Given a set {0,1,2}, is the relation below a equivalence relationship?
{(0, 0),(1, 1),(2, 2)}
It is self evident that this relation is reflexive, as it conforms this rule: ∀x ∈ A, x R x.
Looking at the symmetric rule : R is symmetric if, and only if, ∀x, y ∈ A, if x R y then y R x and the transitive rule R is transitive if, and only if, ∀x, y, z ∈ A, if x R y and y R z then x R z.
I am still confuse on how those 2 rules suit the relation above, please help?
Symmetric:
$R\subset A\times A$ is symmetric if $\forall x,y\in A, (x,y)\in R\implies (y,x)\in R $ $\tag 1$
Its negation will be: $R$ is not symmetric if $\exists x,y\in A$ such that $(x,y)\in R$ and $(y,x)\notin R $ $\tag 2$
Note that $(2)$ is not true for your $R$ and hence its negation must be true that is $R$ is symmetric.
Transitivity:
$R$ is transitive if $\forall x,y,z \in A, xRy \land yRz\implies xRz$ $\tag 3$
Its negation will be: $\exists x,y,z\in A, (xRy\land yRz)$ and $(x\not R z)$ $\tag 4$
Note that $(4)$ is not true for your $R$ and hence its negation must be true that is $R$ is transitive.