[Exercise 1.8 (ii) concerned1^. We repeat that exercise but this time we work in Ж.] Show, by means of an example, that, if an —> a and an > b for all n, it does not follow that a > b. (In other words, taking limits may destroy strict inequality.) Does it follow that a > b? Give reasons.
(i) Show that there does not exist а К бК with К > п for all n 6Z by using Theorem 1.21. (ii) Show the same result directly from the fundamental axiom.
If x (=. Ж, then, given any e > 0, there exists а у 6 Q such that \x — y\ < e.Thus the rationale form a kind of skeleton for the reals. (We say that the rationale are dense in the reals.) The reader will probably already be acquainted with the following definition.
The reader may be interested to see an ordered field containing Z which does not satisfy the axiom of Archimedes. We start by considering polynomials P(X) = ^n=0 anXn with real coefficients an and form the set К of rational functions P(X)/Q(X) where P and Q are polynomials and Q is not the zero polynomial (that is Q(X) = J2m=o bmXm with Ъм ф 0 for some M).
It seems, at first sight, that decimal expansion gives a natural way of treating real numbers. It is not impossible to do things in this way, but there are problems. Here is one of them. Let 0 < a, b < 1 and с = ab. If Ylj=i a-jlO~J —»• a, Ylj=i bjlO~J —»• b, and Ylj=i cj^~J ~^ c fin^ Cj in terms of the various a& and b^. (The reader is invited to reflect on this problem rather than solve it. Indeed, one question is 'what would constitute a nice solution?')