# Example

Suppose $a_1 \equiv b_1 \pmod{m}$ and $a_2 \equiv b_2 \pmod{m}$. Prove the following:

1. $a_1 + a_2 \equiv b_1 + b_2 \pmod{m}$
2. $a_1 - a_2 \equiv b_1 - b_2 \pmod{m}$
3. $a_1 a_2 \equiv b_1 b_2 \pmod{m}$

First note, as a beginning to each of the following arguments, $a_1 \equiv b_1 \pmod{m}$ implies $b_1 = a_1 + mk_1$ for some integer $k_1$, while $a_2 \equiv b_2 \pmod{m}$ implies $b_2 = a_2 + mk_2$ for some integer $k_2$.

Now consider the following:

1. \displaystyle{\begin{align} b_1 + b_2 &= (a_1 + mk_1) + (a_2 + mk_2)\\ &= (a_1 + a_2) + m(k_1 + k_2)\\ \end{align}}

which implies $a_1 + a_2 \equiv b_1 + b_2 \pmod{m}$

2. \displaystyle{\begin{align} b_1 - b_2 &= (a_1 + mk_1) - (a_2 + mk_2)\\ &= (a_1 - a_2) + m(k_1 - k_2)\\ \end{align}}

which implies $a_1 - a_2 \equiv b_1 - b_2 \pmod{m}$

3. \displaystyle{\begin{align} b_1 b_2 &= (a_1 + mk_1) (a_2 + mk_2)\\ &= a_1 a_2 + a_1 m k_2 + a_2 m k_1 + m^2 k_1 k_2\\ &= a_1 a_2 + m(a_1 k_2 + a_2 k_1 + mk_1 k_2) \end{align}}

which implies $a_1 a_2 \equiv b_1 b_2 \pmod{m}$