Does $R=\{(x,y) \in \mathbb{Z}\times\mathbb{Z} : 3|(x+y)\}$ define an equivalence relation?

Question

Given $R=\{(x,y) \in \mathbb{Z}\times\mathbb{Z} : 3|(x+y)\}$, Is $R$ reflexive? Is $R$ symmetric? Is $R$ transitive?

• Reflexivity: Could $(1,1)$ be a counter-example because $3\nmid(1+1)$?

• Symmetry: addition is commutative, so if $3∣(x+y)$, then $3∣(y+x)$ because $x+y=y+x$.

• Transitivity: I can't come up with a counter-example so I assume it must be true, but how do I prove that it is true? Given that $3∣(x+y)$ and $3∣(y+z)$, does $3∣(x+z)$?

Answer 1

Your example works fine to show that the relation is not reflexive.

As to transitivity, a proof approach might notice that if 3 divides both $x+y$ and $y+z$, then it divides $(x+y)-(y+z) = x-z$. However, we want $3$ to divide $x+z$ instead, so it is likely that it might be a good idea to look harder for counterexamples.

Answer 2

• For the reflexivity $(1,1)$ is a good counter example
• For the symmetry your reasoning is fine and correct
• For the transitivity it's false $x=z=1$ $y=2$