Given by Archimedean ordering there is nartual so that . Since by Archimedean ordering there is natural so that thus . Let then . Q.E.D.2. Prove that if a > 0, there is an integer n such that 1/n < a < n.
Assume not, then that means (by definition). Thus by Archimedian ordering with there is natural so that thus which is a contradiction. Q.E.D.3. For real numbers a and b, prove that if a <= b + 1/n for all positive integers n, then a <= b.
The set (non-empty) has an upper bound, i.e. . Argue that this is the least upper bound.4. For each real number a, let the set S_a be the set of all rational numbers less than a. Prove that for each real number a, sup S_a = a.