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How can you prove that the square root of two is irrational?

Can $a^2 = 2b^2$ have a solution where $a, b$ are in $\mathbb{Z}$ but not zero?

$\mathbb{Z}$ = positive and negative whole numbers

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marked as duplicate by Eric Naslund, Gerry Myerson, Hans Lundmark, JavaMan, lhf Nov 7 '11 at 11:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

If you can solve $a^2=2b^2$, then $$2=\left(\frac{a}{b}\right)^2$$ which means that the square root of $2$ is rational. See the linked page for some proofs that this is impossible. – Eric Naslund Nov 7 '11 at 11:34

If you take square root of the both sides you get:

$|a|=\sqrt{2} \cdot |b|$

So the LHS represents an integer while RHS represents an irrational number therefore equality isn't true so there is no solution of this equation in the set of integers without zero.

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