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Prove peicewise function is or is not onto.

  1. Apr 10, 2012 #1
    1. The problem statement, all variables and given/known data
    I am told to prove or disprove whether the following function is 1:1 and onto.

    f(x)= x if x is rational
    x^2 if x is not rational

    The Domain and Range are both given as [0,1]

    I have provided a counter example to show it is not 1:1. x=1/2 and sqrt(0.5) both map to 0.5.

    I am stuck on onto. If I am to disprove it being onto, I need to find/show
    there is a b in the range
    such that there is no a in the range
    for which (a,b) is an element of f(a). I am confused as to where to begin to formulate this.
     
  2. jcsd
  3. Apr 10, 2012 #2
    Why are you trying to disprove it , when you can prove it.
    Take any 'b' in the co-domain and show that there is an 'a' in the domain such that f(a) = b.


    By the way this thread does not belong to the Advanced Physics sub-forum. You should have posted it here - https://www.physicsforums.com/forumdisplay.php?f=156 , i.e in the Calculus and Beyond forum.
     
    Last edited: Apr 10, 2012
  4. Apr 11, 2012 #3
    Thank you.

    Apologies for the mislabeling.
     
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