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This question already has an answer here:

Is there a proof to the irrationality of the square root of 2 besides using the argument that a rational number is expressed to be p/q?

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marked as duplicate by egreg, Dan Rust, Nick Peterson, Daniel Fischer, user1620696 Nov 6 '13 at 1:09

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    $\begingroup$ Pretty much the definition of rational number implies a ratio of two integers. Do you want to propose another definition for rational? $\endgroup$ – Carlos Eugenio Thompson Pinzón Nov 6 '13 at 0:32
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This question has been asked before. Please search Math.SE for an answer to your question before asking.

Regardless, yes, there are lots of ways to prove the irrationality of $\sqrt{2}$. Here are some pertinent resources:

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Here is an overkill proof: $x^2-2$ is irreducible over $\mathbb{Q}$ by Eisenstein's =].

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If $\sqrt 2=p/q$ where $gcd(p,q)=1$ then $2=p^2/q^2$ <=>$p^2=2q^2$ and thus $p^2$ is even and thus $p$ is even let's say $p=2m$. Then $4m^2=2q^2$<=> $q^2=2m^2$ and thus $q$ is also even. So $gcd(p,q)>1$ which is false

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