Divisibility relations on $\Bbb N, \Bbb Z^*$, and properties thereof I have two problems for which I need to find whether it is reflexive, irreflexive, symmetric, antisymmetric, or transitive.


*

*The relation $T$ on $\mathbb{N}$ as defined by $$aTb \iff a \mid b$$

*The relation $U$ on the set $\mathbb{Z}^*$ is defined as $$aUb \iff a \mid b$$

For #1, the way I worked it out was:

*

*Reflexive: Yes, because $a  \mid  a$ for any positive integer $a$


*Irreflexive: No, see above


*Symmetry: No, because $a \mid b$ does not necessarily mean $b \mid a$ example, $4 \mid 12$ but $12$ does not divide $4$


*Antisymmetric: Yes, because if $a  \mid  b$ and $b \mid a$ then $a$ must equal $b$


*Transitive: Yes, because if $a \mid b$ and $b \mid c$ then $a \mid c$
For question #2, however I am lost because the answer is reflexive and transitive.
I get that it is reflexive and transitive and not symmetric for the same reasons as #1 but I don't understand why #2 is not antisymmetric but #1 is. I am guessing that it has something to do with #2 being a relation of $\mathbb{Z}^*$, but that just means a non-zero integer. Can someone help me understand why #1 is antisymmetric and #2 isn't antisymmetric?
 A: It is always best to reference the definitions. Namely, a typical definition is
$$a \mid b \iff \exists k \in \Bbb Z \text{ such that } b = ak$$
Suppose we consider this relation on $\Bbb N$. Let's investigate the properties.

*

*Reflexivity: Trivially true; take $k=1$.

*Irreflexive: Untrue, as $1 \mid 1$ ($k=1$).

*Symmetry: Untrue. $2 \mid 4$ ($k=2$) but $4 \not \mid 2$ ($k = 1/2 \not \in \Bbb Z$).

*Antisymmetry: True. If $a \mid b$ and $b \mid a$, then there are $k,\ell \in \Bbb Z$ such that
$$b = ak \qquad a = b\ell$$
Substitute the latter into the former, and  vice versa, and we have
$$b = b\ell k \qquad a = ak\ell$$
Thus, $\ell k = 1$. The only integers that ensure this are $\ell = k = +1$ or $\ell = k = -1$. Since we're working in $\Bbb N$, though, $a,b > 0$, and thus $\ell = k = +1$. Thus, $a = b$.

*Transitivity: True. Suppose $a \mid b$ and $b \mid c$. Then $\exists p,q \in \Bbb Z$ such that
$$b = ap \qquad c = bq$$
Substitute the former into the latter, and we have
$$c = apq$$
Since $p,q \in \Bbb Z \implies pq \in \Bbb Z$, then $a \mid c$, giving transitivity.

What if we consider things on $\Bbb Z^*$ instead?

*

*Reflexivity: Still holds for the same reason.

*Irreflexive: Same deal - not irreflexive.

*Symmetry: Again, same deal - not symmetric. It doesn't hold for $\Bbb N \subseteq \Bbb Z$ after all.

*Antisymmetry: Now something has changed. Before, we had to completely neglect the case of $k = \ell = -1$ since that would result in $a$ or $b$ not being in $\Bbb N$. However, now that case is permissible, and gives us cases like $2 \mid -2$ and $-2 \mid 2$, but $2 \ne -2$. (Just take $k = -1$ for each divisibility relation.)

*Transitivity: Still holds, nothing fundamentally changed here.

In short, the fundamental difference is that negative integers can divide positive integers, and vice versa!

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.
