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.
Best Answer
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