I need to prove that the function $f:\mathbb{N}\times \mathbb{N}\rightarrow \mathbb{N}$ given by $f(x,y)=2^x\cdot 3^y$ is injective.
Is there an easier method than to use logarithms?
The function certainly looks injective, but I don't know how to show it.
How would I start in this case?
Best Answer
Well assume you have $(x_{1},y_{1})$ and $(x_{2},y_{2})$ with $f(x_{1},y_{1}) = f(x_{2},y_{2})$. W.l.o.g. assume $x_{1}\ge x_{2}$, then:
$$2^{x_{1}}3^{y_{1}} = 2^{x_{2}}3^{y_{2}} \iff 2^{x_{1}-x_{2}}3^{y_{1}} = 3^{y_{2}}.$$
If $x_{1}\neq x_{2}$ then $2$ divides $3^{y_{2}}$ which is absurd. Hence $x_{1}=x_{2}$ and therefore $y_{1}=y_{2}$.