This is a problem from Discrete Mathematics and its Applications enter image description here

Book's definition on bijection enter image description here

Book's definition on onto enter image description here Book's definition on one to one enter image description here

I am trying to do problem 23D. Here is my work so far enter image description here

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.

  • $\begingroup$ You were on the right track with one-one ness. I added my answer to confirm this. $\endgroup$ – 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.

  • $\begingroup$ wait so you don't have to go through all these algebra? If you just find one possible combination of n and m, such that if f(n) = f(m), m!=n, you're done? $\endgroup$ – committedandroider Jan 21 '15 at 4:26
  • $\begingroup$ @committedandroider: That's right! $\endgroup$ – voldemort Jan 21 '15 at 4:27
  • $\begingroup$ Oh that onto part was tricky too. Because f(x) never exceeds 1, not all values in the codomain are mapped to $\endgroup$ – committedandroider Jan 21 '15 at 4:31
  • $\begingroup$ @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. $\endgroup$ – voldemort Jan 21 '15 at 4:33
  • $\begingroup$ I know this is a small detail but shouldn't the inequality be x^2 + 1 < x^2 + 2? To me, any expression added by two should be greater than the result of adding the expression by one $\endgroup$ – committedandroider Jan 28 '15 at 19:36

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