Mapping/Examples/x^3 + y^4 = 1
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Example of Relation which is not a Mapping
Let $R_4$ be the relation defined on the Cartesian plane $\R \times \R$ as:
- $R_4 = \set {\tuple {x, y} \in \R \times \R: x^3 + y^4 = 1}$
Then $R_4$ is not a mapping.
Proof
$R_4$ fails to be a mapping for the following reasons:
$(1): \quad$ For $x > 1$, there exists no $y \in \R$ such that $x^3 + y^4 = 1$.
Thus $R_4$ fails to be left-total.
$(2): \quad$ For $x < 1$, there exist exactly two $y \in \R$ such that $x^3 + y^4 = 1$, for example:
\(\ds 0^3 + 1^4\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds 0^3 + \paren {-1}^4\) | \(=\) | \(\ds 1\) |
So both $\tuple {0, 1}$ and $\tuple {0, -1}$ are elements of $R_4$.
Thus $R_4$ fails to be many-to-one.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 10 \alpha$