m^2=2(2k-1)n^2$$ so $$m$$ is even. Then let $$m=2c$$ for $$c \in \mathbb{N}$$ and let $$2(2k-1)=2d$$ since the product of an even and odd is even.

Then $$(2c)^2=(2d)n^2$$

$$\implies$$ $$2c^2=dn^2$$ so n is even. Since $$\frac{m}{n}$$ not in lowest terms since the numerator and denominator are even the assumption was incorrect that $$r$$ is rational. Are there any flaws in the proof?

Attempted fix Proof #2 Assume $$r$$ rational. Let $$r=\frac{a}{b}$$ where $$\frac{a}{b} \in \mathbb{Q}$$.Also $$a,b$$ share no common factors other than $$1$$.Also $$2\nmid n$$ since n odd.

then $$r^2=\frac{a^2}{b^2}=2n$$

Then $$a^2=2nb^2$$ so $$a$$ is even. Let $$a=2c$$ Then

$$(2c)^2=2nb^2$$ so $$2c^2=nb^2$$, $$b$$ is even.

Since $$a,b$$ both even $$\frac{a}{b}$$ is not reduced ( they both have common factors) a contradiction so r is irrational.

• Welcome to Mathematics Stack Exchange. I think you mean $r^\color{red}2=\frac{m^2}{n^2}$ – J. W. Tanner Jul 9 at 3:06
• @J.W.Tanner Sorry I really need to carefully proofread before I add these thank you – user686544 Jul 9 at 3:11
• I see no major flaws. It would be simpler to say $d=2k-1$ rather than $2(2k-1)=2d$. You probably should say at the beginning that $m/n$ is in lowest terms; technically just because a fraction is not in lowest terms doesn't mean it's irrational (e.g., $\frac 4 2$ is rational) – J. W. Tanner Jul 9 at 3:12
• it looks better now – J. W. Tanner Jul 9 at 4:24
• If you are going into this much detail when you get that $2c^2 = nb^2$ and conclude that $b$ is even, it'd be a good idea to point out that either $2|n$ or $2|b^2$ and as $n$ is odd that $2|b^2$ and so $b$ is even. After all the the statement is not true if $n$ is even and so somewhere your proof must use $n$ is odd. This is where it does. (Note: if $n$ is even $2c^2 = nb^2$ means $nb^2$ is even but that does not mean that $b^2$ is even. – fleablood Jul 9 at 6:23

Yes, There are few flaws in your proof.

First Flaw

You assumed $$r=\frac{m}{n}$$ and at the same time you also assumed that $$r^2=2n$$ This means that $$\frac{m^2}{n^2}=2n$$ which further implies $$m^2=2n^3$$ which invites an unnecessary restriction. I will suggest you to take $$r=\frac{a}{b}$$

Second Flaw

You wrote $$n=2k-1$$ which is not necessary if you assume that $$n$$ is odd means $$2$$ doesn't divide $$n$$. It actually forced to do more unnecessary calculations.

Third Flaw

You concluded that your assumption is wrong because $$m,n$$ has 2 as a common factor, but you never assumed that $$(m,n)=1$$. You just assumed $$r=\frac{m}{n}$$ and went ahead with you proof.

I hope that now you can prove it by yourself by keeping in mind the three points

• I recognize the first flaw and the third but for the second flaw, is it wrong if I don't change this, or does it just require more unnecessary calculation? – user686544 Jul 9 at 3:34
• You may simply state without any attempted justification that since $n$ is odd, it’s indivisible by $2$. No one can complain at that. – Lubin Jul 9 at 3:48
• @ Lubin Is this okay now? – user686544 Jul 9 at 4:01