Prove that $\sqrt{n^2 + 2}$ is irrational 
Suppose $n$ is a natural number. Prove that $\sqrt{n^2 + 2}$ is irrational.

From looking at the expression, it seems quite obvious to me that $\sqrt{n^2 + 2}$ will be irrational, since $n^2$ will be a natural number, and after adding $2$ to it, $n^2 + 2$ will no longer be a perfect square.
From that logic, I tried letting $n^2 + 2$ = ${a^2\over b^2}$ to find a contradiction, but I wasn't able to formulate a proof that worked. Any suggestions as to how I should approach this?

$$\begin{array}{l}
 \sqrt {2 + n^2 }  = \frac{p}{q};\quad and\quad p \wedge q = 1 \\ 
 \sqrt {2 + n^2 }  = \frac{p}{q} \Leftrightarrow 2 + n^2  = \left( {\frac{p}{q}} \right)^2  \\ 
  \Leftrightarrow 2 = \left( {\frac{p}{q}} \right)^2  - n^2  \\ 
  \Leftrightarrow 2 = \left( {\frac{p}{q} - n} \right)\left( {\frac{p}{q} + n} \right) \\ 
  \Leftrightarrow 2q^2  = \left( {p - nq} \right)\left( {p - nq} \right) \cdots \left( * \right) \\ 
 \left( * \right) \Rightarrow \left\{ \begin{array}{l}
 \exists m_0  \in\mathbb{ N} :p - nq = m_0 q \\ 
 or \\ 
 \exists m_1  \in \mathbb{N} :p + nq = m_1 q \\ 
 \end{array} \right. \\ 
  \Rightarrow \left\{ \begin{array}{l}
 \exists m_0  \in \mathbb{N} :p = \left( {n + m_0 } \right)q \\ 
 or \\ 
 \exists m_1  \in \mathbb{N} :p = \left( { - n + m_1 } \right)q \\ 
 \end{array} \right. \\ 
 \left\{ \begin{array}{l}
 p = \left( {n + m_0 } \right)q \Rightarrow q/p \\ 
 or \\ 
 p = \left( { - n + m_1 } \right)q \Rightarrow q/p \\ 
 \end{array} \right. \\ 
 \end{array}$$
contradiction  with $\quad p \wedge q = 1$
