I have a question about how I have to do these exercises for my math study
Let $n \in \mathbb{Z}, n>0$ and $a \in Z$
a) prove: if $n$ is odd, then $\overline{a} = \overline{(-a)}$ if and only if $\bar{a}$ = $\bar{0}$
b) prove: if $n$ is even, then $\bar{a} = \overline{(-a)}$ if and only if $\bar{a}$ = $\bar{0}$ or $\bar{a}$ = $\overline{\frac{n}{2}}$
For (a), I think I solved a part of the left to right arrow:
$\bar{a}$ =$\overline{(-a)}$ , so $n | (a - (-a))$ $\Rightarrow n | 2a$
case 1: $n | a$
Also $n | 0$ , hence $a$ and $0$ do both have a remainder of $0$ by division with $n$, so $\bar{a} = \bar{0}$
case 2: $n$ does not divide $a$, $n | 2a$, so...(this is where I am stuck)
How to I have to prove the rest of these exercises? I'm trying to do it for 2 days now, but it doesn't work.
Thanks in advance!!!
For (a): $\bar{a} = \overline {(-a)} \Rightarrow n | 2a \Rightarrow n|a$ (why?)
For(b): First note that we can choose $a < n,$ because using division algorithm we can write $a = nb + r$ for some $b, r \in \mathbb{Z}$ and $0 \leq r \leq n-1.$ Then $\bar{a} = \bar{r}.$
So let $a < n$ and also $\bar{a} \neq \bar{0}.$ Now $\bar{a} = \overline {(-a)} \Rightarrow n | 2a \Rightarrow \frac{n}{2}|a \Rightarrow a \leq \frac{n}{2} \Rightarrow 2a = n$
I think you can finish it from here.