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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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JohnPhteven
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1 Answers1

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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.

Philip Benj
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chris
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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
  • Ah! Yes, for that approach you do need lowest terms. I prefer this one because the contradiction is more immediate. – chris Dec 17 '12 at 14:58
  • Indeed. +1, by the way. – Cameron Buie Dec 17 '12 at 14:59
  • 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. – Philip Benj Dec 17 '12 at 15:00
  • But what if $b^2=p$? – JohnPhteven Dec 17 '12 at 15:00
  • @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. – Cameron Buie Dec 17 '12 at 15:04
  • Oh wait, I forgot that $a$ and $b$ must be natural, I'm sorry. – JohnPhteven Dec 17 '12 at 15:06
  • @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. – Michael Hardy Dec 17 '12 at 15:16
  • @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. – chris Dec 17 '12 at 15:26