# Bijection/Examples/-1^x of Floor of Half x from Natural Numbers to Integers

## Example of Bijection

Let $f: \N \to \Z$ be the mapping defined from the natural numbers to the integers as:

$\forall x \in \N: f \paren x = \paren {-1}^x \floor {\dfrac x 2}$

Then $f$ is a bijection.

## Proof

Let $x_1, x_2 \in \N$.

Suppose $f \paren {x_1} = f \paren {x_2}$.

Then, for a start:

$\paren {-1}^{x_1} = \paren {-1}^{x_2}$

and so $x_1$ and $x_2$ are either both even or both odd.

Suppose $x_1$ and $x_2$ are both even.

Then:

 $\ds f \paren {x_1}$ $=$ $\ds f \paren {x_2}$ $\ds f \paren {2 m}$ $=$ $\ds f \paren {2 n}$ for some $m, n \in \N$ $\ds \leadsto \ \$ $\ds \paren {-1}^{2 m} \floor {\dfrac {2 m} 2}$ $=$ $\ds \paren {-1}^{2 n} \floor {\dfrac {2 n} 2}$ $\ds \leadsto \ \$ $\ds \floor m$ $=$ $\ds \floor n$ $\ds \leadsto \ \$ $\ds m$ $=$ $\ds n$ $\ds \leadsto \ \$ $\ds 2 m$ $=$ $\ds 2 n$ $\ds \leadsto \ \$ $\ds x_1$ $=$ $\ds x_2$

Suppose $x_1$ and $x_2$ are both odd.

Then:

 $\ds f \paren {x_1}$ $=$ $\ds f \paren {x_2}$ $\ds f \paren {2 m + 1}$ $=$ $\ds f \paren {2 n + 1}$ for some $m, n \in \N$ $\ds \leadsto \ \$ $\ds \paren {-1}^{2 m + 1} \floor {\dfrac {2 m + 1} 2}$ $=$ $\ds \paren {-1}^{2 n + 1} \floor {\dfrac {2 n + 1} 2}$ $\ds \leadsto \ \$ $\ds -1 \floor {m + \dfrac 1 2}$ $=$ $\ds -1 \floor {n + \dfrac 1 2}$ $\ds \leadsto \ \$ $\ds m$ $=$ $\ds n$ $\ds \leadsto \ \$ $\ds 2 m + 1$ $=$ $\ds 2 n + 1$ $\ds \leadsto \ \$ $\ds x_1$ $=$ $\ds x_2$

Thus:

$\forall x_1, x_2 \in \N: f \paren {x_1} = f \paren {x_2} \implies x_1 = x_2$

and so $f$ is an injection by definition.

$\Box$

Let $y \in \Z$ such that $y \ge 0$.

Consider the natural number $x = 2 y$.

Then:

 $\ds f \paren x$ $=$ $\ds \paren {-1}^{2 y} \floor {\dfrac {2 y} 2}$ Definition of $f$ $\ds$ $=$ $\ds \floor y$ $\ds$ $=$ $\ds y$

Thus:

$\forall y \in \Z_{\ge 0}: \exists x \in \N: f \paren x = y$

Let $y \in \Z$ such that $y < 0$.

Consider the natural number $x = 2 \paren {-y} + 1$.

Then:

 $\ds f \paren x$ $=$ $\ds \paren {-1}^{2 \paren {-y} + 1} \floor {\dfrac {2 \paren {-y} + 1} 2}$ Definition of $f$ $\ds$ $=$ $\ds -\floor {\paren {-y} + \dfrac 1 2}$ $\ds$ $=$ $\ds y$

Thus:

$\forall y \in \Z_{<0}: \exists x \in \N: f \paren x = y$

Thus by definition $f$ is a surjection by definition.

$\Box$

Thus $f$ is both an injection and a surjection, and so by definition a bijection.

$\blacksquare$