Consider f: ℝ{1} → ℝ{1} given by f(x) = x/(x-1)
Show that f(x) is one-to-one and onto.
What I have:
If a function is one-to-one then it follows that if f(a) = f(b) then a=b.
If a function is onto then it follows that ∀y∈Y, ∃x∈X such that f(x)=y.
So for one-to-one I need to show that if a/(a-1) = b/(b-1) then a=b. That seems obvious enough, however I end up with ab-a = ab-b, which doesn't mean a = b.
NEED HELP WITH ONTO
Any help is greatly appreciated!
Try dividing $a-1$ into $a$, and do the same for the term involving $b$ (long division, synthetic division...). Then you can rewrite your fractions and state that $$ \begin{align*} \frac{a}{a-1} &= \frac{b}{b-1} \\ \Rightarrow 1+\frac{1}{a-1} &= 1+\frac{1}{b-1} \\ \Rightarrow \frac{1}{a-1} &= \frac{1}{b-1} \\ \Rightarrow a-1 &= b-1 \\ \Rightarrow a &=b. \end{align*} $$
For onto, solve $y=\frac{x}{x-1}$ for $x$ to see that $$y=\frac{x}{x-1} \Rightarrow \ldots \Rightarrow x=\frac{y}{y-1}.$$ Hence a typical member $x$ of the domain $\mathbb{R}\setminus\{1\}$ is of this form. Now when $f(x)=\frac{x}{x-1}$ is evaluated at a typical member of the domain, we get $$f(x)=f\left(\frac{y}{y-1}\right)=\dfrac{\dfrac{y}{y-1}}{\dfrac{y}{y-1}-1}=y.$$ Since $y \in \mathbb{R}\setminus\{1\}$, $f(x)$ is onto.
