Properties of $(+):$ $f:\mathbb{Z} \rightarrow \mathbb{Z}_p$

Question

Let $p$ be natural prime number. For $m \in \mathbb{Z}$, let $\overline{m}$ denote the remainder of $m/p$. For $\mathbb{Z}_p$ = $\{x\in \mathbb{Z}; x \geq0$ and $x

$(+):$ $m(+)n=\overline{m+n}$

Let $f:\mathbb{Z} \rightarrow \mathbb{Z}_P$ such that $f(n)=\overline{n}$.

Prove that $f(m+n)=f(m)(+)f(n)$.

My attempt:

Given $m,n \in \mathbb{Z}$, write $m=ap+\overline{m}$ and $n=bp+\overline{n}$. Therefore, it follows that $m+n=p(a+b)+\overline{m}+\overline{n}$. In particular, $\overline{m}+\overline{n} = m+n-p(a+b)$.

So, $f(m+n)=\overline{m+n} = \overline{p(a+b)+\overline{m}+\overline{n}} = \overline{m}+\overline{n}= m+n-p(a+b)$.

And $f(m)(+)f(n)=\overline{m}(+)\overline{n}=\overline {\overline{m}+\overline{n}} = \overline{m+n-p(a+b)}=m+n$.

I can't see how this can be equal. For me it is only true that $f(m+n)=f(m)+f(n)$.

Answer 1

It's not true that:

$$\overline{\overline{m}+\overline{n}} = \overline{m}+\overline{n}$$

Take, for example, $p=3, m=5, n=2$. Then $\overline{\overline{5}+\overline{2}} = \overline{2+2} = 1$, not $2+2 = 4$ ($4$ is not even a member of $\mathbb{Z}_p$).

Instead, you can prove roughly along the lines you took, that: $$\overline{m+n} = \overline{\overline{m}+\overline{n}} = \overline{m}(+)\overline{n}$$

It may be helpful to note that for all $m \in \mathbb{Z}$, $\overline{m} \in \mathbb{Z}_p$. And also, if $m \in \mathbb{Z}_p$, then $\overline{m} = m$. So $f(m) = f(\overline{m}) = \overline{f(m)}.$

Also, note that $(+)$ is defined for elements of $\mathbb{Z}_p$, not for integers in general.