# Math Help - Proving an "onto" function..

1. ## Proving an "onto" function..

I have to prove that f(n) = 2n is not an onto function when the domain and co-domain are a set of all of the integers.

I graph it and see it isn't an onto function. However, everytime I do the math, it says it is. Can someone step through how to solve this one for me?

This is what I get.

First substitute m for f(n)

m = 2m

Solve for one side

(2/m) = m

If f(n) = m then,

f(n) = f(2/m) = 2(2/m) = m

therefore, f(n) = m.

What am I doing wrong. Thanks...

2. ## Re: Proving an "onto" function..

Suppose you have a group of children numbered 1, 2, 3, ... Suppose also that you are distributing candies to them. If your function from the set of candies to the set of kids' numbers is not onto, you will immediately know this by indignant screams, "What about me? I did not get any!". Now assume you are giving candies only to every other kid starting from 2, i.e., to even-numbered children. Is this function onto? You only have to find just one upset kid to know it's not.

Originally Posted by RatchetTheLombax
I have to prove that f(n) = 2n is not an onto function when the domain and co-domain are a set of all of the integers.

I graph it and see it isn't an onto function. However, everytime I do the math, it says it is. Can someone step through how to solve this one for me?

This is what I get.

First substitute m for f(n)

m = 2m
I assume you want to find the inverse of f. Then you need to solve the equation f(n) = m. If you can solve it for every m, i.e., for every m there exists an n such that f(n) = m, then the function is onto. This equation is 2n = m, not 2m = m.

3. ## Re: Proving an "onto" function..

Originally Posted by RatchetTheLombax
I have to prove that f(n) = 2n is not an onto function when the domain and co-domain are a set of all of the integers.

I think that you may have missunderstood onto.
For $\forall n\in\mathbb{Z}$ the image $2n$ is even.

The problem says "the domain and co-domain are a set of all of the integers". That includes odd integers.

It maybe I am reading co-domain correctly. But the image of that function has no odd integers.