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.

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    $\begingroup$ Welcome to Mathematics Stack Exchange. I think you mean $r^\color{red}2=\frac{m^2}{n^2}$ $\endgroup$ – J. W. Tanner Jul 9 at 3:06
  • $\begingroup$ @J.W.Tanner Sorry I really need to carefully proofread before I add these thank you $\endgroup$ – user686544 Jul 9 at 3:11
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    $\begingroup$ 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) $\endgroup$ – J. W. Tanner Jul 9 at 3:12
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    $\begingroup$ it looks better now $\endgroup$ – J. W. Tanner Jul 9 at 4:24
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    $\begingroup$ 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. $\endgroup$ – 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

  • $\begingroup$ 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? $\endgroup$ – user686544 Jul 9 at 3:34
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    $\begingroup$ You may simply state without any attempted justification that since $n$ is odd, it’s indivisible by $2$. No one can complain at that. $\endgroup$ – Lubin Jul 9 at 3:48
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    $\begingroup$ @ Lubin Is this okay now? $\endgroup$ – user686544 Jul 9 at 4:01

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