Bijective map from $\Bbb Z$ to $\Bbb Q$

The problem is that, while the cardinality of $\mathbb{Z}$ and $\mathbb{Q}$ is the same, the topology is different. Consider the following: Given two integers $a$ and $b$ with $a < b$, can we say how many integers $c$ satisfy $a < c < b$? Now ask the same question for the rational numbers.

This leads to problems. For example, lets say $a$ and $b$ are two such integers with $k$ other integers between them. And we can even assume the $f(a) = r_{1} < r_{2} = f(b)$. But now we can find an increasing sequence of$k+1$ rational numbers (at least) between $r_{1}$ and $r_{2}$, and only $k$ potential integers to use as their preimages.

For A, some $f(n)$ lies between $f(0)$ and $f(1)$, which requires the integer $n$ to lie between $0$ and $1$.For B, let $M$ be the least positive $n$ for which $f(n)<f(0)$. Then $f(M)<f(m)=(f(M)+f(0))/2<f(0)$ for some integer $m$, so $M>m>0$. But by the definition of $M$, if any integer $m$ satisfies $M>m>0$ then $f(m)\geq f(0)$.