Given ($a$ mod $b$) is defined to be $a - b \lfloor \frac{a}{b} \rfloor$. Let $a,b \in \mathbb{Z}$ with $b < 0$. Show that ($a$ mod $b$) $\in (b,0]$.

First, $\lfloor \frac{a}{b} \rfloor$ is a unique integer $m$ such that $m \le \frac{a}{b} \lt m + 1$. Equivalently, $\frac{a}{b} = m + \epsilon$ for some $\epsilon \in [0,1)$.

$$\begin{aligned} a \text{ mod } b &= a - b \lfloor \frac{a}{b} \rfloor \\ &= a - bm \\ &= b(m + \epsilon) - bm \\ &= b\epsilon \end{aligned}$$

Knowing that

$$\begin{aligned} 0 &\le \epsilon < 1 \text{, and given } b < 0 \\ 0 &\ge b\epsilon > b \\ 0 &\ge (a \text{ mod } b) > b \\ \end{aligned}$$

$\Box$

# posted by rot13(Unafba Pune) @ 11:04 AM