1

I'm just trying to understand why the positive rationals do not form a cyclic group under multiplication. Please let me know if my reasoning is correct, and where I made mistakes.

Proof by Contradiction:

If the positive rational numbers form a cyclic group, then that means that $\mathbb{Q}^+$ = $\left\langle\frac{a}{b}\right\rangle$. This implies every single positive rational number can be written in the form of $(\frac{a}{b})^n$. Assume that this is true, and take the rational number $\frac{a}{2b}$. Since $\mathbb{Q}^+$ = $\left\langle\frac{a}{b}\right\rangle$, this means that $\left(\frac{a}{b}\right)^n$ = $\frac{a}{2b}$. However, no such $n$ exists such that this is true. Therefore, there is no generator for $\mathbb{Q}^+$, and is not a cyclic group.

Is my reasoning correct? I'd really like to understand this airtight since the concept will be used in an exam, so I'd like any form of critique or correcting. Thanks for helping me!

Max
  • 1,706

2 Answers2

2

An idea for you to develop, and beginning as you did:

Suppose $\;\Bbb Q=\left\langle\,\frac ab\,\right\rangle\;$ , and suppose $\;p_1,...,p_k\;$ are all the different primes appearing with positive exponent in the prime decomposition of $\;b\;$ .

Check now that for any $\;n\in\Bbb Z\;,\;\;\left(\frac ab\right)^n\;$ is a number such that in its denominator only appear primes in $\;\{p_1,...,p_k\}\;$ , and from here that if $\;q\;$ is some prime different fro $\;p_1,...,p_k\;$ , then $\;\frac1q\notin\left\langle\,\frac ab\,\right\rangle\;$ .

DonAntonio
  • 214,715
  • 1
    +1, this is proper way to answer this since with slight modification one could show that the group is not even finitely generated. – Ennar Sep 25 '16 at 09:13
0

The overall flow of proof is fine. One way is to consider

$(\frac{a}{b})^n=\frac{c}{d}$

where

$gcd(a,c)=1 , gcd(b,d)=1, c,d\neq1$

Then

$a^nd=b^nc$

$c$ divides $a^nd$. As $c$ and $a$ are co-primes, $c$ should divide $d$, which is a contradiction, as we always assume $gcd(c,d)=1$.

Med
  • 2,550