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prove that $a\mid b$ is not a partial order on integers $\mathbb{Z}$

I'm really lost how should I prove that?

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    @Michael, I thought you were referring to this one: http://www.youtube.com/watch?v=IIAdHEwiAy82013-05-13

2 Answers 2

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This is not antisymmetric.

For instance $-1 \mid 1$ and $1 \mid -1$ but $1 \neq -1$.

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If $\sim$ is a partial order, then

$\tag 1 a\sim a$ for every $a$ and $\tag 2 a\sim b,b\sim c \implies a\sim c $

and finally $\tag 3 a\sim b, b\sim a \implies a=b$

Which one of these three properties fail? And which two hold?

Note however that $\;\; \mid \;\;$ is a partial order on $\Bbb N$.