**Re: Discrete Maths Sem 2 2016**
Part a) asked for a Hasse diagram but that's a bit hard to type up. So I'll just write down which lines I had going to what

2->4

2->6

4->12

6->12

6->54

9->27

9->36

12->36

12->60

27->54

36->72

So I've forgotten how to do this properly. My answer was no for g) and 12 for i) (same as back of the book), but it was more out of intuition than proper reasoning. Can someone please explain why they are the correct answers to jog my memory?

(Or rather what is LUB/GLB v.s. the set of upper bounds/lower bounds for a subset)

g) The upper bounds of {2, 9} are by definition numbers in the poset that are multiples of both 2 and 9. These are: 36, 54, 72. So U := {36, 54, 72} is the set of upper bounds of {2, 9} here.

If there is to be a

*least* upper bound of {2, 9}, by definition it must be a u in {36, 54, 72} such that u | v for any upper bound v. But clearly there is no such u (i.e. no element of U will divide all the elements in U). Hence there is no least upper bound.

i) The set of lower bounds of {60, 72} is by definition the set of all elements in the poset that divide both 60 and 72. This is L = {2, 4, 6, 12}. By definition, the GLB is the element m in L such that m is divisible by k for all k in L (if such an element m exists).

Clearly here, such an element does exist, namely 12. So this is the GLB we are after.

(Note that if a GLB or LUB exists, it is unique. You can easily show this as an exercise.)