Confirming my understanding in determining if a relation is reflexive, symmetric, or transitive

Question

I think I have a grasp on how to determine if a relation is reflexive, symmetric, or transitive. Just to make sure I understand it correctly, if I have the following relation:

for $(a,b) \in \mathbb{N}$ we say that $(a,b) \in R$ if $a + b$ is divisible by 3

Then the relation is not reflexive because even though $(3,3)$ exists in the relation and is reflexive, there are many other pairs such as $(1,2)$ and $(2,7)$ that are not reflexive. Is this correct, or does reflexitivity (is that the right word?) only require one pair to be true?

Taking the example just above, it would not be transitive either because $(1,2)$ and $(2,7)$ do not imply $(1,7)$. Same as before, is this correct or does it only require one pair?

The function is symmetric because the elements in each pair can be flipped around, ex: $(1,2)$ is true, and $(2,1)$ is true. This is true, correct?

Answer 1

The relation is not reflexive because $(1,1)$ is not in $R$ because $3$ is not a divisor of $2$. It would have been a reflexive relation if $\forall a,(a,a) \in R$, so you counter examples are false.

Your counter example for transitivity is true. Note: To prove something is false you only need one counter example.

It is indeed symetric. A more formal way to prove it would be : $\forall a,b$ if $(a,b) \in R$ then $3$ divides $a+b$, so $3$ divides $b+a$, so $(b,a) \in R$.