Hw4 Problem 5

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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

(1)
\begin{align} (3^a6^b \times 3^c6^d) \times 3^e6^f =3^{a+c}6^{b+d}\times 3^e6^f =3^{a+c+e}6^{b+d+f}\\ 3^a6^b \times (3^c6^d \times 3^e6^f) =3^a6^b\times 3^{c+e}6^{d+f} =3^{a+c+e}6^{b+d+f} \end{align}

So, it is associative.

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

(2)
\begin{align} 3^a6^b\times 3^06^0 = 3^a6^b \end{align}
(3)
\begin{align} 3^06^0\times 3^a6^b = 3^a6^b \end{align}

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)
\begin{align} 3^a6^b\times 3^{-a}6^{-b} = 3^{(a-a)}6^{(b-b)} = 3^06^0=1 \end{align}
(5)
\begin{align} 3^{-a}6^{-b}\times 3^{a}6^{b} = 3^{(-a+a)}6^{(-b+b)} = 3^06^0=1 \end{align}

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.

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