# Irrationality of $\sqrt{15}$

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.

• That's a great proof. – Yanko Oct 25 '17 at 16:18
• looks correct to me well done – Isham Oct 25 '17 at 16:34
• Agreed and well explained – imranfat Oct 25 '17 at 16:53
• Looks good. One comment though, when writing "Contradiction." it's good to mention what the contradiction is, in this case "Contradiction with $\gcd(p,q)=1$" – rtybase Oct 25 '17 at 18:12

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.