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Proving piece-wise function is one-to-one?

  1. Mar 27, 2012 #1
    f: Z -> Z defined by f(x) = x/2 if x is even, (x-1)/2 if x is odd.

    Proof: If x is even:

    x1 = 2k1
    x2 = 2k2

    Suppose f(x1) = f(x2), then

    2k1/2 = 2k2/2
    k1 = k2

    So if x is even, the function is one to one? Is this an okay proof for the first half of if x is even, then I just do the same for if x is odd correct?

    Not sure if you can use a function to define the independent variable to prove if it's one-to-one or not.
  2. jcsd
  3. Mar 27, 2012 #2
    I'm not sure I follow the above. If a function is piecewise it's sufficient to show that all derivatives are positive or all are negative. Actually, this should be true for all analytic one to one functions. In linear algebra its sufficient to show that only the zero vector is mapped into the null space.
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