Group Action: Group $(\mathbb{Z}, +)$ acting on $\mathbb{R}$

Question

In my group theory notes I have the following:

The Group $(\mathbb{Z}, +)$ acts on $\mathbb{R}$ as follows: $m\in \mathbb{Z}$ and $r\in\mathbb{R}$: $m.x \to (-1)^mr$

in this notation $m.x$ defines the group action.

I am unsure how to obtain the RHS: $(-1)^mr$...

Answer 1

Let me clarify the issue a bit more, since it wasn't very explicit in the comments.

The quote says:

The group ... acts on ... as follows.

This is misleading (or written by a non-native English speaker) since it suggests that there is one known action which is now going to be described. Hence it wasn't surprising that you were asking how the action was derived.

The quote should have said: $\def\zz{\mathbb{Z}}$ $\def\rr{\mathbb{R}}$

The group $(\zz,+)$ acts on $\rr$ via $\bullet$ where $m \bullet r = (-1)^m r$ for any $m \in \zz$ and $r \in \rr$.

This conveys that $\bullet$ is just one action. There are other possible ways that $(\zz,+)$ can act on $\rr$, such as:

The group $(\zz,+)$ acts on $\rr$ via $\diamond$ where $m \diamond r = m r$ for any $m \in \zz$ and $r \in \rr$.

Now in both cases we would have a statement that is either true or false. If we wish to prove it, then we could expand the definition of actions and verify the conditions one by one. Alternatively we could prove some general theorem about actions. For example the following is always true:

The group $(\zz,+)$ acts on any ring $R$ via $\cdot$ where $0 \cdot r = 0_R$ and $(z+1) \cdot r = z \cdot r + r$ for any $z \in \zz$ and $r \in R$. (Here $\cdot$ is inductively defined.)