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Surjectivity of a function

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data

    I'm trying to show that the function below is surjective

    [tex]f: Z^2 \rightarrow Z, f((n,m))=nm[/tex]

    2. Relevant equations



    3. The attempt at a solution

    1. Suppose [tex]z\in Z[/tex].

    2. Then [tex]z=nm[/tex].

    3. So [tex]n=\frac{z}{m}[/tex] and [tex]m=\frac{z}{n}[/tex]

    4. So [tex]f((n,m))=\frac{z^2}{mn}=\frac{n^2m^2}{mn}=nm=z[/tex]

    I'm mostly wondering about step 2... Am I allowed to assume that? (z=nm?)
     
  2. jcsd
  3. Oct 14, 2009 #2
    Why are you solving for z? Why not just use the fact that f(m, n) = mn = z?
     
  4. Oct 14, 2009 #3
    Right.. but what's the (n, m) in that case?

    I'm trying to prove that for any arbitrary z in the codomain, there exists some (n, m) in the domain that satisfies the function. I'm trying to find (n, m) in terms of z.

    Sorry if I didn't get your point.
     
  5. Oct 15, 2009 #4
    I meant solving for n and m in terms of z in my previous post.



    If you know that f(a, b) = ab and you know that z = mn then you have f(m, n) = z. You dont need it more complicated than that.

    And you can assume that there's always a factorization z=mn, just choose z = 1*z.
     
  6. Oct 15, 2009 #5

    HallsofIvy

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    Uhh, Given any integer z, aren't both (z,1) and (1,z) mapped into z? Isn't that enough?
     
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