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

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:

\(\ds \map f {x_1}\)

\(=\)

\(\ds \map f {x_2}\)

by supposition

\(\ds \leadsto \ \ \)

\(\ds 2 x_1 + 1\)

\(=\)

\(\ds 2 x_2 + 1\)

Definition of $f$

\(\ds \leadsto \ \ \)

\(\ds x_1\)

\(=\)

\(\ds x_2\)

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 $2.5$