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Prove or disprove that if $R^2$ is transitive then $R$ is also transitive.

I tried to prove $(R\circ R)^2\subseteq (R\circ R)\implies R^2\subseteq R$

this way

$(R\circ R)\circ (R\circ R)\subseteq (R\circ R)\implies R^2\subseteq R$

$R\circ (R\circ R) \circ R\subseteq (R\circ R)\implies R^2\subseteq R$

$R\circ R\subseteq (R\circ R)\implies R^2\subseteq R$

I saw this "pattern" somewhere so I tried to use it but it doesnt seem to be right way

lllook
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  • What if $R={(1,2),(2,3),(1,1),(2,2),(3,3)}$ on the set ${1,2,3}$? Wouldn't that be a counterexample? – Arthur Nov 24 '15 at 13:34

1 Answers1

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$R:=\{(1,2),(2,1)\}$ is not transitive. $R\circ R=\{(1,1),(2,2)\}$ is transitive.

drhab
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