Exercise 3.1.2 (Quotient as integral part)

Question

With reference to the notation of Theorem 1.2 prove that $q = \lfloor b/a \rfloor$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} The pair quotient-remainder  of the integers $b,a$ is given by
$$b = aq +r,\qquad 0 \leq r < a.$$
Then $\frac{b}{a} = q + \frac{r}{a}$, where $ 0 \leq  \frac{r}{a}< 1$, thus
$$q \leq \frac{b}{a} < q + 1.$$
Therefore
$$q = \left \lfloor \frac{b}{a} ight floor.$$
\end{proof}

Exercise 3.1.2 (Quotient as integral part)

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Exercise 3.1.2 (Quotient as integral part)

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