The irrationality of the square root of 2

Question

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?

Pretty much the definition of rational number implies a ratio of two integers. Do you want to propose another definition for rational? — Carlos Eugenio Thompson Pinzón, Nov 06 '13 at 00:32
Using the rational root theorem, you can prove that $x^2-2=0$ has no rational roots. Since $\sqrt 2$ is a root of this quadratic equation, you can claim that it's irrational — , Dec 30 '21 at 19:58

score 1 · Answer 1 · answered Nov 06 '13 at 00:30

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

answered Nov 06 '13 at 00:30

LASV

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score 0 · Accepted Answer · edited Apr 13 '17 at 12:58

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:

https://mathoverflow.net/questions/32011/direct-proof-of-irrationality/32017#32017 (That's a really neat proof of the type for which you are looking.)
http://en.wikipedia.org/wiki/Square_root_of_2#Proofs_of_irrationality
Direct proof for the irrationality of $\sqrt 2$.
Irrationality proofs not by contradiction

score 0 · Answer 3 · answered Nov 06 '13 at 00:47

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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

answered Nov 06 '13 at 00:47

Haha

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The irrationality of the square root of 2

3 Answers3