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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 $nq^2 = p^2$.2012-12-26
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    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.2012-12-26
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    @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.2012-12-26
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    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.2012-12-26

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