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.

  • 2
    $\begingroup$ 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? $\endgroup$ – Did Feb 7 at 12:27
  • $\begingroup$ $G$ is an additive subgroup of the real numbers. $\endgroup$ – ncmathsadist Feb 7 at 12:50
  • $\begingroup$ 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. $\endgroup$ – Arthur Feb 7 at 13:11
  • $\begingroup$ @Jac Frall About which operation we say? $\endgroup$ – Michael Rozenberg Feb 7 at 13:26
  • $\begingroup$ @MichaelRozenberg under addition $\endgroup$ – 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 $\mathbb Z$.

  • $\begingroup$ Exactly what I was looking for. $\endgroup$ – 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 .


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.


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