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

  • Thread starter Norm850
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  • #1
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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.
 

Answers and Replies

  • #2
2,161
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Not sure if you can use a function to define the independent variable to prove if it's one-to-one or not.

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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