Possible Duplicate:
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
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4$\begingroup$ 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. $\endgroup$ – Eric Naslund Nov 7 '11 at 11:34
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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.
