Introduction to Abstract Algebra, Solutions Manual, 4th Edition

2.1 Binary Operations

1.

a. This is not commutative: 1 ∗ 2 = − 1 while 2 ∗ 1 = 1. It is not associative: (2 ∗ 1) ∗ 3 = 1 ∗ 3 = − 2, while 2 ∗ (1 ∗ 3) = 2 ∗ (− 2) = 4. There is no unity: If e ∗ a = a for all a, then e − a = a so e = 2a for all a. This is impossible.

c. This is commutative: a ∗ b = a + b − ab = b + a − ba = b ∗ a. It is associative:

and, similarly, this equals (a ∗ b) ∗ c. The unity is 0:

Every a ≠ 1 has an inverse

e. This is not commutative (p, q) ∗ (p′, q′) = (p, q′) while (p′, q′) ∗ (p, q) = (p′, q) . It is associative:

There is no unity: If (a, b) ∗ (p, q) = (p, q) for all p, q, then a = p for all p.

g. This is commutative: gcd (n, m) = gcd (m, n). It is associative: Write d = gcd (k, m), d′ = gcd (m, n). Then d₁ = (k ∗ m) ∗ n = gcd (d, n), so d₁ d and d₁ n. But then, d₁ k, d₁ m and d₁ n. It follows that d₁ k and d₁ d′, so d₁ gcd (k, d′) = k ∗ (m ∗ n