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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.

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    $\begingroup$ I think you could flesh that out a little more - specifically, what property of $\sqrt{2}$ makes it such that no such $n$ exist? (Moreover, you're assuming a fair bit about cosets earlier in your proof; depending on where you're at it might be good to explain explicitly why $(\mathbb{Z}+\sqrt{2})n=\mathbb{Z}$ implies $\sqrt{2}\cdot n\in\mathbb{Z}$ or even to work with explicit elements of the coset, though that may well not be necessary.) $\endgroup$ Commented May 12, 2014 at 17:49
  • $\begingroup$ $\sqrt 2 \in \mathbb R$ and $\sqrt 2 \not \in \mathbb Z$? $\endgroup$
    – Stoof
    Commented May 12, 2014 at 17:52
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    $\begingroup$ Stefan, that's not quite enough - for instance, what's the order of $\mathbb{Z}+\frac57$? $\endgroup$ Commented May 12, 2014 at 17:53
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    $\begingroup$ 7. So would $\sqrt 2 \in \mathbb R$ and $\sqrt 2 \not \in \mathbb Q$? $\endgroup$
    – Stoof
    Commented May 12, 2014 at 17:55
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    $\begingroup$ That should do it! Write it out explicitly, but that's the core missing piece at least to my mind. $\endgroup$ Commented May 12, 2014 at 17:57

3 Answers 3

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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

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You might find the proof clearer in congruence language

$\qquad\quad\begin{eqnarray} &&\ \ x &{\rm has}\!\!\!& {\rm\ \ \ order}\ \, n\,\ {\rm in}\,\ \Bbb R\!\!\!\!\pmod{m\Bbb Z}\, =\, \Bbb R/m\Bbb Z\\ &\Rightarrow\ &nx&\equiv\,& 0\!\!\pmod{\!m\Bbb Z}\\ &\Rightarrow\ &nx &=& 0 + km\,\ \ {\rm for\ some}\ \ k\in\Bbb Z\\ &\Rightarrow\ &\ \ x &=& k(m/n)\ \ {\rm for\ some}\ \ k\in\Bbb Z\\ &\Rightarrow\ &\ \ x &\in\Bbb Q\!\!\!\!\!\\ \end{eqnarray}$

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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

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