Quotient group $\mathbb{R}/\mathbb{Z}$, prove element has infinite order 
In the quotient group $\mathbb{R}/\mathbb{Z}$ prove that the coset $\mathbb Z + \sqrt 2$ has infinite order.

Here is what I have so far:
Assume that $\mathbb Z + \sqrt 2$ has finite order. Then there exists an $n$ such that $(\mathbb Z + \sqrt 2) n = \mathbb Z$. Thus, $\sqrt 2 n \in \mathbb Z$. But no such $n$ exists, which contradicts the fact that $\mathbb Z + \sqrt 2$ has finite order. Therefore, $\mathbb Z + \sqrt 2$ has infinite order.
Is this a correct way to prove this statement? I feel like I might be missing something in the step $\sqrt 2 n \in \mathbb Z \implies$ no such $n$ exist.
 A: Assume the coset $\mathbb Z + \sqrt 2$ has finite order.
Then $\exists n$ such that $n(\mathbb Z + \sqrt 2) = \mathbb Z$. So, $\mathbb Z + \sqrt 2 n = \mathbb Z \implies \sqrt 2 n \in \mathbb Z$. So $\sqrt 2 = \frac{a}{n}$ for some $a \in \mathbb Z$. But $\sqrt 2 \not \in \mathbb Q$. So no such number exists. This contradicts the fact that $\mathbb Z + \sqrt 2$ has finite order. So $\mathbb Z + \sqrt 2$ has infinite order
A: You might find the proof clearer in congruence language
$\qquad\quad\begin{eqnarray} &&\ \  x &{\rm has}\!\!\!& {\rm\ \ \ order}\ \, n\,\ {\rm in}\,\ \Bbb R\!\!\!\!\pmod{m\Bbb Z}\, =\, \Bbb R/m\Bbb Z\\
&\Rightarrow\ &nx&\equiv\,& 0\!\!\pmod{\!m\Bbb Z}\\
&\Rightarrow\ &nx &=& 0 + km\,\ \ {\rm for\ some}\ \ k\in\Bbb Z\\
&\Rightarrow\ &\ \ x &=& k(m/n)\ \ {\rm for\ some}\ \ k\in\Bbb Z\\
&\Rightarrow\ &\ \ x &\in\Bbb Q\!\!\!\!\!\\
\end{eqnarray}$
A: If $\exists \ n \ \in  \ \mathbb{N}, n\neq 0 $ such that $ n (\mathbb Z + \sqrt 2) = 0$ then $$\mathbb Z + n \sqrt 2 = 0  \Rightarrow n\sqrt{2} = m \in \mathbb{Z}$$ and so $$\sqrt{2} = \frac{m}{n} \in \mathbb{Q}$$ which is absurd
