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Prove that the square root of all non-square numbers $n \in \mathbb{N}$ is irrational

I have made an attempt to prove this, I don't know if it's correct though:

Take a non-square number $n \in \mathbb{N}$, and we'll assume that $\sqrt{n}$ is rational.

$\sqrt{n} = \dfrac{p}{q}$ , $p,q \in \mathbb{N}$ and they have no common factors.

$$nq^2=p^2$$ Lets say that $z$ is a prime factor of $q$, it must also be a prime factor of $q^2$. However, it then must ALSO be a prime factor of $p^2$ because of the equality above, and this is a contradiction.

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It should be $nq^2 = p^2$. –  ybungalobill Dec 26 '12 at 22:36
Good, need to add a little more. From your argument you can conclude that $q^2=1$. Then need to say a few words about that. Need also some number theory to show that if $z$ is prime and divides $ab$, then $z$ divides $a$ or $b$ or both. In informal enough contexts, perhaps this can be taken for granted. –  André Nicolas Dec 26 '12 at 22:45
@AndréNicolas: Not sure what your last sentence means. If you mean that this answer assumes the fundamental theorem of arithmetic, and this might be acceptable in this proof, then: yes, obviously that theorem is assumed (the use of "prime factor" suggests this), and there is nothing wrong with that. –  Marc van Leeuwen Dec 26 '12 at 22:52
Well, not quite Fundamental Theorem, just the fact any number $\gt 1$ is divisible by some prime (which can probably be taken for granted), and $ab$ stuff. As I indicated, it is not clear to me, since I don't know full context, whether this is needed. –  André Nicolas Dec 26 '12 at 22:57

2 Answers 2

Check carefully what you've done so far and you'll realize that you've "proved" that $\,\sqrt n\,$ is always irrational, which of course is false.

What thus is lacking? For example, take

$$\sqrt {25}=\frac{p}{q}\Longrightarrow 25q^2=p^2 \,\,,\,\,\text{ so...what?!}$$

True, if there's some prime factor of $\,q\,$ then it must also be a prime factor of $\,q^2\,$ and thus of $\,p^2\,$ (see Andre comments!) and we get a contradiction...unless...Can you see what little correction goes here?

In general, your direction is pretty good.

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Your proof is on the right lines, but you've stopped half way through. We can prove a similar proposition by using proof by contradiction.

Consider $p$, a positive, prime.

Let us assume that $\sqrt{p}$ is rational. If $n$ is rational then there exist natural number $a$ and $b$ with $b \neq 0$ such that $\sqrt{p} = a/b$. Moreover, without loss of generality we may assume that $a$ and $b$ have no common factor. (If they did have common factors then we could cancel them, e.g. $4/6$ would become $2/3$.) We shall show that this initial assumption leads to a contradiction.

If $\sqrt{p} = a/b$ then $p = a^2/b^2$ and so $a^2 = pb^2.$ Since $p$ is a factor of the right-hand side then $p$ must also be a factor of the left-hand side, i.e. a factor of $a^2$. If $p$ is a factor of $a^2$ then it must also be a factor of $a$. If $p$ is a factor of $a$ then, by definition, there exists an integer $k$ for which $a = pk$.

Using the fact that $a^2 = pb^2$ and $a = pk$ we see that $(pk)^2 = pb^2$. Expanding the left-hand side gives $p^2k^2=pb^2$. Dividing both sides by $p$ yields $pk^2 = b^2$. Clearly, $p$ divides the left-hand side, and so $p$ must also divide the right-hand side. If $p$ divides $b^2$ then $p$ must also divide $b$.

It follows that $p$ divides both $a$ and $b$, but this contradicts out previous assumption that $a$ and $b$ had no common factors. Thus, the must be a flaw in our argument. Every step, besides out original assumption, has been pure logic and could not possibly be incorrect. It follows that our mistake must have been our assumption that there exist $a$ and $b$ for which $\sqrt{p} = a/b$, i.e. that $\sqrt{p}$ is rational.

Since a number is either rational or irrational, it follows that $\sqrt{p}$ must be irrational.

Note: To complete your understanding, try following the same argument through with $n = 4$. You will find that you quickly become stuck.

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