# Surjective (onto) function proof

### Surjective (onto) function proof

Hi,
I need some help with proving that this function is surjective(onto):

f: Z$$\rightarrow$$Z
$$f= \begin{cases} x + 1 \text{\ \ x is even} \\ x - 1 \text{ \ \ x is odd} \end{cases}$$

thanks!
Guest

### Re: Surjective (onto) function proof

Saying that a function f: X->Y is surjective (onto) means that, for any y in Y, there exists x in X such that f(x)= y. That is, we can "get" any member of Y using some member of X.

Here, f(x)= x+ 1 if x is an even integer, f(x)= x- 1 if x is an odd integer. The crucial point is that if an integer, n, is even then both n+ 1 and n- 1 are odd and if n is odd then both n+ 1 and n- 1 are even.

Suppose y is even. Then x= y+ 1 is odd. f(x)= x- 1= (y+ 1)- 1= y.
Suppose y is odd. Then x= y- 1 is even. f(x)= x+ 1= (y- 1)+ 1= y.

Since every member of Z is either even or odd, for any y in Z, there exist x in Z such that f(x)= y.

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