Well ordered set.....confusing I read the wiki article on it, but still don't get what it is. I have a question regarding it: 
A set is well ordered if every nonempty subset of this set has a least element. Determine which of these sets are well ordered.
Answer:
a)  the set of positive integers
b)  the set of integers less than -100
c)       the set of positive rationals
d)  the set of positive rationals with denominator less than 200
I guess option a and c are correct because when we inverse them, they are well ordered. Correct me if I am wrong.
 A: A. is correct; it's one of the paradigm wellordered sets.
C. is not correct. This has no least member, since for any $x>0$ one can take $x/2$ and get a smaller number greater than $0$, and hence a positive rational.
I'm not sure what you mean by "inversing", but the inverse order has no direct bearing on whether something is wellordered. The only thing that matters is if there are any subsets without a least member. For instance, the integers are not wellordered because $\mathbb{Z}$ itself has no least member; the negative numbers keep going down forever. Likewise the positive rationals keep getting smaller and smaller forever.
To show that a set is not wellordered, you just need to show a non-empty subset such that for every $x$, there's a $y$ with $y<x$. To show that it is wellordered,  you need to show that there are no such sets.
A: To test if a set is well ordered, ask yourself first this question:  Does it have a least element?  Already (b) and (c) fail the test, since (b) has all the negative integers (which has no least element) and (c) has positive fractions $1/2, 1/3, 1/4, \ldots$ which have limit $0$ and yet $0$ is not in the set.
If the set passes this first test, then it has a chance of being well-ordered.  However we must also consider all subsets of the set, and apply the same question to each of them.  For example, if we consider the set of nonnegative rationals -- this set has a least element (namely $0$), yet a subset of this is the set (b) which fails the least-element test, and hence the set of nonnegative rationals is not well-ordered (at least in the usual way $\le$).
