Introduction to Abstract Algebra, Solutions Manual, 4th Edition

3.5 Ordered Integral Domains

1.

a. (b + c) − (a + c) = b − a R⁺.

c. (− a) − (− b) = b − a R⁺.

e. Since a < b and c > 0, ac < bc by Lemma 1. Similarly bc < bd because b > 0, again by Lemma 1. Finally then ac < bd by Lemma 1.

2.

a. a ≤ a because a = a.

c. Let a ≤ b and b ≤ c. If a = b then b ≤ c is a ≤ c; if b = c then a ≤ b is a ≤ c. Otherwise a < b and b < c so a < c by Lemma 1.

3.

a. If a ≥ 0 then |a| = a ≥ 0. If a < 0 then −a = 0 − a R⁺ so |a| = − a > 0.

c. If either a = 0 or b = 0 then |ab| = 0 = |a||b|.

If a > 0, b > 0, then ab > 0 so |ab| = ab = |a||b|.

If a > 0, b < 0, then ab < 0 so |ab| = − ab = a(− b) = |a||b|.

If a < 0, b > 0, as above.

If a < 0, b < 0 then ab > 0 so |ab| = ab = (− a)(− b) = |a||b|.

4. If a is such that b ≤ a for all b then a + 1 ≤ a, whence 1 ≤ 0, a contradiction.

5.

a. and would be a contradiction.

7. and , so ru R⁺ and sv R⁺. Thus (using ...