Question

Please, help

Answer 1

Question?

Answer 2

How to prove f is an onto function?
\[f(x)=\begin{cases}x-1 ~~if ~~x~~is even\\x+1~~if~~ x ~~is~~odd\end{cases}\]

Answer 3

@zepdrix

Answer 4

\[
f(x) = \begin{cases}
(2n)-1 = 2m+1\\
(2n'+1)+1 = 2m'
\end{cases}
\]

Answer 5

I got you so far. Please, continue

Answer 6

What is the domain and codomain?

Answer 7

Oh, I am sorry, f: Z --> Z

Answer 8

If \(z\in \mathbb Z\) then it is either odd or even.

If it is odd, then \[
\exists n \in \mathbb N\quad z=2n+1
\]We can let \(x = 2n+2\). Since \(x\) is even:\[
f(x) = (2n+2)-1 = 2n+1 = z
\]
If it is even, then\[
\exists n \in \mathbb N\quad z=2n
\]We can let \(x = 2n-1\). Since \(x\) is odd:\[
f(x) = (2n-1)+1 = 2n = z
\]

Therefore \[
\forall z\in\mathbb Z\;\exists x \in \mathbb Z\quad f(x) =z
\]

Answer 9

Thank you so much. I got it.
One more question, why do we have to use N here?

Answer 10

Well, we should use \(\mathbb Z\) I guess.