# How to show $12^a \cdot 18^b$ is injective

We are told that $$f: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$$

I know one method is to prove that a inverse exists, but I'm not 100% sure how to do that in this case. so instead I decided to create two functions $$f(a,b)$$ and $$f(x,y)$$ and set them equal to each other. $$f(a,b)=f(x,y)$$ $$12^a \cdot 18^b = 12^x \cdot 18^y$$ How exactly would be the next step to get $$a,b = x,y$$?

Also, if you could explain how to take the inverse of the function that would be much appreciated.

• Hint: The order of $2$ on the left is $2a+b$, the order of $3$ is $a+2b$ – lulu Apr 4 at 23:21
• Context clues suggest your function $f$ is a binary operation on the naturals $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$, but it's important to specify the domain and codomain explicitly, since the extension of $f$ to the reals is not injective. – K B Dave Apr 4 at 23:39

## 1 Answer

$$f(a,b)=(2^2\times 3)^a(2\times 3^2)^b=2^{2a+b}\times 3^{a+2b}$$.

$$f(a,b)=f(x,y)$$ is equivalent to $$2a+b=2x+y$$ and $$3a+2b=3x+2y$$

We deduce that $$4a+2b=4x+2y$$

$$3a+2b=3x+2y$$ You substract this two last equations you deduce that $$a=x$$ and $$b=y$$.