# Root of prime numbers [duplicate]

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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Can someone explain to me how to centre the part in bold? I know how to do it with formulas and such ($$signs at each end), I don't know how to do it in text though? – JohnPhteven Dec 17 '12 at 14:37 See this post. – David Mitra Dec 17 '12 at 14:37 In between double dollar signs put "\bf\text{ blah}" – David Mitra Dec 17 '12 at 14:38 @DavidMitra That is a bit over my head if I may be honest. – JohnPhteven Dec 17 '12 at 14:38 Do you mean the square root? The square root of any integer that is not a perfect square is always irrational, and primes are never perfect squares. – Henning Makholm Dec 17 '12 at 14:38 ## 1 Answer Assuming you mean the square root, you could proceed by contradiction. Assume \sqrt{p} is rational for prime p. Then$$ \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.
I approved the edit, but since I specified natural numbers (which don't include $0$) I'm not entirely sure it was necessary. – chris Dec 17 '12 at 14:48
Aha! I see the approach now. I was deriving a different one--namely, that $p$ will end up dividing both $a$ and $b$, which is impossible b/c of lowest terms. – Cameron Buie Dec 17 '12 at 14:58