# Show that a relation is equivalent if it is both reflexive and cyclic.

A relation $R$ on set $X$ is called cyclic if whenever both $xRy$ and $yRz$ then $zRx$ where $x,y,z\in X$. Show that a relation on $X$ is an equivalence relation if and only if it is both reflexive and cyclic.

Since an equivalence relation is reflexive, symmetric and transitive and we already know the relation is reflexive I assume you prove that symmetric and transitive is the same as reflexive and cyclic. I honestly don't know where to start though or if my intuition is correct so any help would be great!

• What? Obviously that isn't true! Cyclic seems to mean whenever two elements are related symmetrically then every element is related to both the elements. Consider the identity relationship. a = a means a=a but that does not mean that z = a for all z in X. So the identity relationship is not cyclic. – fleablood Nov 2 '15 at 17:46
• I defined cyclic wrong, It is xRy and yRz, sorry! – Countable Nov 2 '15 at 17:52
• Did you mean circular? – user265675 Nov 2 '15 at 17:53
• Ah, that makes all the difference! :) – fleablood Nov 2 '15 at 17:55
• Hint: you only really need to prove symmetry. What can you say if xRy? Note that you also know yRy. – Wojowu Nov 2 '15 at 17:57

1) reflexive + cyclic => equivalence

a) reflexive => reflix

b) reflexive + cyclic => symmetry

xRy => xRy and yRy => yRx.

c) symmetry + cyclic => transitivity

xRy + yRz => zRx => xRz

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2) Equivalence => reflixe + cyclic

a) reflexive => reflexive

b) transitive + symmetry => cyclic

xRy and yRz => xRz => zRx.