# 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 d

_{1}= (k ∗ m) ∗ n = gcd (d, n), so d_{1}d and d_{1}n. But then, d_{1}k, d_{1}m and d_{1}n. It follows that d_{1}k and d_{1}d′, so d_{1}gcd (k, d′) = k ∗ (m ∗ nGet *Introduction to Abstract Algebra, Solutions Manual, 4th Edition* now with O’Reilly online learning.

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