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$\begingroup$

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

I'm really lost how should I prove that?

  • 1
    You should start by writing down what "|" means, and what "partial order" means.2012-10-29
  • 0
    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.2012-10-30
  • 0
    @Michael, Romanes eunt domus.2013-05-12
  • 0
    In about the second half of this video a cop lecture a graffiti artist about spelling: http://www.youtube.com/watch?v=gZg7tQ7KykM2013-05-12
  • 0
    @Michael, I thought you were referring to this one: http://www.youtube.com/watch?v=IIAdHEwiAy82013-05-13

2 Answers 2

4

This is not antisymmetric.

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

3

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 $\Bbb N$.