Is $R$ symmetric and transitive?

Question

Let $R$ be a relation defined on the set $Z$ by $aRb$ if $a \neq b$. Is it true that $R$ is symmetric and transitive? If not, give an example.

1. It is symmetric because $a \neq b$ and $b \neq a$.
2. It is transitive because $a\neq b$ and $b \neq c$ so therefore $a\neq c$.

Answer 1

Hint:

Do $2\ne 3$ and $3\ne 2$ imply that $2\ne 2$?

Answer 2

A relation is symmetric if and only if: $$\mathbf{\forall a \forall a ((a,b)\in R \implies (b,a)\in R)}$$ where the domain of $a$ and $b$ is understood to be $\mathbb {Z}$. For transitivity you have: $$ \mathbf{\forall a \forall b \forall c ((a,b)\in R \land (b,c)\in R \implies (a,c)\in R)}$$ where again the domain of discourse is understood be $\mathbb{Z}$. From this rigorous development of the definitions you can easily see that the relation posited is symmetric since $a\neq b \implies b\neq a$. However, it is not transitive since you can have $a\neq b $ and $b \neq c$ but $a = c$ and hence the implication fails.