How can I prove that when a relation $R$ is symmetric, the $k^{th}$power of the relation $R$ ($R^k$ ($k\gt0$)) is also symmetric?
If a relation R is symmetric, so is $R^k$ for any $k\gt0$
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discrete-mathematics
relations
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1By induction on k. – 2017-01-18
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0Start by showing, as a lemma, that if $R_1,R_2$ are symmetric relations on a set $S$, then $R_1 \circ R_2$ is a symmetric relation on $S$. Then apply this lemma to your problem (by induction on k). – 2017-01-18