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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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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 – 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
up vote 1 down vote accepted

It looks like this question has been dealt with in the comments, so if you're happy with it now, please accept this community wiki answer so that it doesn't stay 'unanswered'. (Community wiki means nobody gets reputation from it.)

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