A function $$f \colon \mathbb{N} \rightarrow \mathbb{Z}$$ is defined by $$f(n) = \frac{n}{2}$$ if $$n$$ is even, and $$f(n) = -\frac{n - 1}{2}$$ if $$n$$ is odd. Is $$f$$ a bijection?

PLACEHOLDER ANSWER - the answer is yes with a proof. Even $$n$$ map to the positive integers $$1, 2, 3, \ldots$$ and odd $$n$$ map to $$0, -1, -2, \ldots$$, with no value repeated, so $$f$$ is one-one. Every integer $$m$$ is reached, since $$f(2m) = m$$ for $$m > 0$$ and $$f(1 - 2m) = m$$ for $$m \le 0$$, so $$f$$ is onto. Hence $$f$$ is a bijection.