Is this an identity in modular arithmetic or does there exist a proof for this?

Question

I'd like to call this modular multiplication but all the proofs I can find prove the following :

If $a_1 \equiv b1 \pmod{n}$ and $a_2 \equiv b_2 \pmod{n}$, then $a_1a_2 \equiv b_1b_2 \pmod{n}$

How do you go about proving this?

I want to try to show the following, but I am not sure how to do so: $$ab \equiv (a \bmod{c})(b \bmod{C}) \pmod{C}$$

Answer 1

Go by the definition. If $a_1\equiv b_1 \equiv r_1 \pmod{n}$ then we have $a_1=Q_1n+r_1$ and $b_1=q_1n+r_1$ for integers $Q_1$ and $q_1$. Define similarly for $a_2$ and $b_2$. We have: $$a_1a_2 = (Q_1n+r_1)(Q_2n+r_2) = (Q_1Q_2n+Q_1r_2+Q_2r_1)n+r_1r_2 \implies a_1a_2 \equiv r_1r_2 \pmod{n}$$ by distributive property. Similarly $b_1b_2 \equiv r_1r_2 \pmod{n} \implies a_1a_2 \equiv b_1b_2 \pmod{n}$. Essentially, this is the modular version of distributive property.

Answer 2

It is equivalent to proving that if $a_1-a_2$ and $b_1-b_2$ are divisible by $n$, then $a_1a_2-b_1b_2$ is too.

This is easy: just rewrite the last difference: $$a_1a_2-b_1b_2=(a_1-b_1)a_2+b_1a_2-b_1b_2=(a_1-b_1)a_2+b_1(a_2-b_2),$$ wherefrom it is obvious.