# Is the relation $R:=\{(1,2),(1,3)\}$ transitive on $M=\{1,2,3\}$ with $R\subseteq M\times M$?

Is the relation $$R:=\{(1,2),(1,3)\}$$ transitive on $$M=\{1,2,3\}$$ with $$R\subseteq M\times M$$?

I think it's transitive, because we don't have elements that satisfy $$xRy \land yRz$$ and therefore $$\forall x,y,z \in M: xRy \land yRz \implies xRz$$ is always true. Is this right?

• Yes, that’s correct. In this kind of situation you can say that it’s vacuously true. Oct 19, 2020 at 19:12
• Alternatively: There are no such $x,y,z$ of $M$ where $x\operatorname R y\wedge y\operatorname R z\wedge \neg(x\operatorname R z)$. Oct 20, 2020 at 0:13