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> Let $a,b,N\in\mathbb{Z}$, where $N>0$. Prove that $[ab]=[[a][b]]$, where $[x]$ denotes the remainder of $x$ after division by $N$.
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Here's my attempt:

*Proof.* Since $x\equiv{[x]}\_c\pmod c$, we know $a\equiv{[a]} \pmod{N}$ and $b\equiv{[b]} \pmod{N}$. By Proposition 1.3.4 in the book,
$$ab\equiv{[a]}{[b]} \pmod N$$
Applying the first theorem we used to each side of the equation, we have
$${[ab]}={[{[a]}{[b]}]}$$
QED.

I think the last line in my proof is wrong. I tried combining these two statements:

1. $ab\equiv[ab]\pmod{N}$
2. ${[a]}{[b]}\equiv[{[a]}{[b]}]\pmod{N}$

But what makes me hesitate is that the statements have *equivalence*, not equality. I'm new to number theory and do not know how to handle equivalence.

Can you offer any help? I am looking for hints, not complete answers.

Thank you.