#### Answer

a. Assume $d|n$. Then by the definition of "divides," $n=kd$ for some integer $k=\frac{n}{d}$. But for every integer $m$, we have $\lfloor m\rfloor=m$, so $\lfloor \frac{n}{d}\rfloor \cdot d=\frac{n}{d} \cdot d=n$. Since $d$ and $n$ were arbitrarily chosen, we conclude that $d|n$ implies $n=\lfloor \frac{n}{d}\rfloor \cdot d$ for all integers $n$ and $d$ such that $d \neq 0$.
b. Assume $n=\lfloor \frac{n}{d}\rfloor \cdot d$. Dividing both sides by $d$, we see that $\frac{n}{d}=\lfloor \frac{n}{d}\rfloor$. But for every real number $m$, we have $m=\lfloor m\rfloor$ if and only if $m$ is an integer. Hence, we must have $\frac{n}{d}=k$ for some integer $k$. Multiplying both sides by $d$, we get $n=kd$, so by the definition of "divides," $d|n$. Since $d$ and $n$ were arbitrarily chosen, we conclude that $n=\lfloor \frac{n}{d}\rfloor \cdot d$ implies $d|n$ for all integers $n$ and $d$ such that $d \neq 0$.
c. An integer $n$ is divisible by a nonzero integer $d$ (i.e., $d|n$) if and only if $n=\lfloor \frac{n}{d}\rfloor \cdot d$.

#### Work Step by Step

The assertion in parts (a) and (b) that $m=\lfloor m\rfloor$ if and only if $m$ is an integer is justified several places in the text and exercises, including, for example, exercise 11 of this section. Recall also for part (c) that "if and only if" is a synonym for "is a necessary and sufficient condition for."