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Proving something is onto

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine whether f: ZxZ->Z is onto if

    (a) f(m, n) = 2m-n
    (b) f(m, n) = n^2 - m^2
    (c) f(m, n) = m + n + 1
    (d) f(m, n) = |m| - |n|
    (e) f(m, n) = m^2 - 4

    2. Relevant equations

    A function is onto if for every y in the codomain there is at least one X in the domain st f(x) = y

    3. The attempt at a solution

    I have no idea how to start proving this. I know I'm supposed to prove that for every integer y, there is at least 1 (m, n) pair, but that requires solving for another variable m or n doesn't it? This leaves a lot of m,n to test for yes?

    For example, I can go about the first one by setting m = (y+n)/2 and if y = 0 and n is odd then m is not an integer. However, I know from reading ahead that the first one is onto.

    For the last one I can do

    sqrt(y+4) = m
    If y = 1, then m is not an integer and so y = 1 is not in the range

    Am I supposed to prove the cases where m is odd, n is even, both are even, both are odd, m is even, n is odd?
    Last edited: Sep 14, 2008
  2. jcsd
  3. Sep 14, 2008 #2
    find what the range is. Clearly, it is the set of integers. Thus for every integer a, we should have f(x, y) = a

    for the first part [tex]2m - n = a => 2m = a + n[/tex]. Since m and n are also integers. find how many such pairs satisfy. In fact, dont find all pairs, just prove that atleast one answer exists...

    Remember, that a is an integer and can be negative
  4. Sep 14, 2008 #3
    Heres a try at the first one

    m = (a+n)/2 is an integer only if m is even which => (a+n) must be even.

    (a+n) is even if a & n are both even or both odd which covers a for both even and odd integers
  5. Sep 14, 2008 #4
    ok, why should m be even??? a + n should be even, because it is being divided by two, and the result must be an integer. However, m need not be.

    9 = (11 + 7)/2
  6. Sep 18, 2008 #5


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    No, m does not have to be even, only a+ n does.

    For a any integer, 2m-n= a implies m= (a+n)/2. Okay, if a is even, say a= 2k, take n= 0 so m= a/2= k: 2k- 0= 2k= a. If a is odd, say a= 2k+1, take n= 1 so m= (2k+ 2)/2= k+1; 2(k+1)- 1= 2k+ 2- 1= 2k+ 1= a. For any a, there exist m and n such that 2m- n= a. The function is "onto" Z.

    (Note: this function is not "one to one". f(3,5)= 2(3)- 5= 1 and f(2, 3)= 2(2)- 3= 1.)
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