
Denumerable Set
My book shows that the set of integers, $\displaystyle \mathbb{Z}, $ is denumerable by expressing the elements of the set as infinite sequence
$\displaystyle \mathbb{Z}=\{0, 1,1,2,2,...\}$, and then shows that the function
$\displaystyle f: \mathbb{N}\rightarrow \mathbb{Z}$ is bijective, where
$\displaystyle f$ is defined as
$\displaystyle f(n)=\frac{1+(1)^n(2n1)}{4}$
The picture shows that there is oneone correspondence, such as this:
$\displaystyle 1 \rightarrow 0$
$\displaystyle 1 \rightarrow 1$
$\displaystyle 2 \rightarrow 1$
$\displaystyle 3 \rightarrow 2$
$\displaystyle 4 \rightarrow 2$
.
.
.
Now I want to prove that $\displaystyle f(n)=\frac{1+(1)^n(2n1)}{4}$ bijective, but $\displaystyle (1)^n$ in the expression is making it very clumsy.
To prove the onetoone isn't too bad, since $\displaystyle f(a_1)=f(a_2)$ seems to look clean, but now when I get to the part where I need to prove it an onto function, I got a huge mess, so I am wondering whether it be alright to split the proof in two parts such that
$\displaystyle f(n) = g(s) \cup h(t)$ and define $\displaystyle g(s) \subset \mathbb{Z}^$ and $\displaystyle h(t) \subset \mathbb{Z}^+$ and by so doing, I plan to prove that
there exists an odd positive integer $\displaystyle p\in \mathbb{N}$ such that $\displaystyle g(s)=\frac{1+(1)^s(2s1)}{4}$ where $\displaystyle s=2n1, n \in \mathbb{N}$
and there also exists an even positive integer $\displaystyle q\in \mathbb{N}$ such that $\displaystyle h(t)=\frac{1+(1)^t(2t1)}{4}$ where $\displaystyle t=2n, n \in \mathbb{N}$
Questions:
1. Is there a simpler way than to prove the ontofunction in two parts?
2. Is there a better way with handling $\displaystyle (1)^n$ expression ?

If $\displaystyle x=0$ then $\displaystyle f(1)=0$.
If $\displaystyle x>0$ then $\displaystyle f(2x)=x$.
If $\displaystyle x<0$ then $\displaystyle f(2x+1)=x$.
So it is onto.