Show that, given an element $\bar{x}\in\mathbb{Q}/\mathbb{Z}$, there is an integer $n \ge 1$ such that $n\bar{x} = 0$.

Question

Consider $\mathbb{Z}$ as a subgroup of the additive group $\mathbb{Q}$ of rational numbers. >Show that, given an element $\bar{x}\in\mathbb{Q}/\mathbb{Z}$, there is an integer $n \ge 1$ such that $n\bar{x} = 0$.

I know that $\mathbb{Q}/\mathbb{Z} = \left \{ \frac{a}{b} + \mathbb{Z} \mid a,b \in Z, b \neq 0 \right \}$. From that, $ \bar{x} = \frac{a}{b} + c$ where $a,b, c \in Z$ and $ b \neq 0 $, so $n\left ( \frac{a}{b} + c \right ) = 0$. But when I try to solve $n\left ( \frac{a}{b} + c \right ) = 0$ the result is always $n=0$. Any help in solving this is welcome.

Answer 1

Your definition of $\overline x$ isn't quite right - the coset is actually

$$\frac a b + \mathbb{Z}$$

for some integers $a, b$. Now a coset $\alpha + \mathbb{Z}$ is zero in the quotient if and only if $\alpha \in \mathbb{Z}$, so what you actually need is an $n$ so that

$$n \cdot \frac a b \in \mathbb{Z}$$

Can you think of such an $n$?