# Proving a function is onto?

Let $f: \mathbb{R}\setminus \{3\} \to \mathbb{R}\setminus \{1\}$ be defined by $f(x)=\dfrac{x+3}{x-3}$

Prove that $f$ is onto:

Okay, here is the deal. I just started my first abstract algebra course. I know the definition of onto and understand it.

What I want to do is something along these lines:

Let $y \in \mathbb{R}\setminus \{1\}$ and note that (*) is a real-valued function for all $y \in \mathbb{R}\setminus \{1\}$

then:

$$\frac{*+3}{*-3}=y$$

I want to show this holds. Since it holds for arbitrary $y$ in our codomain then it must be true of all $y$. I think this will prove that $f$ is onto.

I am having trouble finding (*). I am getting stuck with the algebra when trying to express $f$ as $x=...$

Could anyone give me a solid hint here?

• Hint, instead of the answers given: If, when I give you any $y$ in the codomain, you can find an $x$ in the domain such that $f(x)=y$, then $f$ is onto. – imallett Jan 26 '15 at 3:48
• Style question: Why switch to $*$ from $x$? Often $*$ is an operator. – Paul Draper Jan 26 '15 at 7:26

$\dfrac{* + 3}{* - 3} = y$

$\implies * + 3 = (* - 3)y$

$\implies * + 3 = *y - 3y$

$\implies * - *y = -3 - 3y$

$\implies *(1 - y) = -3- 3y$

$\implies * = \dfrac{-3 - 3y}{1 - y}$

And $*$ is defined for all $y$ not equal to $1$, i.e., for all $y \in \Bbb R - \{1\}$.

So, given $y \in \Bbb R - \{1 \}$, let $* = \dfrac{ - 3 - 3y}{1 - y}$. Then $f(*) = y$. At the end, you should definitely check that $f(\dfrac{ - 3 - 3y}{1 - y}) = y$.

Is this what you are looking for?

• Yes - this is what I was looking for. For some reason I just couldn't see the simple algebra. – 123 Jan 26 '15 at 3:05

Let $y \in \mathbb{R}$, and $y \neq 1$. We solve for $x$: $\dfrac{x+3}{x-3} = y \Rightarrow x+3=xy-3y \Rightarrow x(y-1)=3(y+1)\Rightarrow x = \dfrac{3(y+1)}{y-1}$. Check that $x \neq 3$. Thus $f$ is onto.