Return to Homework 4, Homework Problems, Glossary, Theorems
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.