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.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $4$: Mappings: Exercise $5 \ \text{(b)}$