1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Math Proof/Logic Injection/Surjection Problem

  1. Feb 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine which of the following statements are true. Give proofs for the true statements and counterexamples for the false statements.

    2. Relevant equations
    B. Every non-decreasing function from R to R is injective.

    3. The attempt at a solution
    So I think I know a function that would suit, but I can't just give a drawing of a graph as a proof. I think it would be a function that would be increasing, eventually go monotone for a little while, and then continue to increase to infinity. Does anybody know a function that would represent this well?
  2. jcsd
  3. Feb 13, 2008 #2
    non-decreasing function? Just remember that non-decreasing doesn't necessarily imply increasing. I don't really want to give any more away than that, but a trivial counterexample would work fine; No need to construct some bizarre graph.
  4. Feb 13, 2008 #3
    do you mean something like y=5 would work?
  5. Feb 13, 2008 #4
    Exactly. It's non-decreasing and it is not injective.
  6. Feb 13, 2008 #5
    I got some more. I'd appreciate some feedback.

    C. Every injective function from R to R is monotone.

    I think this one's true.

    D. Every surjective function from R to R is unbounded.

    Would arctan(x) disprove this one? I'm not sure if surjection is limited to the range of a function or for all reals.

    E. Every unbounded function from R to R is surjective.

    I think this one's true.
  7. Feb 13, 2008 #6
    C (Dick explains, I failed at the definition of monotone)

    Your D also does not work because for every y in R there is not an x in R such that arctan(x) = y. For example, for y = 2 in the codomain, there is no corresponding x.

    E I think is the trickiest to construct, but I believe you can make a counterexample by manipulating an obvious unbounded surjection like f(x)=x. See if you can't make that function "skip" a number in the range.
    Last edited: Feb 13, 2008
  8. Feb 13, 2008 #7


    User Avatar
    Science Advisor
    Homework Helper

    C) Suppose the function is discontinuous. D) arctan(x) isn't surjective. E) I think you are again thinking of only continuous functions, and not even all of those.
  9. Feb 13, 2008 #8


    User Avatar
    Science Advisor
    Homework Helper

    For E) how about f(x)=x^2? C) isn't easy to show because it's not true, they didn't say the function is continuous. D) is on the mark, because there is no counterexample. Surjection means the range is all reals.
    Last edited: Feb 13, 2008
  10. Feb 13, 2008 #9
    I understand the example you give for E.

    For C, I don't get how that can be false. How can the function be discontinuous if it's from R to R?

    I understand why arctan doesn't work for D. So is it safe to assume that one's true?
  11. Feb 13, 2008 #10


    User Avatar
    Science Advisor
    Homework Helper

    A surjection from R->R covers all reals, it's pretty safe to assume it's unbounded. For C) let f(x)=x for x<0. f(x)=1-x for 0<=x<=1. f(x)=x for x>1. It's injective but it's not monotone. It's also not continuous. It would have to be.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Math Proof/Logic Injection/Surjection Problem