# Denumerable Set

#### novice

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

$$\displaystyle f(n)=\frac{1+(-1)^n(2n-1)}{4}$$

The picture shows that there is one-one correspondence, such as this:

$$\displaystyle 1 \rightarrow 0$$

$$\displaystyle 1 \rightarrow 1$$

$$\displaystyle 2 \rightarrow -1$$

$$\displaystyle 3 \rightarrow 2$$

$$\displaystyle 4 \rightarrow -2$$

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Now I want to prove that $$\displaystyle f(n)=\frac{1+(-1)^n(2n-1)}{4}$$ bijective, but $$\displaystyle (-1)^n$$ in the expression is making it very clumsy.

To prove the one-to-one 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 $$\displaystyle g(s)=\frac{1+(-1)^s(2s-1)}{4}$$ where $$\displaystyle s=2n-1, n \in \mathbb{N}$$

and there also exists an even positive integer $$\displaystyle q\in \mathbb{N}$$ such that $$\displaystyle h(t)=\frac{1+(-1)^t(2t-1)}{4}$$ where $$\displaystyle t=2n, n \in \mathbb{N}$$

Questions:

1. Is there a simpler way than to prove the onto-function in two parts?

2. Is there a better way with handling $$\displaystyle (-1)^n$$ expression ?

#### Plato

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

• novice