Answer
True. We prove the result in three cases.
Let $n$ be an integer such that $n\ mod\ 3=0$. Then by the definition of the modulus function, $n=3k$ for some integer $k$. Hence, we have $\lfloor n/3\rfloor=\lfloor (3k)/3\rfloor=\lfloor k\rfloor$$=k=n/3$, since the floor of any integer is that integer (cf. exercise 11).
Now let $n$ be an integer such that $n\ mod\ 3=1$. Then by the definition of the modulus function, $n=3k+1$ for some integer $k$. Substituting $3k+1$ for $n$, we get $\lfloor n/3\rfloor=\lfloor (3k+1)/3\rfloor$$=\lfloor k+\frac{1}{3}\rfloor=k+\lfloor \frac{1}{3}\rfloor=k$ by Theorem 4.5.1 ($\lfloor m+x\rfloor=m+\lfloor x\rfloor$ for all integers $m$). But $k=(n-1)/3$, so we have the desired result.
Finally, let $n$ be an integer such that $n\ mod\ 3=2$. Then by the definition of the modulus function, $n=3k+2$ for some integer $k$. Substituting $3k+2$ for $n$, we get $\lfloor n/3\rfloor=\lfloor (3k+1)/3\rfloor$$=\lfloor k+\frac{2}{3}\rfloor=k+\lfloor \frac{2}{3}\rfloor=k$ by Theorem 4.5.1. But $k=(n-2)/3$, so we have the desired result.
Since these three are the only possible cases, we conclude that the proposition holds for all integers.
Work Step by Step
Although the directions say not to appeal to Theorem 4.5.1, there is no obvious way to prove this result without the theorem, and even the hint at the back of the text uses the theorem without naming it. If your instructor insists on avoiding Theorem 4.5.1, consider justifying the same step as a corollary to the proposition expressed in exercise 11.
