Is $h(x)=\frac{f(x)}{g(x)}$ injective? Let $f$ and $g$ be injective functions from $\mathbb{N}$ to $\mathbb{N}$. Define $h$, a function from $\mathbb{N}$ to $\mathbb{Q}$ by $h(x)=\frac{f(x)}{g(x)}$ for all $x \in \mathbb{N}$.
Is $h$ injective?
My attempt:
By the definition of injectivity, $f(x)=f(y) \implies x=y$ and $g(x)=g(y) \implies x=y$
Thus, $\frac{f(x)}{g(x)}=\frac{f(y)}{g(y)} \implies \frac{x}{x}=\frac{y}{y} \implies 1=1$. Therefore, $h(x)$ isn't injective if $f=g$.
 A: No: let $f$ be injective, if you take $g=\alpha f$,$\alpha\ne0,1$. $g$ is of course injective and $g\ne f$, but $$\frac{f}{g}=\frac{f}{\alpha f}=\frac{1}{\alpha}$$ is constant, hence it is not injective.
A: Recall that a function $w: \mathbf{N}\to \mathbf{Q}$ is injective if "$w(x)=w(y) \implies x=y$, for all $x,y \in \mathbf{N}$".
You are assuming that $\frac{f(x)}{g(x)}=h(x)=h(y)=\frac{f(y)}{g(y)}$ implies $f(x)=f(y)$ and $g(x)=g(y)$. This is a sufficient condition, but it is not necessary. Indeed, if the function $f$ is equal to the function $g$ (whatever $f$ is) then $f/g$ is the constant function $1$, which is the opposite of being injective.
If you want a less (?) trivial example, $h(x)=h(y)$ if and only if the reduced fraction of $\frac{f(x)}{g(x)}$ is equal to the reduced fraction of $\frac{f(y)}{g(y)}$. So pick for instance $f(1)=1$, $g(1)=2$, $f(2)=10$, $g(2)=20$, $f(n)=n$ and $g(n)=n+20$ for all $n\ge 3$. Then both $f,g$ are injective. But $h(1)=1/2=h(2)$, hence $h$ is not injective.
