Tell if the following order relation is total and find the minimum, maximum, minimals and maximals: $$\forall a,b \in\mathbb Z,\ \ a\ \rho\ b \iff a \leq b\ \text{ and }\ \pi(a) \subseteq \pi(b)$$ where $\pi$ is defined as follows: $\forall n \in\mathbb Z: \pi(n) = \{p \in P\ |\ \ p|n\}$ where $P$ is the set of positive prime numbers.
My attempt at solving this exercise:
- $\rho$ is not total. In fact, $2\ \not\rho\ 5$ and $5\ \not\rho\ 2$.
- I am confused because we are solving this exercise in $\mathbb Z$ so I guess $1$ can't be minimum because $-1\ \rho\ 1$. So $-1$ is minimum of $\rho$.
- I think there is maximum/maximals and it is $0$ because also negative numbers are in relation with $0$: $-1\ \rho\ 0$. Correct?
Thank you.
I think you have not yet got a hold of the difference between minimum and minimal and maximum/maximal. A minimum of a partially ordered set is an element smaller than all other elements. An element is minimal if there is no element that is strictly smaller than it.
For a totally ordered set, these are the same. But for a partially ordered set there can be many minimal elements which are not comparable to each other. In this case, there would be no minimum.
This is correct.
The number 1 is indeed not minimal, for the reason you said. $-1$ is minimal since nothing can be strictly smaller -- any other negative number will have some prime factors. -1 is not the minimum though since $-1 \!\!\not\operatorname{\!\rho} -2$.
0 is not larger than every other element so it is not the maximum. $1 \!\!\not\operatorname{\!\rho} 0$. So 0 might be maximal but it is not the maximum.
To complete this exercise, you'll want to have proof that you've found all of the maximal and minimal elements. You are on the right track to single out -1, 0, and 1 and pay careful attention to them.
Then you might want to ask yourself if a regular old number like 3 or 22 or -12 could be maximal or minimal.