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:
\(\ds \map f {x_1}\) | \(=\) | \(\ds \map f {x_2}\) | 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 $2.5$