a + (-a) = (-a) + a = 0. * **Commutativity:** For all a, b ∈ R, a + b = b + a. 2. **(R, ⋅) is a monoid:** * **Closure:** For all a, b ∈ R, a ⋅ b ∈ R. * **Associativity:** For all a, b, c ∈ R, (a ⋅ b) ⋅ c = a ⋅ (b ⋅ c). * **Identity:** There exists an element 1 ∈ R such that for all a ∈ R, a ⋅ 1 = 1 ⋅ a = a. 3. **Distributive laws:** For all a, b, c ∈ R: * a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c) (left distributive law) * (a + b) ⋅ c = (a ⋅ c) + (b ⋅ c) (right distributive law) ## Showing that Positive Integers are not a Ring Let R be the set of positive integers, denoted as ℤ⁺ = {1, 2, 3, ...}. We consider the usual addition (+) and multiplication (⋅) operations. We will show that (ℤ⁺, +, ⋅) is not a ring by demonstrating that (ℤ⁺, +) is *not* an abelian group. Specifically, the axiom requiring the existence of an additive inverse fails. For any positive integer *a* ∈ ℤ⁺, there is no element -a ∈ ℤ⁺ such that a + (-a) = 0. The additive inverse of a positive integer is its negative counterpart, which is not a positive integer. For example, the additive inverse of 5 is -5, and -5 ∉ ℤ⁺. Since (ℤ⁺, +) is not an abelian group, the first condition for a ring is not satisfied. Therefore, the set of positive integers with respect to standard addition and multiplication is **not** a ring. 29:T461,To show that (Q⁺, *) forms a group, we need to verify four properties: closure, associativity, the existence of an identity element, and the existence of inverses for every element. **1. Closure:** Let a, b ∈ Q⁺. Then a * b = ab². Since a and b are positive rational numbers, their product ab² is also a positive rational number. Therefore, a * b ∈ Q⁺, and the operation * is closed on Q⁺. **2. Associativity:** Let a, b, c ∈ Q⁺. We need to show that (a * b) * c = a * (b * c). (a * b) * c = (ab²) * c = (ab²)c²