Math Problem Solver

The inverse relation is not a function, because of the following inverse relation pairs:

\[\begin{align*} & -5 & \mapsto & \left\{4, 6\right\}\\ \end{align*}\]

A function maps every possible input in its domain to exactly one output in its range.

The inverse of the relation is the set of ordered pairs $\{ \left( -5, \ 4\right), \left( 0, \ 0\right), \left( 3, \ -6\right), \left( -6, \ -4\right), \left( -5, \ 6\right) \}$.

The domain of the relation is:\[\left\{-6, -5, 0, 3\right\}\]

The range of the relation is:\[\left\{-6, -4, 0, 4, 6\right\}\]

This relation has the following properties:\[\begin{array}{c|c|l} \textbf{Property} & \textbf{Value} & \textbf{Description} \\ \hline \text{Universal} & \text{No} &\text{For any two elements } a \text{ and } b \text{ in the domain, there exists a pair } (a,b)\text{.}\\ \hline \text{Identity} & \text{No} &\text{For each element } a \text{ in the domain, there is exactly one pair } (a,a)\text{.}\\ \hline \text{Constant} & \text{No} &\text{For each element } a \text{ in the domain, the only pair is } (a,k) \text{ for some constant } k\text{.}\\ \hline \text{Symmetric} & \text{No} &\text{For each pair } (a,b) \text{ in the relation, there exists a pair } (b,a)\text{.}\\ \hline \text{Reflexive} & \text{No} &\text{For each element } a \text{ in the domain, there exists a pair } (a,a)\text{.}\\ \hline \text{Transitive} & \text{No} &\text{If both } (a,b) \text{ and } (b,c) \text{ are in the relation, there exists a pair } (a,c)\text{.}\\ \hline \text{One~to~One} & \text{N/A} &\text{Both the relation and the inverse relation are functions.}\\ \end{array}\]