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$A=\{2,3,4,6,8,12\},\ x,y \in A, x\le y \leftrightarrow x^2 \mid y$

Is this a partial order and draw the Hasse diagram.

I know to be a partial order it needs to be reflexive, anti symmetric and transitive. I have the solutions to this problem but I do not seem to understand why it is not reflexive and how to draw the Hasse diagram for $A$.

Solution: not a partial order( since not reflexive, not anti symmetric, and transitive).

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I edited your post. Check whether the edited version represents what you intended, since the original could have been interpreted as in my edit or as $xx\;|\;y$. –  Rick Decker Oct 15 '12 at 15:39
    
But.. plainly $|$ is a partial order, I think indeed $x^2|y$ was meant.. –  Berci Oct 15 '12 at 15:43
    
@Berci. You're right of course, $|$ is indeed a partial order, but look at the source of the original. –  Rick Decker Oct 15 '12 at 15:47

1 Answer 1

It is not reflexive because for example, putting $x:=y:=2\in A$, they won't be in relation to each other as $2\cdot 2 \nmid 2$. Same for anti-symmetric, use $2$ and $4$.

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For future reference: \nmid produces a significantly better result than \not|: $\nmid$ vs. $\not|$. –  Lord_Farin Jun 11 '13 at 19:17

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