prove that $a\mid b$ is not a partial order on integers $\mathbb{Z}$
I'm really lost how should I prove that?
prove that $a\mid b$ is not a partial order on integers $\mathbb{Z}$
I'm really lost how should I prove that?
This is not antisymmetric.
For instance $-1 \mid 1$ and $1 \mid -1$ but $1 \neq -1$.
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 $\Bbb N$.