Introduction to Abstract Algebra, Solutions Manual, 4th Edition

0.1 Proofs

1.

a.

1. If n = 2k, k an integer, then n² = (2k)² = 4k² is a multiple of 4.

2. The converse is true: If n² is a multiple of 4 then n must be even because n² is odd when n is odd (Example 1).

c.

1. Verify: 2³ − 6 · 2² + 11 · 2 − 6 = 0 and 3³ − 6 · 3² + 11 · 3 − 6 = 0.

2. The converse is false: x = 1 is a counterexample. because

2.

a. Either n = 2k or n = 2k + 1, for some integer k. In the first case n² = 4k²; in the second n² = 4(k² + k) + 1.

c. If n = 3k, then n³ − n = 3(9k³ − k); if n = 3k + 1, then

if n = 3k + 2, then n³ − n = 3(9k³ + 18k² + 11k + 2).

3.

a.

1. If n is not odd, then n = 2k, k an integer, k ≥ 1, so n is not a prime.

2. The converse is false: n = 9 is a counterexample; it is odd but is not a prime.

c.

1. If then , that is a > b, contrary to the assumption.

2. The converse is true: If then , that is a ≤ b.

4.

a. If x > 0 and y > 0 assume . Squaring ...