# Bijection/Examples/2x+1 Function on Real Numbers

Jump to navigation
Jump to search

## Example of Bijection

Let $f: \R \to \R$ be the mapping defined on the set of real numbers as:

- $\forall x \in \R: \map f x = 2 x + 1$

Then $f$ is a bijection.

## Proof

Let $x_1$ and $x_2$ be real numbers.

Then:

\(\displaystyle \map f {x_1}\) | \(=\) | \(\displaystyle \map f {x_2}\) | by supposition | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 2 x_1 + 1\) | \(=\) | \(\displaystyle 2 x_2 + 1\) | Definition of $f$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x_1\) | \(=\) | \(\displaystyle x_2\) |

Hence by definition $f$ is an injection.

$\Box$

Let $y \in \R$.

Let $x = \dfrac {y - 1} 2$.

We have that:

- $x \in \R$
- $\map f x = y$

Hence by definition $f$ is a surjection.

$\Box$

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

$\blacksquare$

## Sources

- 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $2$: Maps and relations on sets: Example $2.5$