Is a total order compatible with a partial order?

Question

I was given the following multipart problem.

Part 1: Consider the poset ({2,4,6,9,12,18,27,36,48,60,72},|), with the indicated integers and the divides relation. Find the following, if they exist; if they don’t exist, explain why they don’t.

Maximal and minimal elements? Max: 27, 48, 60, 72. Min: 2,9.

Least upper bound of {2,12}? 12.

Greatest lower bound of {60,72}? 12.

I drew out a Hasse diagram and think I solved those 3 questions correctly, although I would appreciate if someone could confirm.

Then comes the next part:

Suppose we impose the following total order on the set: 2,4,18,6,27,12,36,9,72,60,48. Is this compatible with the partial ordering of the divides relation? Why or why not?

I just don't even get what this is asking at all. I understand what a poset and total ordering are, but what does is it mean "is it compatible?". The only thing I can think of is that it's not actually a total order. Given the Hasse diagram, the order given doesn't seem to work. Any help would be appreciated, thanks.

Answer 1

This answer basically just confirms what it appears from the comments that you’ve already understood.

Let $\langle P,\le\rangle$ be a partial order, and let $\preceq$ be a linear (total) order on $P$. The order $\preceq$ is compatible with the partial order $\le$ if the following condition is satisfied:

for all $x,y\in P$, if $x\le y$, then $x\preceq y$.

In other words, $\le$ and $\preceq$ never assign opposite orders to any pair of elements of $P$; either they assign the same order, or $\le$ doesn’t assign any order to that pair.

In your problem, $\preceq$ is given by

$$2\prec4\prec18\prec6\prec27\prec12\prec36\prec9\prec72\prec60\prec48\;.$$

As you noted in the comments, this disagrees with the order $\le$ on the pairs $\{6,18\}$, $\{9,18\}$, and $\{9,27\}$, so it’s not compatible with the partial order $\le$.