How to show that $G=\{a+b\sqrt 2\mid a,b\in \mathbb{Z} \}$ is not a cyclic group? 
Show that $G=\{a+b\sqrt2\mid a,b\in\mathbb{Z}\}$ is not cyclic under addition.

Trying to show that $G$ is not a group, and without showing that $G$ is isomorphic to $\Bbb{Z}\times\Bbb{Z}$, it seems somewhat strange to show that $G$ cannot be generated by an element of $G$
Edit: I realized I left off some pertinent information. I am supposed to show it is not cyclic. 
 A: Hint : The group G is generated by the set $\{1, \sqrt{2} \}. $ Cyclic group is generated by single element of the group .
Or 
Let if possible group is cyclic then $G=c\mathbb{Z}$ which implies that $c$ is irrational number and $G=c\mathbb{Z}$, a contradiction.
A: Suppose it were cyclic. Then, there exists a $c \in G$ such that every $d \in G$ is of the form $cn$ for some $n \in \mathbb Z$.
Now, note that $1 = cm$ for some $0 \neq m \in \mathbb Z$ and $\sqrt 2 = cn$ for some $0 \neq n \in \mathbb Z$. Therefore, $c$ is rational, as $c = \frac 1m$, but then $\sqrt{2} = \frac nm$ is rational, a contradiction.

Alternately, you may also generalize this idea to the following result :

Let $\{1,m_1,...,m_n\} \subset \mathbb R$ be linearly independent over $\mathbb Q$ (that is, if $a_0,a_1,...,a_n \in \mathbb Q$ are such that $a_0+\sum_{i=1}^n  a_im_i = 0$ then $a_i = 0$ for all $i$), with $n \geq 1$. Then, the group $$G' = \mathbb Z[m_1,...m_n] := \left\{\sum a_im_i : a_i \in \mathbb Z\right\}$$ cannot be cyclic.

It is a more difficult exercise to show that  the above cannot be generated even by any $n-1$ elements over $\mathbb Z$.
