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There's a famous proof that $\sqrt{2}$ is irrational by assuming $\sqrt{2}=p/q$ for relatively prime $p$ and $q$ and then proving that this leads to $p$ and $q$ being both even which contradicts with them being coprime.

Now there's something I noticed that may make another easier proof:

Assume for the sake of contradiction that there exists positive integers $p$ and $q$ such that $\gcd(p,q)=1 \implies \gcd(p^2,q^2)=1$, and $$\sqrt{2}=\frac{p}{q} \implies 2q^2=p^2 \implies q^2|p^2$$ Now the last result contradicts with $p^2$ and $q^2$ being coprime, except if $q^2=1 \implies q= 1 \implies p^2=2$, but $2$ is not a perfect square so there's no such $p$

Is this proof correct?

Edit (adding a comment): this can be generalized that for any positive integer $n>1$, the $n$th root of a non-$n$th-powered integer is irrational.

For example: let $k\ne m^n$ be a positive integer for any positive integer $m$, we have $\sqrt[n]{k}$ is irrational, because assume for the sake of contradiction (for coprime positive integers $p,q$, we have $\gcd(p,q)=1 \implies \gcd(p^n,q^n)=1$ because $p$ and $q$ have different prime factors and thus so is $p^n$ and $q^n$, considering the fundamental theorem of arithmetic) $$\sqrt[n]{k}=p/q \implies kq^n=p^n \implies q^n|p^n$$ this, similarly, contradicts with $\gcd(p^n,q^n)=1$ and implies $q^n=1$ and so $q=1$ so $k=p^n$, a contradiction.

Anas A. Ibrahim
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    to me, the proof seems almost identical to the first one, just from the equation $2q^2 = p^2$ you derive $q^2 | p^2$, instead of $2|q^2$. – Peter Franek Jun 04 '20 at 06:49
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    your might want to prove that $\gcd(p,q)=1 \implies \gcd(p^2,q^2)=1$ instead of assuming it. – razivo Jun 04 '20 at 06:50
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    Also note that you can replce $2$ by any prime. – caffeinemachine Jun 04 '20 at 06:51
  • @razivo but don't you think it's a bit obvious? Or do I have to say that $p$ and $q$ have different prime factors and so will $p^2$ and $q^2$ – Anas A. Ibrahim Jun 04 '20 at 06:54
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    @AnasA.Ibrahim lot of "obvious" facts are also proven in the original – razivo Jun 04 '20 at 06:55
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    The original proof only needs easy facts about odd and even numbers. This proof might be a bit more elegant, but I do not think it is a significant simplification. – Peter Jun 04 '20 at 07:11
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    However, this approach is more suitable for the more generalized fact that the square root of a non-square positive integer is irrational. Therefore (+1) – Peter Jun 04 '20 at 07:15
  • @Peter or the $n$th root of a non-$n$th-powered integer is irrational, we can have (for a positive integer $k \ne m^n$ for a positive integer $m$) $$\sqrt[n]{k}=p/q \implies kq^n=p^n \implies q^n=1 \implies k=p^n$$ and this is a contradiction. – Anas A. Ibrahim Jun 04 '20 at 07:20
  • @AnasA.Ibrahim yes, this is even more generalized. – Peter Jun 04 '20 at 07:21
  • @Peter I added it in the question to be useful for others, thanks – Anas A. Ibrahim Jun 04 '20 at 07:30

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Yes, it's valid, provided you address @razivo's point. Indeed, any prime factor dividing both $p^2$ and $q^2$ divides both $p$ and $q$.

In case you're interested, there are several other well-known proofs. (It looks like that list omits the proof by the rational root theorem.)

I'll leave it to others to say when "two proofs" are different enough to be different proofs.

J.G.
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