# Thread: Prove a function is onto

1. ## [Solved]Prove a function is onto

I need to prove $\displaystyle f(x) = ((x+1)/x)$ is an onto function. I'm kind of hazy here, am I supposed to subsitute and solve for x? Such as, $\displaystyle y=((x+1)/x)$. IF that is the case, I can't get it to add up to prove that function is onto. I know answer to problem, its just I think I'm jackng up my algebra and want to be able to do all the steps involved.

Thanks!

2. Onto what?
What is the domain?
What is the image set?

3. Doh! Sorry all real numbers, domain is x not equal to 0 and co-domain not equal to 1.

4. Here's what you do:
Suppose f is onto. Then there exists an x such that f(x) = (x+1)/(x) = y. Solve for x to get x = 1/(y-1). Now plug this value back into the equation, like this:
f(1/(y-1)) = (x+1)/(x) = [(1/(y-1))+1]/[1/(y-1)]. Now solve this equation to end up with f(1/(y-1)) = y. There, now you have proved that f(x) is onto.

5. Originally Posted by spearfish
Here's what you do:
Suppose f is onto. Then there exists an x such that f(x) = (x+1)/(x) = y. Solve for x to get x = 1/(y-1). Now plug this value back into the equation, like this:
f(1/(y-1)) = (x+1)/(x) = [(1/(y-1))+1]/[1/(y-1)]. Now solve this equation to end up with f(1/(y-1)) = y. There, now you have proved that f(x) is onto.
That is what I was orignally doing. I know the problem is I'm messing up my algebra. I can't solve for x, I always end up cancelling my x-variables. Sorry, my algebra is rusty.

y=((x+1)/x) ...multiply both sides by x
xy=x+1 ...subtract 1
xy-1=x ...divide by (y-1)
x=x/(y-1)

Stuck here, not even sure if I did those steps properly.

6. Yeah, I hear you, it happens to me too!

y = (x+1)/x
y = x/x + 1/x
y = 1 + 1/x
-1/x = 1-y
1/x = -1+y
1 = x(-1+y)
1/(y-1) = x

Hope that helps.

7. Originally Posted by spearfish
Yeah, I hear you, it happens to me too!

y = (x+1)/x
y = x/x + 1/x
y = 1 + 1/x
-1/x = 1-y
1/x = -1+y
1 = x(-1+y)
1/(y-1) = x

Hope that helps.
Thats is what I was forgetting right there, thanks alot!