Proving $\{9^n|n\in \mathbb Q\} = \{3^n|n\in \mathbb Q\}$ The statement is: $\{9^n|n\in \mathbb Q\} = \{3^n|n\in \mathbb Q\}$. 
I am more familiar with traditional proofs. Do I just split up $9^n$ into $3^{n}\times 3^{n}$?

Let $B$=$\{9^n|n\in Q\}$ and $A= \{3^n|n\in Q\}$.
Let $x \in A$ $\implies$ $x$=$3^n$ $\implies$ $x$=$ 9^ \frac{n} {2}$ and n $\in$ $Q$ $\implies$ $\frac {n}{2}$ $\in Q$ $i.e. x \in B$ So $A\subset B$. 
Again if $y \in B$ $\implies$ $y$=$9^m$ $\implies$ $y$=$ 3^{2m}$ and $m$ $\in$ $Q$ $\implies$ $2m $ $\in Q$ $i.e. y\in A$ so $B\subset A$. Hence we can say $A=B$.
