Let A = {a, b} and list the four elements of the power set P (A). We consider the operations + to be ∪, · to be ∩, and complement to be set complement. Consider 1 to be A and 0 to be ∅.
a). Explain why the description above defines a Boolean algebra.
b). Find two elements x,y in P(A) such that xy =0, x̸=0 andy̸=0.
a). To explain why the given description defines a Boolean algebra, we need to show that it satisfies the three defining properties of a Boolean algebra: closure, complementation, and distributivity.
1. Closure: In this context, closure means that for any two elements x, y in P(A), the operations ∪ (union), ∩ (intersection), and set complementation will also produce elements in P(A).
- Let's say x and y are subsets of A. The union of x and y, denoted as x ∪ y, is the set that contains all elements that are in either x or y or both. Since x and y are subsets of A, the union x ∪ y will also be a subset of A.
- Similarly, the intersection of x and y, denoted as x ∩ y, is the set that contains all elements that are in both x and y. Again, since x and y are subsets of A, the intersection x ∩ y will be a subset of A.
- The complement of x, denoted as x̸, is the set that contains all elements in A that are not in x. Since x is a subset of A, the complement x̸ will be a subset of A.
Therefore, the operations of union, intersection, and complementation maintain closure in this context, satisfying the first property of a Boolean algebra.
2. Complementation: Complementation means that for every element x in P(A), there exists an element x̸ in P(A) such that x ∪ x̸ = 1 and x ∩ x̸ = 0.
- The element 1 in this context represents the set A, which contains all elements of the universal set. The union of any subset x of A and its complement x̸ will result in A itself (1), as x̸ contains all elements not in x.
- The element 0 represents the empty set ∅. The intersection of any subset x of A and its complement x̸ will result in the empty set (0), as x̸ contains all elements not in x.
Therefore, for every element x in P(A), there exists a complement x̸ in P(A) satisfying the second property of a Boolean algebra.
3. Distributivity: Distributivity refers to the property that for any elements x, y, and z in P(A), the following equalities hold: x ∩ (y ∪ z) = (x ∩ y) ∪ (x ∩ z) and x ∪ (y ∩ z) = (x ∪ y) ∩ (x ∪ z).
- Let's verify the first equality. Suppose x, y, and z are subsets of A. The left side of the equation represents the intersection of x with the union of y and z. This means taking the elements that are common to x, y, and z together. The right side of the equation represents the union of the intersection of x with y and the intersection of x with z. This also means taking the elements that are common to x, y, and z together. By the definition of sets, these two expressions are equal, satisfying the first distributive property.
- Similarly, we can verify the second distributive property in the same way.
Therefore, the given description satisfies all three defining properties of a Boolean algebra - closure, complementation, and distributivity.
b). To find two elements x and y in P(A) such that xy = 0, x̸ ≠ 0, and y̸ ≠ 0, we need to consider the possible subsets of A and their corresponding operations.
Given that A = {a, b}, the power set P(A) consists of the following subsets: ∅ (empty set), {a}, {b}, and {a, b}.
To find the desired elements, we can evaluate the operations directly:
- Let x = {a} and y = {b}
- xy represents the intersection of x and y. In this case, {a} ∩ {b} = ∅, which is the empty set (0).
- x̸ represents the complement of x. In this case, x̸ = {b}.
- y̸ represents the complement of y. In this case, y̸ = {a}.
Therefore, x and y are two elements in P(A) that satisfy xy = 0, x̸ ≠ 0, and y̸ ≠ 0.