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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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marked as duplicate by m_t_, Cameron Buie, froggie, Davide Giraudo, Martin Sleziak Dec 17 '12 at 15:50

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.

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
up vote 5 down vote accepted

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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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
Check out the third paragraph here to see why someone may have suggested the edit. – Cameron Buie Dec 17 '12 at 14:52
You'll want to specify lowest terms. Otherwise, you'll not get the desired contradiction. – Cameron Buie Dec 17 '12 at 14:53
@CameronBuie Thanks. I'm just used to using natural numbers in the number theory sense. Also, I think since the contradiction lies in the fact that the left hand side has an odd number of terms in prime factorization, it doesn't matter if the fraction is reduced or not. – chris Dec 17 '12 at 14:56
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

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