Mapping/Examples/root x + root y = 1
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Example of Relation which is not a Mapping
Let $R_5$ be the relation defined on the Cartesian plane $\R \times \R$ as:
- $R_5 = \set {\tuple {x, y} \in \R \times \R: \sqrt x + \sqrt y = 1}$
Then $R_5$ is not a mapping.
Proof

Graph of $\sqrt x + \sqrt y = 1$
$R_5$ fails to be a mapping because, for example, $\sqrt x$ does not exist for $x < 0$.
Thus $R_5$ is undefined for $x <0$.
Thus $R_5$ fails to be left-total.
We have:
\(\ds \sqrt x + \sqrt y\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {\sqrt x + \sqrt y}^2\) | \(=\) | \(\ds 1\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x + y + 2 \sqrt {x y}\) | \(=\) | \(\ds 1\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x + y - 1}^2\) | \(=\) | \(\ds 4 x y\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^2 + y^2 - 2 x y - 2 x - 2 y + 1\) | \(=\) | \(\ds 0\) |
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 10 \alpha$