A dense ordering on a set is one such that for any two elements there's another element between them, i.e for all a<b in X there exists c such that a<c<b.

So the rationals are densely ordered, between any two rationals p and q we have (p+q)/2 which is also rational. The reals are also densely ordered. The integers, though, are not: there is no integer between 1 and 2.

This seems obvious, the integers are discrete and only come in 'lumps', while the rationals can be as small as you like. However, this intuition is wrong, it is possible to put a dense ordering on the integers, and a non-dense one on the rationals!

This is how we do it: the rationals are countable, so we can list them q_{1}, q_{2},... Then we just define the ordering "{" by q_{i} { q_{j} if i<j. Then q_{n} { q_{n+1}, but there is no rational p such that q_{n} { p { q_{n+1}. So this is a non-dense ordering.

Similarly, we define "«" on the integers by i « j if q_{i} < q_{j}. Then for any i « j there's always a rational q_{k} with q_{i} < q_{k} < q_{j}, so i « k « j. So this is a dense ordering.

This illustrates that all countable sets have the same level of 'discreteness', you have to go to an uncountable set (like the reals) to get continuity.