Dichotomy of function mapping and its inverse.

Let there be a set of values, denoted by $S = \{2^a3^b : a, b \in \mathbb{Z}^+\}$.

I have two functions that are inverse of each other:

1. function $f$ with $\mathbb{Q}^+ \to S$ to be $f(\frac{a}{b}) = 2^a3^b$, where $\frac{a}{b} \in \mathbb{Q}^+$ iff $(a,b)=1$.

2. function $g: S \to \mathbb{Q}^+$ defined by $g(2^a3^b) = \frac{a}{b}$.

There is dichotomy as in $g$ there are multiple numbers (unique) that map to a single range element (i.e., with values $ka, kb, \forall k \in \mathbb{Z^+}$), but in map $f$ there are “not” multiple range elements — there is only one mapping in the range for a given domain element. So, $g$ is not an injective function, while $f$ is a bijective mapping.

Either I am wrong, or there is some reason behind this behavior of the function and its inverse.

• Can you please provide an example of different elements of $\mathbb{Q}^+$ which are mapped (by $g$) to the same element? – José Carlos Santos Jan 12 '18 at 9:21
• @JoséCarlosSantos One (out of infinitely many) example is $3$ (again out of infintely many) numbers $2^33^2, 2^63^4, 2^93^6$ all mapping to $\frac{3}{2}$. Also, I hope you meant "different elements of $S$ which are mapped by $g$ to the same element of $\mathbb{Q}^+$". – jiten Jan 12 '18 at 9:30
• Yes, I meant $S$. Sorry for the question; it was silly. However, if we define $S$ as$$\{2^a3^b\,|\,a,b\in\mathbb{Z}^+\wedge\gcd(a,b)=1\},$$then all works fine, right?! – José Carlos Santos Jan 12 '18 at 9:38
• If that's what you wanted, then I will post it as an answer. – José Carlos Santos Jan 12 '18 at 9:41

If we define $S$ as$$\left\{2^a3^b\,\middle|\,a,b\in\mathbb{Z}^+\wedge\gcd(a,b)=1\right\},$$then $f$ and $g$ will be bijections and $g=f^{-1}$.
• Sir, does it make it mandatory to interpret the given condition (here, $(a,b)=1$) so as to have the inverse ($g$) also to be a bijection, iff a function (here, $f$) is a bijection. – jiten Jan 12 '18 at 9:56