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### Problem 5

Prove that the set of all rational numbers of the form $3^m6^n$, where $m,n \in \mathbb{Z}$ is a group under multiplication.

**Solution**

$\mathscr{G}_1:$ (Associativity) Let $a,b,c,d,e,f \in \mathbb{Z}$, then

So, it is associative.

$\mathscr{G}_2:$ (Identity) $0\in \mathbb{Z}$ so,

(2)So, there exists a left and right identity under multiplication.

$\mathscr{G}_3:$ (Inverse) $a,b\in \mathbb{Z}$ and $-a,-b \in \mathbb{Z}$ by the properties of multiplication:

(4)So, there exists a left and right inverse under multiplication.

Since all axioms are followed the set of all rational numbers of the form $3^m6^n$, where $m,n \in \mathbb{Z}$ is a group under multiplication.