Proof of $\mathbb{Q}$ is not cyclic

Question

So i'm trying to understand the proof of:

$\mathbb{Q}$ is not cyclic. So this is the proof:

We proceed by contradiction.

Suppose $\mathbb Q$ is cyclic then it would be generated by a rational number in the form $\frac{a}{b}$ where $a,b \in \mathbb{Z} $ and $a$, $b$ have no common factors. Also, $a,b \neq 0$.

The set $\langle\frac{a}{b}\rangle$ consists of all integer multiples of $\frac{a}{b}$.

Therefore, if $\mathbb{Q}=\langle\frac{a}{b}\rangle$, then $\frac{a}{2b}$ is an integer multiple of $\frac {a}{b}$

PROBLEM: Why is $\frac{a}{2b}$ an integer multiple? or how is it an integer multiple. I'm not seeing it because isn't $\frac{a}{b} \times \frac{a}{b}=\left(\frac{a}{b}\right)^2$

Anyways, here is the rest of the proof:

but if

$c \times \frac{a}{b}=\frac{a}{2b}$ then $c=\frac{1}{2}$ is not an integer.

Thus, $\mathbb{Q}$ cannot be generated by a single rational number and is not cyclic.

If anyone can clarify that would be great. Also, another problem I have is doesn't this show that $\mathbb{Q}-\{0\}$ is not cyclic because I thought $\mathbb{Q}$ under the operation multplication is not a group unless zero is removed.

Answer 1

This is a proof by contradiction. You want to show $\mathbb Q$ is not generated by $a/b$, i.e., it is absurd that every rational number is an integral multiple of $a/b$. You show its absurdity by observing under this assumption $a/2b$, being a rational number, should be an integral multiple of $a/b$, which it clearly isn't. Hence the assumption that $\mathbb Q$ is generated by $a/b$ cannot be true. Since $a/b$ is arbitrary, this shows $\mathbb Q$ is not generated by any single element, i.e., $\mathbb Q$ is not cyclic.

Answer 2

The result is referring to the additive group $(\mathbb{Q}, +)$, not the multiplicative group, $(\mathbb{Q}-\{0\}, \times )$. You should be able to work out what is going on from there.