# Can someone point out why these directed graphs aren't equivalence relations?

As far as I can tell, these two directed graphs are reflexive, symmetric and transitive.

In the first one we have c ~ a and a ~ d but not c ~ d, so the relation is not transitive. A similar problem exists for the other graph. See if you can find the issue yourself.

Then ask yourself what the graph of an equivalence relation looks like in general.

• I thought its transitive if you can leave any point and be able to get back to that point, ill reread the definition. – Luke D Feb 4 at 5:13
• @LukeD No, transitive is "Whenever there's an edge $a\to b$ and an edge $b\to c$, there's also an edge $a\to b$." – David Richerby Feb 4 at 10:44
• Ah ok that makes a lot of sense. Thank you. The reason why the second one isn't transitive is because there isn't diagonal relations, as in there's no a->c so c->d and a->d don't complete its transitivity. – Luke D Feb 5 at 3:56