I feel like I am being too brief and maybe incorrect on my proof by contradiction for transitivity/antisymmetry. So is this proof flawed in any way?
A relation R on the set of positive integers is defined by $x \geq y \rightarrow (x,y) \in R$.
Note that $R \subseteq \mathbb{Z}^+\times\mathbb{Z}^+$
$R$ is reflexive because $\forall x,y \in \mathbb{Z}^+$ if $x = y$ then $(x,y) \in R$,
$\therefore R$ is reflexive.
$R$ is not symmetric, consider $(5,3) \in R$ because $5 \geq 3$, and $(3,5) \notin R$ because $3 \ngeq 5$
$\therefore R$ is not symmetric
$R$ is antisymmetric, because if $(x,y) \in R$ and $(y,x) \in R$ then $x = y$. We show this by contradiction, assume $(x,y) \in R$ and $(y,x) \in R$ and $x \neq y$. Then $x > y$ and $y > x$. $\implies\impliedby$
So then $x=y$
$\therefore R$ antisymmetric.
EDIT: (Thanks Mohan)
$R$ is also transitive: Assume $(x,y) \in R$ and $(y,z) \in R$. Then $x \geq y$ and $y \geq z$. Then $x \geq z$.$\square$
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