Drawing a lattice for a set partially ordered by divisibility I am a little bit confused as to how I would draw a lattice for the set $S= \{1,2,3,4,6,9,12,18\} $when it is partially ordered by divisibility. 
I was able to prove that it has a greatest lower bound, and a least upper bound for all elements $x,y \in S$. I believe my proof is correct, and by having these two bounds, the set is therefore a lattice. 
I am just a bit confused as to how I would draw it. 
 A: Just how are you confused about drawing the lattice? Here's how I did it. I started by putting 1 at the bottom, since it's the least element of the lattice, it divides everything. 
Since 2 and 3 are divisible by 1, but not by each other, I put 2 one cm northwest of 1, I put 3 one cm northeast of 1, and draw lines from 2 to 1 and from 3 to 1.
Since 6 is divisible by everything so far, I put 6 one cm NE of 2 (and one cm NW of 3) and draw lines from 6 to 2 and from 6 to 3. Of course I don't draw a line from 6 to 1.
I now have a diamond, with 1 at the bottom, 2 and 3 on the sides, 6 at the top.
Next I look at 4: divisible by 1 and 2, not divisible by 3 or 6. So I put 4 one cm NW of 2, and draw a line from 4 to 2.
12 is divisible by everything so far. I put 12 one cm NE of 4 (and NW of 6), and draw lines from 12 to 4 and from 12 to 6.
9 is divisible by 1 and 3, but not divisible by 2, 4, 6, or 12. I put 9 one cm NE of 3, and draw a line from 9 to 3.
18 is divisible by 6 and 9, and therefore also by 1, 2, and 3; but 18 is not divisible by 4 or 12. I put 18 one cm NE of 6 (NW of 9) and draw lines from 18 to 6 and 9.
That's the diagram. I'd draw it for you if I knew how. 
Looking at the diagram, it's easy to see that it's a lower semilattice, meaning that any two elements have a greatest lower bound.
It's also easy to see that it's not an upper semilattice (and therefore not a lattice); the pairs {4,9}, {4,18}, {9,12}, and {12,18} do not have least upper bounds.
By the way, what did you think was the least upper bound of 12 and 18? Maybe there was a typo in your question, and 36 was supposed to be an element of S?
A: "Ordered by divisibility" means that you have $y$ somewhere above $x$ in the lattice whenever $y$ is a multiple of $x$.  So for example, there should be an upward path in the lattice from 4 to 12, because 12 is a multiple of 4.  But there should not be a path upward from 3 to 4, or from 4 to 3, because neither is a multiple of the other.
A good way to proceed is:


*

*identify the maximal elements of the lattice, which are the elements that have nothing else above them.

*Put those in a top row. 

*Then cross those out from your list of elements 

*Find the next highest elements and put them in a row underneath the maximal elements.  

*Draw lines from those to the appropriate maximal elements. 

*Repeat this until you have added all the elements in the proper places.


Does this help?
