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I came across this question, and when it comes to proofs, I'm not very good.

Prove by contradiction that $\sqrt{7}$ is irrational. You may assume that for any integer $x$ and any prime number $p$ we have ($p$ divides $x$)$ \Leftrightarrow $($p$ divides $x^2$).

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    Assume it is rational and takes the rational form of $\frac{a}{b}$ where $a$ and $b$ are integers... Then you should come to a contradiction if you find the right play. – 2017-01-19
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    The proof will mirror that of the [classical greek proof of the irrationality of $\sqrt{2}$ via infinite descent](https://en.wikipedia.org/wiki/Square_root_of_2#Proofs_of_irrationality) by Aristotle and Euclid. – 2017-01-19
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    See here (the image in the question is your answer) : http://math.stackexchange.com/questions/83745/prove-the-square-root-of-7-is-irrational-using-the-division-algorithm-and-case-r – 2017-01-19
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    According to rational root theorem the only rational solutions to $x^2-7=0$ can be $\pm 7$ which is clearly nonsense. – 2017-01-19
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    That's a very odd assumption to "allow" since it's trivially obvious, implied by the definition of "divides." – 2017-01-19
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    @Wildcard did you not notice the assumption is worded as an *if and only if*, the reverse direction is not nearly as immediate (*though admittedly still trivial using the definition of prime*). – 2017-01-19
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    @JMoravitz, ah, thanks for pointing that out. Indeed, the reverse direction is not so trivial. – 2017-01-19

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Assume that we can write $\sqrt{7}$ as a ratio of two numbers, $p$ and $q$. Assuming that this is in simplest form, we have $p/q=\sqrt{7}$. This can be squared to give: $$\begin{align*}\frac{p}q &= \sqrt{7}\\ \left(\frac{p}q\right)^2&=\frac{p^2}{q^2}=7\\p^2&=7q^2 \end{align*}$$ However, this would imply that one of the factors of $p$ would square to get $7$, which is clearly nonsense as only integer values are allowed in the prime factorization. Thus, this is a contradiction, and the proof is false. Thus, $\sqrt{7}$ is irrational.

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    I don't understand why necessarily "this would imply that one of the factors of $p$ would square to get $7$". – 2017-01-19
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    Well, if $p^2=7q^2$, then obviously the prime factorization of $p^2$ is the same as that of $7q^2$. All of the factors of $p$ get squared, correct? Thus, we have that one of the factors squared yields $7$, and the others squared yield $q^2$. Now, since $7$ is not a square number, that factor cannot exist and thus, $\sqrt{7}$ is irrational. – 2017-01-19
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    Honestly I would prefer the same argument used in the case of the irrationality of $\sqrt{2}$. If $p^2=7q^2$, then $7\mid p$, so $p=7r$ and hence $7r^2=q^2$, but this means that $7\mid q$, contradiction. – 2017-01-19
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    If you prefer it, I can edit my post to match that. – 2017-01-19
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    I think you should do it. IMO it's a clearer argument. – 2017-01-19