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

[Norm850]

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.

 

[SW VandeCarr]

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.