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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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You should start by writing down what "|" means, and what "partial order" means. –  Chris Eagle Oct 29 '12 at 22:49
    
I changed $a|b$ to $a\mid b$, coded as a \mid b in $\LaTeX$. The spacing is different. If you're caught defacing monuments with graffiti, then the same police office who will give you a stern lecture about correct spelling will tell you that the latter form is correct. –  Michael Hardy Oct 30 '12 at 3:00
    
@Michael, Romanes eunt domus. –  Gerry Myerson May 12 '13 at 6:59
    
In about the second half of this video a cop lecture a graffiti artist about spelling: youtube.com/watch?v=gZg7tQ7KykM –  Michael Hardy May 12 '13 at 17:57
    
@Michael, I thought you were referring to this one: youtube.com/watch?v=IIAdHEwiAy8 –  Gerry Myerson May 13 '13 at 10:36
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2 Answers

up vote 4 down vote accepted

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 $\Bbb N$.

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