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Quirino27's question at Yahoo! Answers (R symmetric implies R^2 symmetric)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Quirino27,

If $U$ is a relation from $A$ to $B$ and $V$ a relation from $B$ to $C$, i.e. $U\subset A\times B$ and $V\subset B\times C$, then the relation $V\circ U$ from $A$ to $C$ is defined in the following way: $$(a,c)\in V\circ U\Leftrightarrow \exists b\in B:(a,b)\in U\mbox{ and } (b,c)\in V$$ In our case, suppose $(x,y)\in R^2=R\circ R$, then exists $y\in A$ such that $(x,y)\in R$ and $(y,z)\in R$. But $R$ is symmetric, so $(y,x)\in R$ and $(z,y)\in R$ and by definition of composition of relations, $(z,y)\in R^2$. That is, $R^2$ is symmetric. $\qquad \square$