Divisibility question: if $a=be+r$, then $e$ $= ⌊\frac{a}{b}⌋$

Question

If $a,b \in \Bbb Z$, then I know that $ a=be+r$, where $e\in \Bbb Z$ and $r$ is the remainder. How can I prove that $e$ is equal to $⌊\frac ab⌋$?

I'm missing this step in another proof and I really don't know how to prove it.

Thanks.

Answer 1

$a = be+r$ and $r<b$ so $$ be\le a = be+r<be+b = (e+1)b;\\ e\le \frac ab< e+1 $$ and as $e\in\Bbb Z$, $$ e = [\frac ab] $$

Answer 2

$\begin{eqnarray}{\bf Hint}\,\ &&0 \le\overbrace{ a-be}^{\textstyle r}\, <\, b\\ \overset{\div\, b} \iff && 0 \le a/b-e < 1\\ \overset{+\,e}\iff && e \le a/b < e+1\\[.3em] \iff && e =\lfloor a/b\rfloor \end{eqnarray}$