<s>[INST] Proof irrationality $n\sqrt{11}$ 
Prove that $\sqrt{11}$ is irrational, subsequently prove that $n\sqrt{11}$ is also irrational for every $n \in \mathbb{N}$. You are allowed to use that if $p$ is prime, and $p | a^2$, then $p|a$.  

Can't you also prove that $n\sqrt{11}$ is irrational simply by saying:
Assume $n\sqrt{11}$ is rational, then $n\sqrt{11} = \dfrac{p}{q}$, so $\sqrt{11} = \dfrac{p}{qn}$, which would mean that $\sqrt{11}$ is rational, which is false (assume we already proved that it's false). 
So we know $n\sqrt{11}$ is irrational by contradiction.
 [/INST] I guess that your post also asked for a proof that $\sqrt{11}$ is irrational? Here it is. 
Assume, for contradiction, that $\sqrt{11}$ is rational. Let $a$ and $b$ be relatively prime, positive integers. Then
\begin{align*}
\sqrt{11} &= \frac{a}{b} \\
11 &= \frac{a^2}{b^2}\\
11 b^2 &= a^2 .
\end{align*}
This means that $11 \vert a^2$ and so $11 \vert a$ as 11 is prime. So $a=11c$ and
\begin{align*}
11 b^2 &= (11c)^2 \\
b^2 &= 11c^2. 
\end{align*}
This means that $11 \vert b^2$ and so $11 \vert b$ as 11 is prime. But that both $11 \vert a$  and $11 \vert b$ contradicts $a$ and $b$ being relatively prime, so $\sqrt{11}$ is not rational, which means that it is irrational. Q.E.D.
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