Question

Prove that (Z,⋅,1) is a commutative monoid using the formal definition of multiplication from the video. Furthermore, prove that multiplication distributes over addition in the integers. Hint: You first need to show that givenu,v∈N×N,d(uv) =d(u)d(v) (this is similar to how, in the video, we proved d(u+v)=d(u)+d(v)).Doing this will make your life dramatically easier in the proof. Hint 2: Let u _0=(n,n)n∈N. What is d(vu 0) ? Ploof n=m+ ^kLet w=(m,n) and u _0=(m,m). Th, d(w+n _0)=d(m+m,n+n)= d(m+m ,(m+k)+m )=d(m+m⋅,m+(k+m ' ))== d(m+n,m+(m+k))−

Answer 1

The set of integers with multiplication forms a commutative monoid because it exhibits closure, associativity, has an identity element (1), and is commutative. Additionally, multiplication inherently distributes over addition in the integers, following the distributive property.

Step-by-step explanation:

To prove that (Z,⋅,1) is a commutative monoid, we need to address three primary properties of a monoid: closure, associativity, and the existence of an identity element within the context of multiplication on integers (Z). To be commutative in addition, the operation must be such that the order of the operands does not affect the result.

Closure: For all integers a and b, the product ab is also an integer, which is the definition of closure under multiplication.

Associativity: For all integers a, b, and c, (ab)c = a(bc), which is true by the associativity of multiplication over integers.

Identity element: The integer 1 serves as the identity element since for every integer a, a⋅1 = 1⋅a = a, fulfilling the necessary condition for a monoid.

Since multiplication of integers is inherently commutative—ab = ba—we can conclude that (Z,⋅,1) is indeed a commutative monoid.

To prove that multiplication distributes over addition in the integers, we must show that for all integers a, b, and c, a(b+c) = ab + ac. This follows directly from the distributive property of multiplication over addition in the set of integers.