# Inverse of a set of ordered pairs.

An exam ask me the following question.

Let $r=\{(x,y) \ | \ x \in [-1,1] \ \text{and} \ y=x^2\}$, is the following statement true?

$$r^{-1}=\{(x,y) \ | \ x \in [0,1] \ \text{and} \ y=\pm\sqrt{|x|} \}$$

I have to answer the question by explaining the reason using some referable definition. At first, I was going to use the definition of function and inverse function because I noticed that $r$ is a function from $[-1,1]$ to $[0,1]$, but when I looked at it again, $r$ is not one-to-one and $r^{-1}$ is not a function because it maps $x$ to $-\sqrt{|x|}$ and $\sqrt{|x|}$.

I have been searching for a definition of inverse of a set of ordered pairs to see if it can be useful, but could not find one.

I am wondering what the $r^{-1}$ means in this case. If the question clearly state that $r$ is a function and $r^{-1}$ is the inverse of $r$, I would answer "No" already.

• May 24 '15 at 5:04
• That is what I am looking for, thank you :D. May 24 '15 at 5:16

The set $$r$$ is an example of a relation, which is a simply a subset $$R \subseteq X \times Y$$ of the Cartesian product of two given sets $$X, Y$$. For $$r$$, the sets $$X$$ and $$Y$$ are not specified, but we could certainly take $$X = [-1, 1], \quad Y = [0, 1].$$
We say that a relation $$R \subseteq X \times Y$$ is a function $$R: X \to Y$$ iff for all $$x \in X$$ there is exactly one $$y \in Y$$ such that $$(x, y) \in R$$. (Note that our choice of $$X$$, $$Y$$ makes $$r$$ a function.) If $$R$$ also satisfies the reverse, or more precisely that for all $$y \in Y$$ there is exactly one $$x \in X$$ such that $$(x, y) \in R$$, then by definition, the relation $$R^{-1} \subseteq Y \times X$$ defined by $$\phantom{(\ast)} \qquad R^{-1} := \{(y, x) : (x, y) \in R\} \qquad (\ast)$$ is a function $$R^{-1}: Y \to X$$; we say that $$R^{-1}$$ is the inverse of $$R$$ and that $$R$$ (regarded as a function) is invertible.
On the other hand, the definition $$(\ast)$$ makes perfectly good sense for any relation $$R$$, not just those relations that are functions; so, for any relation $$R$$, we define the inverse relation $$R^{-1}$$ by that formula.