10 1. THE CONSTRUCTION OF REAL AND COMPLEX NUMBERS Also recall that the absolute value on Q satisfies the following three proper- ties. (1) For any a Q, |a| 0, and |a| = 0 if and only if a = 0. (2) For any a, b Q, |ab| = |a||b|. (3) For any a, b Q, |a + b| |a| + |b| (triangle inequality). Exercise 1.5.1. Show that, for any a, b Q, we have ||a|−|b|| |a−b|. Definition 1.5.2. A sequence (ak)k∈N of rational numbers is a Cauchy sequence in Q if, given any rational number r 0, there exists an integer N such that if n, m N, then |an am| r. Definition 1.5.3. A sequence (ak)k∈N converges in Q to a Q if, given any rational number r 0, there exists an integer N such that if n N, then |an a| r. Sometimes, we just say that the sequence (ak)k∈N converges in Q without mentioning the limit a. Exercise 1.5.4. If a sequence (ak)k∈N converges in Q, show that (ak)k∈N is a Cauchy sequence in Q. In addition, show also that the limit a of a convergent sequence is unique. Definition 1.5.5. Let (ak)k∈N be a sequence of rational numbers. We say that (ak)k∈N is a bounded sequence if the set {ak | k N} is a bounded set in Q. Lemma 1.5.6. Let (ak)k∈N be a Cauchy sequence of rational numbers. Then (ak)k∈N is a bounded sequence. Proof. Let (ak)k∈N be a Cauchy sequence of rational numbers. Pick N N such that |an am| 1 for n, m N. Then |an aN| 1 for all n N, so that |an| 1 + |aN| for all n N. Let M be the max of |a1|, |a2|,..., |aN−1|, 1 + |aN|. Then (|ak|)k∈N is bounded by M. Let C denote the set of all Cauchy sequences of rational numbers. We define addition and multiplication of Cauchy sequences termwise that is, (ak)k∈N + (bk)k∈N = (ak + bk)k∈N and (ak)k∈N(bk)k∈N = (akbk)k∈N. Exercise 1.5.7. Show that the sum of two Cauchy sequences in Q is a Cauchy sequence in Q. Theorem 1.5.8. The product of two Cauchy sequences in Q is a Cauchy sequence in Q. Proof. Let (ak)k∈N and (bk)k∈N be Cauchy sequences in Q. Then |anbn ambm| = |anbn anbm + anbm ambm| |an||bn bm| + |bm||an am| A|bn bm| + B|an am|, where A and B are upper bounds for the sequences (|ak|)k∈N and (|bk|)k∈N. Since (ak)k∈N and (bk)k∈N are Cauchy sequences, the theorem now follows.
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