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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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    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?2012-12-17
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    See [this post](http://math.stackexchange.com/questions/145065/use-contradiction-to-prove-that-the-square-root-of-p-is-irrational?rq=1).2012-12-17
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    In between double dollar signs put "\bf\text{ blah}"2012-12-17
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    @DavidMitra That is a bit over my head if I may be honest.2012-12-17
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    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.2012-12-17
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    Really? I never thought about that, I always thought some non perfect squares had rational roots.. so that renders my question obsolete I guess?2012-12-17
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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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    I approved the edit, but since I specified natural numbers (which don't include $0$) I'm not entirely sure it was necessary.2012-12-17
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    Check out the third paragraph [here](http://en.wikipedia.org/wiki/Natural_numbers) to see why someone may have suggested the edit.2012-12-17
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    You'll want to specify lowest terms. Otherwise, you'll not get the desired contradiction.2012-12-17
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    @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.2012-12-17
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    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.2012-12-17
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    Ah! Yes, for that approach you do need lowest terms. I prefer this one because the contradiction is more immediate.2012-12-17
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    Indeed. +1, by the way.2012-12-17
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    anyway, sir @CameronBuie and sir chris, lets move on. Our aim here is to make things clarified. Though I made some edit, its was not really my intention. but I learned also that it is indeed a need to specify, since natural numbers may include zero, I also thought it just consist of positive numbers. Anyway, I learned from this.2012-12-17
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    But what if $b^2=p$?2012-12-17
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    @ZafarS: An important property of prime numbers (in fact, the *definition* of prime numbers) is: If $p$ is a factor of a product of integers $m\cdot n$, then $p$ must be a factor of one of $m,n$. In particular, then, we **can't** have $p=b^2$, for then $p$ is a factor of $b\cdot b$, and necessarily a factor of $b$. Thus, $b=mp$ for some integer $m$, whence $$p=b^2=m^2p^2,$$ which is quite impossible, since all values there are integers.2012-12-17
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    Oh wait, I forgot that $a$ and $b$ must be natural, I'm sorry.2012-12-17
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    @chris : Your initial assumption that _square_ roots were intended, if it meant as opposed to cube roots, fourth roots, etc., seems not to be needed.2012-12-17
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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