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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.

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    $\begingroup$ Is the statement "$\frac a {2b}$ is an integer multiple of $\frac a b$" right? $\endgroup$
    – Newb
    Mar 11, 2015 at 3:34
  • $\begingroup$ Certainly, $\frac{a}{2b}$ is not an integer multiple of $\frac{a}{b}$, which is the point of mentioning it in the first place. $\endgroup$ Mar 11, 2015 at 3:46
  • $\begingroup$ Yes, it is right. and @Travis, I got the idea. thanks! $\endgroup$
    – Justin
    Mar 11, 2015 at 23:16

2 Answers 2

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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.

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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.

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    $\begingroup$ Perhaps this is better suited to be a comment than an answer. $\endgroup$ Mar 11, 2015 at 3:48
  • $\begingroup$ the word multiple threw me off for some reason. $\endgroup$
    – Justin
    Mar 11, 2015 at 23:17
  • $\begingroup$ He want votes i guess but still an important thing which may clear op doubts so +1 $\endgroup$
    – Cloud JR K
    Sep 13, 2018 at 17:38
  • $\begingroup$ doesnot same proof works in case of multiplicative Q group ? $\endgroup$ Apr 9, 2023 at 16:35
  • $\begingroup$ you have to modify the proof a little, ie its not integer multiples, its powers. But for the multiplicative group its easier to just ask under what conditions there is some power of a number equal to $-1$. $\endgroup$
    – user24142
    Apr 11, 2023 at 5:50

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