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

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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

$\blacksquare$


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