- #1

- 11

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Proof: If x is even:

x

_{1}= 2k

_{1}

x

_{2}= 2k

_{2}

Suppose f(x

_{1}) = f(x

_{2}), then

2k

_{1}/2 = 2k

_{2}/2

k

_{1}= k

_{2}

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.