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(Real Analysis) Show the function is Bijection

  1. Aug 28, 2010 #1
    1. The problem statement, all variables and given/known data

    The problem and my attempt are attached.

    I am unable to solve the function for x.

    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Aug 28, 2010 #2
    hmm, if you want to show injection, you need to show for all the values in the domain, now you only showing [tex]-1\neq1[/tex] how about [tex]-1\neq-1.1[/tex]?

    so if you want to show an injection first see the definition of injection

    whenever [tex]x\neq y \Rightarrow f(x) \neq f(y)[/tex]

    or the contrapositive [tex]f(x)=f(y) \Rightarrow x=y[/tex]

    normally i use the second one, so, for any for any x,y in R [tex] f(x)=f(y)\Rightarrow....... \Rightarrow x=y[/tex]

    that will complete the injection
     
    Last edited: Aug 28, 2010
  4. Aug 28, 2010 #3
    hmm i just learn the latex, i don't know if there's a bug or its just me who did it wrong, i'm sorry, if i convey the different thing, try to refresh it ;P and sorry for my english
     
  5. Aug 29, 2010 #4

    Mark44

    Staff: Mentor

    I don't understand what you're trying to do in 1). You say "Injection proof -1 [itex]\neq[/itex]1, and then go on to set f(-1) equal to f(1). ???

    The idea here is the if f(a) = f(b), and f is one-to-one (i.e., f is an injection), then a = b.

    The function is one-to-one.

    To solve for x in the equation y = x/sqrt(x^2 + 1), square both sides of the equation, then move all terms in x to one side, and then factor.

    It's a little tricky, because squaring both sides introduces extraneous solutions, so the equation for x does not seem to be a function. You will need to simplify the equation somewhat to make x a one-to-one function of y.
     
  6. Aug 29, 2010 #5
    The definition for a bijection I usually use is:

    Given [tex] f: A \rightarrow B, [/tex] f is a bijection if and only if

    EDIT (thanks annoymage): [tex] \forall y \in B , \exists !x \in A \ni f(x) = y. [/tex]

    "For all y in B, there exists a unique x in A such that f(x) = y."
     
    Last edited: Aug 29, 2010
  7. Aug 29, 2010 #6
    hey, aren't that the definition of a function?
     
  8. Aug 29, 2010 #7
    Yep, I have it backwards. Sorry, I'm a bit sleepy. It should be:

    [tex]
    \forall y \in B , \exists !x \in A \ni f(x) = y.
    [/tex]
     
  9. Aug 29, 2010 #8
    but that's the definition of a surjection
     
  10. Aug 29, 2010 #9

    Mark44

    Staff: Mentor

    annoymage, I think you're right.
     
  11. Aug 29, 2010 #10
    nope, the unique x part makes it a bijection. It's subtle but it's there! A surjection only guarantees there exists an x in A. A bijection tells us it's a unique x.
     
  12. Aug 29, 2010 #11
    i see i see
     
  13. Aug 29, 2010 #12
    The process is still of course the same:
    Solve for x in terms of y and show that's it's a unique solution.
     
  14. Aug 29, 2010 #13
    Based on your feedback. This is my bijection thing.
     

    Attached Files:

  15. Aug 29, 2010 #14
    I would rather you do as Mark44 suggested and solve for x in terms of y.
     
  16. Aug 29, 2010 #15
    this complete the injection, and now as mark44 suggested to do for the surjection
     
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