ANSWERS AND SOLUTIONS TO SELECTED EXERCISES^379
have b < a or a > b, as desired. The converse follows by reversing each of the
preceding steps. Next, we note that x < 0 o [(a, b)] < [(I, I)] o a + 1 <
b + 1 o a < b, where the final step follows from Exercise qa), Article 10.1, and
Theorem 8(g), Article 10.1. Finally, x = 0 - [(a, b)] = [(I, l)] e a + 1 =
b + 1 9 a = b. Note that the last equivalence follows from the well-definedness
of addition (in the direction e) and additive cancellation (for *). 0
Article 10.3
- (a) Proof Let {p,: rn = 1,2,.. .) be a Cauchy sequence of rational numbers. '
Corresponding to the positive rational number 1, there exists a positive integer
N such that Ip, - pNl < 1, whenever m 2 N. Since 1x1 - lyl I lx - yl for any
x and y, we have lpml - lpNl I lpm - pNI < 1, SO that Ip,( I 1 + (pNl for all
m 2 N. Letting B = max(lpll, lp,J,... , IP,-~(, 1 + (pNl), we have lpml < B, as
desired. 0