Find the number(s) $k$ such that $k\mathbb{Z} = \mathbb{Z}$ I am just starting out with Discrete Math and there is a question in my book that is 

Find the number(s) $k$ such that $k \mathbb{Z} = \mathbb{Z}$.

The answer is -1 and 1. I understand why it is 1 because any number multiplied by 1 is itself, what I don't understand is why the other answer is -1.
For example, if $k = -1$ then  $(-1)\mathbb{Z}$ would be all the values of $\mathbb{Z}$, but if that is the case, then why can't $k$ could also equal 2 or any other integer?
 A: Yanko resolved it already, but I'll flesh it out a bit more as you seemed sure that $(-1)\mathbb{Z} = \mathbb{Z}$ but unsure whether, for example, $2\mathbb{Z} \ne \mathbb{Z}$. This makes me think that $k\mathbb{Z}$ is slightly confusing to you. 
So by $k\mathbb{Z}$ we just mean the set of all integers multiplied by $k$ (or equivalently, the set of all multiples of $k$):
$$k\mathbb{Z} = \{ kn \mid n \in \mathbb{Z} \} = \{\dots,-2k,-k,0,k,2k,\dots\}$$
Thus,
$$1\mathbb{Z} = \{ 1n \mid n \in \mathbb{Z} \} = \{\dots,-2,-1,0,1,2,\dots\} = \mathbb{Z} \tag 1$$
$$(-1)\mathbb{Z} = \{ (-1)n \mid n \in \mathbb{Z} \} = \{\dots,2,1,0,-1,-2,\dots\} = \mathbb{Z} \tag 2$$
$$2\mathbb{Z} = \{ 2n \mid n \in \mathbb{Z} \} = \{\dots,-4,-2,0,2,4,\dots\} \ne \mathbb{Z}$$
Also, remember that with sets, order doesn't matter, which is why $(1)$ and $(2)$ are true.
I hope this makes it a bit clearer. 
A: If $k\mathbb{Z} = \mathbb{Z}$ then $1\in k\mathbb{Z}$ (in fact this is an if and only if) but then $\frac{1}{k}\in\mathbb{Z}$. This means that $k\in\{-1,1\}$.
