Solution Found!
Refer to Exercise 2.4. Use the identities A = A S and S = B B and a distributive law to
Chapter 2, Problem 2.5(choose chapter or problem)
Refer to Exercise 2.4. Use the identities \(A=A \cap S\) and \(S=B \cup \bar{B}\) and a distributive law to prove that
a. \(A=(A \cap B) \cup(A \cap \bar{B})\).
b. If \(B \subset A\) then \(A=B \cup(A \cap \bar{B})\).
c. Further, show that \((A \cap B)\) and \((A \cap \bar{B})\) are mutually exclusive and therefore that A is the union of two mutually exclusive sets, \((A \cap B)\) and \((A \cap \bar{B})\).
d. Also show that B and \((A \cap \bar{B})\) are mutually exclusive and if \(B \subset A\), A is the union of two mutually exclusive sets, B and \((A \cap \bar{B})\).
Questions & Answers
QUESTION:
Refer to Exercise 2.4. Use the identities \(A=A \cap S\) and \(S=B \cup \bar{B}\) and a distributive law to prove that
a. \(A=(A \cap B) \cup(A \cap \bar{B})\).
b. If \(B \subset A\) then \(A=B \cup(A \cap \bar{B})\).
c. Further, show that \((A \cap B)\) and \((A \cap \bar{B})\) are mutually exclusive and therefore that A is the union of two mutually exclusive sets, \((A \cap B)\) and \((A \cap \bar{B})\).
d. Also show that B and \((A \cap \bar{B})\) are mutually exclusive and if \(B \subset A\), A is the union of two mutually exclusive sets, B and \((A \cap \bar{B})\).
ANSWER:Step 1 of 5
(a)
The union of A and B, denoted by \(A \cup B\), is the set of all points in A or B or both. That is, the union of A and B contains all points that are in at least one of the sets. The set \(A \cup B\) is the shaded region consisting of all points inside either circle (or both).
\(A \cup B \Rightarrow\) means A or B or both.