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Possible Duplicate:
use contradiction to prove that the square root of $p$ is irrational

I was sitting at school bored, and I suddenly thought about prime numbers and an interesting question popped up in my head:

$\bf\text{Is the root of every prime number irrational?}$

My intuition told me yes, and I wonder if there exists a simple proof proving this statement (or a counter-example)?

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    Not obsolete, though it is a duplicate. Yes, it's a theorem that if $n$ is a natural number, then $\sqrt{n}$ is rational if *and only if* it is an integer. Thus, only the perfect squares have rational roots.2012-12-17

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Assuming you mean the square root, you could proceed by contradiction. Assume $\sqrt{p}$ is rational for prime $p$. Then $ \sqrt{p}=\frac{a}{b} $ for some natural numbers $a$ and $b$, $b\neq 0$. Then $ p \cdot b^2=a^2 $ Do you see a contradiction? Try considering the prime factorization of both sides.

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    @MichaelHardy Very true. That's$a$useful generalization I hadn't thought of. For sake of my current notation though, the assumption is necessary. Thanks.2012-12-17