Question

Let A and B be the multisets \(\left\{ {3.a,2.b,1.c} \right\}\) and \(\{ 2.a.3.b,4.d\} \), respectively. Find

a) \(A \cup B\)

b) \(A \cap B\)

c) \(A - B\)

d) \(B - A\)

e) \(A + B\)

Short answer

(a) The value \(A \cup B\)is \(\left\{ {3.a,3.b.1.c,4.d} \right\}\).

(b) Thevalue \(A \cup B\) is \(\left\{ {2.a,2.b} \right\}\).

(c)The value \(A - B\)is\(\left\{ {1a,1c} \right\}\).

(d) The the value of \(B - A\) is \(\left\{ {1.b,4.d} \right\}\).

(e) The solution is \(\left\{ {5.a,5.b,1.c,4.d} \right\}\).

Step-by-step solution

Explanation

Given in the question,

\(A = \left\{ {3.a,2.b,1.c} \right\}\)and\(B = \{ 2.a.3.b,4.d\} \).

Here we need to find\(A \cup B\).

Concept

Allow A and B to be multiple sets. multi-set union A and B multiple sets where the multiplication of an object is the maximum of its multiplication in A and B.

Explanation

Now we find here , \(A \cup B\).

S0,

\(A \cup B\)= \(\left\{ {3.a,2.b,1.c} \right\} \cup \{ 2.a.3.b,4.d\} \)

\( = \left\{ {3.a,3.b.1.c,4.d} \right\}\)

Hence,

\(A \cup B = \left\{ {3.a,3.b.1.c,4.d} \right\}\)

Explanation

Given in the question,

\(A = \left\{ {3.a,2.b,1.c} \right\}\)and\(B = \{ 2.a.3.b,4.d\} \).

In the question, we need to find\(A \cap B\).

Concept

the interaction of A and B is a multiple set where the multiplication of an element is the minimum of its recurrence in A and B.

calculation

Here we solve our question,

\(A \cap B = \left\{ {3.a,2.b,1.c} \right\} \cap \{ 2.a.3.b,4.d\} \)

\( = \left\{ {2.a,2.b} \right\}\)

\(A \cap B = \left\{ {2.a,2.b} \right\}\)

Hence, \(A \cap B = \left\{ {2.a,2.b} \right\}\)

Explanation

Given in the question is,

\(A = \left\{ {3.a,2.b,1.c} \right\}\)and\(B = \{ 2.a.3.b,4.d\} \).

Here we need to find the differences between A and B.

Concept

The difference between A and B is a multiset where the multiplication of an element is the multiplication of the element A under its multiplication of B unless this difference is negative, where the multiplication is 0.

Explanation

Here we solve our question,

\(A = \left\{ {3.a,2.b,1.c} \right\}\)and\(B = \{ 2.a.3.b,4.d\} \).

\(A - B = \left\{ {3.a,2.b,1.c} \right\} - \{ 2.a.3.b,4.d\} \)

\( = \left\{ {1a,1c} \right\}\)

Hence , \(A - B = \left\{ {1a,1c} \right\}\).

Explanation

In the given question,

\(A = \left\{ {3.a,2.b,1.c} \right\}\)and\(B = \{ 2.a.3.b,4.d\} \).

Here we find\(B - A\).

Concept

the difference between A and B is a multiset where the multiplication of an element is the multiplication of the element A under its multiplication of B unless this difference is negative, where the multiplication is 0.

Calculation

Now we solve our question,

\(A = \left\{ {3.a,2.b,1.c} \right\}\)and\(B = \{ 2.a.3.b,4.d\} \).

\(B - A = \{ 2.a.3.b,4.d\} - \left\{ {3.a,2.b,1.c} \right\}\)

\( = \left\{ {1.b,4.d} \right\}\)

Hence, our solution \(B - A = \left\{ {1.b,4.d} \right\}\)

Explanation

Given in the question, \(A = \left\{ {3.a,2.b,1.c} \right\}\) and\(B = \{ 2.a.3.b,4.d\} \).

Concept

The sum of A and B is a multiple set where element multiplication is the sum of multiples in A and B.

Calculation

Here we solve our question,

\(A + B = \{ 2.a.3.b,4.d\} + \left\{ {3.a,2.b,1.c} \right\}\)

\( = \left\{ {5.a,5.b,1.c,4.d} \right\}\)

Hence, our solution is \(A + B = \left\{ {5.a,5.b,1.c,4.d} \right\}\).