Prove that $\mathbf{R} = \{ (x,y) \in \mathbb{N^2} : x \ge y \}$ is an equivalence relation

Question

For the binary relation R over $\mathbb{N}$ described below,

establish whether R is a reflexive, symmetric or transitive relation.

$\mathbf{R} = \{ (x,y) \in \mathbb{N^2} : x \ge y \}$

How can I prove that?

my try:

$$ \text{reflexive}: x \ge x \ \text{(true)}\\ \text{symmetric}: x \ge y \iff y\ge x \ \text{(??That's not true)} \\ \text{transitive}: x \ge y, \ y \ge z \implies x \ge z \ \text{(true)} $$

symmetric propriety doesn't seem true if $x \neq y$.

Answer 1

It is not an equivalence relation because, as you see in the comments, symmetry fails (just pick the point $(2,1)$). The $\ge$ relation on, say, the natural numbers is, instead, satisfying the antisymmetry property.