2

How can I prove that $f,g : \mathbb{Z}/(m)\times\mathbb{Z}/(m)\rightarrow\mathbb{Z}/(m)$ defined as $f([r],[s])=[r+s], g([r],[s])=[r \cdot s]$ are well defined functions?

What I'm reasoning is that if we pick two representatives of the same classes, $[r_1]=[r_2], [s_1]=[s_2]$, meaning that $r_1 \equiv r_2 \mod{m}, s_1 \equiv s_2 \mod{m}$, then $f([r_1],[s_1])=f([r_2],[s_2])$ but I don'w know how to prove this, since what I'm proving is already that last equation. It's like a circular argument.

Any help?

Caleb Stanford
  • 45,895
Allonsy
  • 659

2 Answers2

1

Let $r_1,r_2,s_1,s_2\in\mathbb{Z}_m$ such that $[r_1]=[r_2]$ and $[s_1]=[s_2]$. So there exists integers $i,j$ such that $r_1=r_2+i\cdot m$ and $s_1=s_2+j\cdot m$. Thus: \begin{align*} f([r_1],[s_1])&=[r_1+s_1]=[(r_2+i\cdot m)+(s_2+j\cdot m)]=[(r_2+s_2)+(i+j)\cdot m] \\ &=[r_2+s_2]=f([r_2],[s_2])\\ g([r_1],[s_1])&=[r_1\cdot s_1]=[(r_2+i\cdot m)\cdot(s_2+j\cdot m)]\\ &=[r_2\cdot s_2+r_2\cdot j\cdot m+s_2\cdot i\cdot m]=[r_2\cdot s_2+m(r_2\cdot j+s_2\cdot i)]=[r_2\cdot s_2]\\ &=g([r_2],[s_2]). \end{align*} Thus, $f$ and $g$ are well defined.

Alex Schiff
  • 396
0

It's not circular. We have $r_1 \equiv r_2 \mod m$ and $s_1 \equiv s_2 \mod m$. Then show that $r_1 + s_1 \equiv r_2 + s_2 \mod m$. Conclude $[r_1 + s_1] = [r_2 + s_2]$.

The same thing works for $g$.

Caleb Stanford
  • 45,895