# Is this a correct proof for this relation?

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 $x \geq z$.$\square$

-
Tip: Use \times to get $\times$ instead of saying "cross". – Jim Feb 26 '13 at 8:23
@Jim Thanks, I also cant find a good contradiction symbol. – Leonardo Feb 26 '13 at 8:25
detexify.kirelabs.org/classify.html – Jim Feb 26 '13 at 8:25
@Jim I was actually just using that, and that is where I found \lightning but alas it did not work. – Leonardo Feb 26 '13 at 8:26
@Leonardo $\mathbb{Z}\times\mathbb{Z}$ is more readable than $\mathbb{Z}cross\mathbb{Z}$, but too much symbols is also bad. For example, I would write "therefore" or "contradiction" instead of $\therefore$ or some lightning arrows. Your question and its answers should be available to wider audience, including those who do not use those special symbols, e.g. $\times$ is common, but $\therefore$ is not. – dtldarek Feb 26 '13 at 8:44