# Thread: [SOLVED] Proving a function is one-to-one

1. ## [SOLVED] Proving a function is one-to-one

Hey guys,

I am trying to prove the following function is one-to-one and onto.
D=Q\{4}, R=\{2} and F-->R is f(x) = (2x + 3)/(x-4) for all x in D.

Here is my work so far:

One-to-One:
Let f(x1) = f(x2)
(2X1 + 3)/(X1 + 4) = (2X2 + 3)/(X2 + 4)
Ok, so I know I have to end up with x1=x2, but from here, I get stuck. (seems more like an algebra problem from here, but can somebody show me how to end up with x1=x2 from here)

Onto:
Let y be an element of Q.
Suppose f is onto.
Then there exist some x in D s.t. f(x)=y.
f(x) = (2x+3)/(x-4) = y
Now I have to solve for x, and once again, I don't know how to solve for x. Seems I missed this part in Algebra class. Please show me how to solve for x.

Also, am I on the right track to proving these

2. Use simple notation: $f(a)=f(b)$.
$\begin{gathered}
\frac{{2a + 3}}
{{a - 4}} = \frac{{2b + 3}}
{{b - 4}} \hfill \\
2ab - 8a + 3b - 12 = 2ab - 8b + 3a - 12 \hfill \\
11\left( {b - a} \right) = 0 \hfill \\
a = b \hfill \\
\end{gathered}$

3. Thanks, will work on that.

4. Ok, I was able to get the one-to-one part, x= (4y+3)/(y-2). However, now I am having a bit of trouble on the onto part. If I try to show that f((4y+3)/(y-2))= y ,plugging in for x in the original equation (f(x)=(2x+3)/(x-4)) and solving, I get 11/8 = y, but it needs to be just y, so that I can prove onto. Can anybody see where I am going wrong? Thanks

5. Found my error: I was missing a (-) sign in there. Thanks for looking.