0
$\begingroup$

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

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).

  • 0
    @Berci. You're right of course, $|$ is indeed a partial order, but look at the source of the original.2012-10-15

1 Answers 1

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 $4$.

  • 0
    For future reference: `\nmid` produces a significantly better result than `\not|`: $\nmid$ vs. $\not|$.2013-06-11