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Suppose for the sake of contradiction that $\sqrt{3}$ is rational. Then $\sqrt{3} = p/q$ for $p,q \in \mathbb{Z}$. Also assume that $p$ and $q$ have no common divisor, thus $p$ and $q$ can't both be even.

So, $\sqrt{3} = p/q \implies 3 = p^2/q^2 \implies p^2 = 3q^2 \implies 2p^2 = 6q^2 \implies 2p^2 = 2(3q^2).$

Since $2p^2$ is even, it follows that $p$ is even. Thus, $p=2k$ for some $k\in \mathbb{Z}$.

So we have $2p^2 = 2(3q^2) \implies 2(2k)^2 = 2(3q^2) \implies 8k^2 = 6q^2 \implies 2(4k^2) = 6q^2.$

Since $6q^2$ is even, it follows that $q$ is even. But $p$ and $q$ can't both be even, thus we have a contradiction.

lopan
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    Nope, $2(3^2)=18$ is even which does not imply that $3^2=9$ is even. – n3u1r0n Jan 20 '23 at 19:35
  • "Since $2p^2$ is even, it follows that $p$ is even" and "Since $6q^2$ is even, it follows that $q$ is even" are both seriously wrong. If $p$ is an odd integer then $2p^2$ and $6p^2$ must be even – Henry Jan 20 '23 at 19:37
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    Your argument seems to work just as well to prove that $\sqrt 9$ is irrational. – lulu Jan 20 '23 at 19:37
  • Henry, if $p$ is an even integer then $2p^2$ and $6p^2$ must also be even right? – lopan Jan 20 '23 at 19:44
  • $2\cdot 3^2$ is even but $3$ isn't even. – Thomas Andrews Jan 20 '23 at 19:44
  • In the proof that $\sqrt 2$ is irrational, you get that $2p^2=q^2,$ and you deduce first that $q$ is even (because $q^2$ is even.) From that, you conclude that $p$ is even. You can't conclude that $p$ is even just because $2p^2$ is even - indeed $2p^2$ is always an integer when $p$ is any integer. – Thomas Andrews Jan 20 '23 at 19:48
  • If we had $p^2=2(3q^2)$ we say $p^2$ is even by the definition of an even number which is $2k$ for some $k\in\mathbb{Z}$. So we have $p^2=2k$ for $k=3q^2$. Thus if $p^2$ is even, $p$ must be even. So if this is true, then $2p^2$ must be even. – lopan Jan 20 '23 at 19:53
  • Really, go through your argument using $\sqrt 9$ instead of $\sqrt 3$. Almost nothing changes. You get $p^2=9q^2$ which you can multiply by $2$ to get $2p^2=18q^2$ so $2p^2$ is even which, according to you at least, somehow implies that $p$ is even, and so on. But of course $\sqrt 9$ is rational. – lulu Jan 20 '23 at 19:57
  • Thanks, I know I'm wrong, just trying to pinpoint it. Will need to reflect on this. – lopan Jan 20 '23 at 19:58
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    Never use "solution verification" as the only tag for your question. – amWhy Jan 20 '23 at 22:39

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