# How to interpret algebraic relationship/ next step to take to prove function is onto?

This is a problem from Discrete Mathematics and its Applications Book's definition on bijection Book's definition on onto Book's definition on one to one I am trying to do problem 23D. Here is my work so far First I tried showing that the function is a one to one function. I set the equation that if two function outputs of different variables, m, and n have the same output, m and n are actually the same variable. Through some i expression, I got to the equation m^2 = n^2. I wasn't sure how to interpret this. I could have square rooted both sides to get m=n, but at the same time I also thought that if m = -2 and n = 2, they would both evaluate to the same function output, breaking the rule of one to one. Is that the right way to think about it or is m=n if you take the square root of both sides?

Because I was unsure about this, I moved onto to seeing if the function was onto. I know that to do this, you have to solve for the input x to show that yes indeed for every y in the output codomain, there is an x that evaluates to it. How would you do this in this situation? I couldn't find a way to isolate the x because of the squared term and the other term has a y attached to it.

• You were on the right track with one-one ness. I added my answer to confirm this. – voldemort Jan 21 '15 at 4:23

Hint: compute $f(-1)$ and $f(1)$. In fact, show that $f(-x)=f(x)$ for all $x$. So, this function isn't one to one.
Also, note that $x^2+1 \leq x^2+2$ for all $x$. Then $0 \leq f(x) \leq 1$ for all $x$. Not onto either!
So, $f$ is not a bijection.
• @committedandroider: Since the question only asked you to find if the function was a bijection or not, you could skip verification of the onto part, noting that since $f$ is not one one- it's not a bijection. – voldemort Jan 21 '15 at 4:33