Problem: Define $\mathbb{Q}/\mathbb{Z}$ by the natural projection $\pi \, : \, \mathbb{Q} \longrightarrow \mathbb{Q}/\mathbb{Z}$, where $\pi(x) = \text{frac}(x) = x - \lfloor x \rfloor$ for all $x \in \mathbb{Q}$. Show that $\pi$ is a homomorphism.
Question: Is $\pi$ really a homomorphism by this definition?
$$\pi\left(\frac{5}{3} + \frac{7}{4}\right) = \pi\left(\frac{41}{12}\right) = \frac{41}{12} - \left \lfloor \frac{41}{12}\right \rfloor = \frac{41}{12} - 3 = \frac{5}{12}$$
but
$$\pi\left(\frac{5}{3}\right) + \pi\left(\frac{7}{4}\right) = \left[\frac{5}{3} - \left \lfloor \frac{5}{3}\right \rfloor\right] + \left[\frac{7}{4} - \left \lfloor \frac{7}{4}\right \rfloor\right] = \left[\frac{5}{3} - 1\right] + \left[\frac{7}{4} - 1\right] = \frac{2}{3} + \frac{3}{4} = \frac{17}{12}.$$
This is a homomorphism. Another way of expressing the map is by the rule $x\mapsto x+\Bbb Z$.
$\Bbb Z\triangleleft\Bbb Q$, so the quotient $\Bbb Q/\Bbb Z$ is a group.
$π$ is the canonical projection onto the quotient by a normal subgroup, which is always a homomorphism.