Trying to show that the set of all $2$-element subsets of a denumerable set is denumerable Suppose $A$ is denumerable and put $X = \{ B : B \subset A, \; \; |B| = 2 \} $. I want to show that $X$ is denumerable as well.
My try: Let $f$ be bijection from $\mathbb{N}$ to $A$. 
We know any $B \in X$ is of the form $B = \{a,b \} $ for unique $a,b \in A $. We know there exist elements $n,m \in \mathbb{N}$ such that $a = f(n) $ and $b = f(m) $. 
We define $F: X \to \mathbb{N} $ by $F( \{ f(n), f(m) \} ) = 2^{f(n)}3^{f(m)} $. 
To show this is injective it is enough to show that if $2^k3^r = 1$, then $k=r=1$ 
But I am stuck here. I mean it is obvious but how can we prove this rigorously ?

Instead, simply define
$$F(\{f(n),f(m)\})=2^n+2^m.$$
Suppose $$F(\{f(n),f(m)\})=F(\{f(k),f(r)\}),$$
i.e., $$2^n+2^m=2^k+2^r.$$
We may assume that $n\lt m,\ $ $k\lt r,\ $ and $m\le r.$
Assuming that $m\lt r,$ we have
$$2^n+2^m\le2^{r-2}+2^{r-1}\lt2^r\lt2^k+2^r,$$
contradicting the assumption that $2^n+2^m=2^k+2^r.$ Hence $m=r$ and $n=k.$
