Let $f: \mathbb{Z} \to \mathbb{Z}_n$ be the map given by $f(a)$ = the remainder of the division of $a$ by $n$. Show that $f$ is a homomorphism.
I know that $f$ is a homomorphism if $f(x*y)=f(x)·f(y)$ for all $x,y$ in $ \mathbb{Z} $.
My approach is that I set up an equation using $f(a)$, $Q·n+f(a)=a$, where $Q$ is the quotient of the division.
But what should I do next?
By division algorithm, write $x*y:=x+y=q_3 n+r_3, x=q_1 n+r_1 (*), y=q_2 n + r_2(**)$, where $0\leq r_1, r_2, r_3<n$. Then $f(x*y)=r_3*n \mathbb{Z}, f(x)=r_1 *n\mathbb{Z}, f(y)=r_2*n\mathbb{Z}$.
Want to show: $f(x*y)=r_3*n \mathbb{Z}=(r_1*n \mathbb{Z})\cdot (r_2*n \mathbb{Z})=f(x)\cdot f(y)$. i.e, $r_3*n \mathbb{Z}=(r_1 * r_2) * n \mathbb{Z}$. i.e, $r_3*(r_1 * r_2)^{-1}\in n\mathbb{Z}$, i.e, $r_3 -(r_1+r_2)=k n$ where $k\in \mathbb{Z}$.
Now multiply (*) and (**) to conclude: $x*y=(q_1 n+r_1)*(q_2 n+r_2)=(q_1+q_2)n+(r_1+r_2)=q_3n+r_3$, hence $r_3 -(r_1+r_2)=(q_1+q_2 -q_3)n$, as desired.