Mapping/Examples/x^2 + y^2 = 1
Jump to navigation
Jump to search
Example of Relations which is not a Mapping
Let $R_1$ be the relation defined on the Cartesian plane $\R \times \R$ as:
- $R_1 = \set {\tuple {x, y} \in \R \times \R: x^2 + y^2 = 1}$
Then $R_1$ is not a mapping.
Proof

Graph of $x^2 + y^2 = 1$
$R_1$ fails to be a mapping for the following reasons:
$(1): \quad$ For $x < -1$ and $x > 1$, there exists no $y \in \R$ such that $x^2 + y^2 = 1$.
Thus $R_1$ fails to be left-total.
$(2): \quad$ For $-1 < x < 1$, there exist exactly two $y \in \R$ such that $x^2 + y^2 = 1$, for example:
\(\ds \paren {\dfrac 1 2}^2 + \paren {\dfrac {\sqrt 3} 2}^2\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \paren {\dfrac 1 2}^2 + \paren {-\dfrac {\sqrt 3} 2}^2\) | \(=\) | \(\ds 1\) |
So both $\tuple {\dfrac 1 2, \dfrac {\sqrt 3} 2}$ and $\tuple {\dfrac 1 2, -\dfrac {\sqrt 3} 2}$ are elements of $R_1$.
Thus $R_1$ fails to be many-to-one.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 10 \alpha$