Mapping/Examples/root x + root y = 1

From ProofWiki
Jump to navigation Jump to search

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