# 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.

• Several misconceptions at work here: 1. G is a group. 2. G has a closure (!). 3. You are asked to show that G is not equal to its closure. 4. The usual way to do this is to show that G is dense. Can you do that? – Did Feb 7 at 12:27
• $G$ is an additive subgroup of the real numbers. – ncmathsadist Feb 7 at 12:50
• The "closure" here is most likely the closure as a subspace of the (topological or metric) space $\Bbb R$, not some sort of algebraic or arithmetic closure. – Arthur Feb 7 at 13:11
• @Jac Frall About which operation we say? – Michael Rozenberg Feb 7 at 13:26
• @MichaelRozenberg under addition – Jac Frall Feb 7 at 14:02

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 $$\mathbb Z$$.

• Exactly what I was looking for. – Jac Frall Feb 8 at 18:50

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.