Mapping/Examples/x^3-x on Real Numbers
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Example of Mapping
Let $f: \R \to \R$ be the mapping defined on the set of real numbers as:
- $\forall x \in \R: \map f x = x^3 - x$
Then $f$ is a surjection but not an injection.
Proof
Let $y \in \R$.
As $x \to \infty$, we have that $y \to \infty$.
Similarly, as $x \to -\infty$, we have that $y \to -\infty$.
From Real Polynomial Function is Continuous, $f$ is continuous on $\R$.
It follows from the Intermediate Value Theorem that:
- $\forall y \in \R: \exists x \in \R: y = \map f x$
Thus, by definition, $f$ is a surjection.
$\Box$
We have that:
\(\ds \map f 0\) | \(=\) | \(\ds 0^3 - 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \map f 1\) | \(=\) | \(\ds 1^3 - 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
demonstrating that $f$ is not an injection.
$\blacksquare$
Sources
- 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Appendix $\text{A}.4$: Functions: Problem Set $\text{A}.4$: $22$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $2$: Maps and relations on sets: Example $2.5$