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Could someone verify the correctness of this proof for the irrationality of $\sqrt{15}$?

Assume $\sqrt{15}\in\mathbf{Q}$, then $\sqrt{15}=\frac{p}{q}$ with $p,q\in\mathbf{Z}$ ($q\ne0$ and $\gcd(p,q)=1$). $\implies 15q^2=p^2 \implies 15\mid p^2 \implies 3\mid p^2 \implies 3\mid p$ (Euclid's Lemma)

Now we write $p=3k$ for $k\in\mathbf{Z}$, then we have $15q^2=9k^2 \implies 5q^2=3k^2 \implies 3\mid 5q^2$. Since $\gcd(3,5)=1$ (this lemma: if $a\mid bc$ and $\gcd(a,b)=1$ then $a\mid c$) this gives $3\mid q^2 \implies 3\mid q$ (Euclid's Lemma).

Therefore $3\mid p$ and $3\mid q$. Contradiction.

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    $\begingroup$ That's a great proof. $\endgroup$ – Yanko Oct 25 '17 at 16:18
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    $\begingroup$ looks correct to me well done $\endgroup$ – Isham Oct 25 '17 at 16:34
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    $\begingroup$ Agreed and well explained $\endgroup$ – imranfat Oct 25 '17 at 16:53
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    $\begingroup$ Looks good. One comment though, when writing "Contradiction." it's good to mention what the contradiction is, in this case "Contradiction with $\gcd(p,q)=1$" $\endgroup$ – rtybase Oct 25 '17 at 18:12
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I like your proof. As a further observation, you could show that your proof generalizes: you could have replaced $15$ with any product $p_0 r$ where $p_0$ is prime and $r$ is relatively prime to $p_0$. So you have really proven the following:

Proposition. Let $n$ be an integer and let $p_0$ be a prime number such that $p_0$ divides into $n$ exactly once (that is, $p_0$ divides $n$ but $p_0^2$ does not divide $n$). Then $\sqrt{n}$ is irrational.

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