Mapping/Examples/x^4 + y^3 = 1
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Example of Mapping
Let $R_3$ be the relation defined on the Cartesian plane $\R \times \R$ as:
- $R_3 = \set {\tuple {x, y} \in \R \times \R: x^4 + y^3 = 1}$
Then $R_3$ is a mapping.
Proof
\(\ds x^4 + y^3\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y^3\) | \(=\) | \(\ds 1 - x^4\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \sqrt [3] {1 - x^4}\) |
We have that:
- $\forall x \in \R: \exists! y \in \R: \sqrt [3] {1 - x^4}$
and so $R_3$ is both left-total and many-to-one.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 10 \alpha$