Set Theory Practice Problems

Introduction to Set Theory

Set theory is the branch of mathematics that learned about the sets, which are the collections of objects. Even though any type of objects can be collected into a set, set theory is applied most often to objects that are related to mathematics. Here we will see the examples and practice problems of set theory.(Source: Wikipedia).

Examples

1)Proof the A∪(B∩C)=(A∪B) ∩(A∪C) condition for the following sets.

A={2,3,6,7,8} B={1,2,5,7,8} C={1,3,7,8,10}

Solution

The given sets are A={2,3,6,7,8} B={1,2,5,7,8} C={1,3,7,8,10}

A∪(B∩C)=(A∪B) ∩(A∪C)

First solve the left side condition.

A∪(B∩C)

B∩C

It means common values of the sets B and C.

B∩C={1,7,8}

A∪(B∩C)

Now joining the A set values.

A∪(B∩C)={ 1,2,3,6,7,8}

Solve right side condition.

(A∪B) ∩(A∪C)

A∪B={1,2,3,5,6,7,8}

A∪C={1,2,3,6,7,8,10}

(A∪B) ∩(A∪C)={1,2,3,6,7,8}

So A∪(B∩C)=(A∪B) ∩(A∪C).

2) Proof the A∪(B∩C)=(A∪B) ∩(A∪C) condition for the sets A={2,4,7,8} B={2,7,9,10} C={1,7,8,10}

Solution

The given sets are A={2,4,7,8} B={2,7,9,10} C={1,7,8,10}

Take left hand side condition.

A∩(B∪C)

B∪C={1,2,7,8,9,10}

A∩(B∪C)={2,7,8}

Now take the right hand side condition.

(A∩B) ∪ (A∩C)

A∩B={2,7}

A∩C={7,8}

(A∩B) ∪ (A∩C)={2,7,8}

So A∩(B∪C)=(A∩B) ∪ (A∩C).

3)What is the A∪(B∪C)=(A∪B) ∪C for the following sets? A={3,7,9,10,11} B={2,4,7,8,11} C={3,5,7,9,12}

Solution

Given sets are A={3,7,9,10,11} B={2,4,7,8,11} C={3,5,7,9,12}

A∪(B∪C)=(A∪B) ∪C

Take left hand side condition

A∪(B∪C)

B∪C={2,3,4,5,7,8,9,11,12}

Grouping the set of A values with this B∪C

A∪(B∪C)={2,3,4,5,7,8,9,10,11,12}

Right hand side.

(A∪B) ∪C

A∪B={2,3,4,7,8,9,10,11}

Joining the values is the set C with this A∪B.

(A∪B)∪C={2,3,4,5,7,8,9,10,11,12}

So the condition is proofed.

4)A={4,8,9,10} B={3,4,6,7} and C={1,2,7,9}. Find the A-(B∪C).

Solution

The given sets are A={4,8,9,10} B={3,4,6,7} and C={1,2,7,9}.

B∪C

B∪C={1,2,3,4,6,7,9}

A-( B∪C)

The condition is a difference of set. It means we select the values from the A set. But that values is not present in the B∪C set.

So A- B∪C={8,10}

These are the practice problems of set theory. Please express your views of this topic how many lines of symmetry does a square have by commenting on blog.

Practice Problems

Practice Problems 1

A={7,8,9,10} B={6,8,9,14} C={5,9,20,12,15}. Proof the associative law in set theory.

Practice Problems 2

A={1,2,5,8,9} B={1,3,4,5,6} C={5,6,8,9,11}. Proof the distributive law in set theory.

Practice Problems 3

X={2,4,10,11,12,13} Y={3,5,6,7,22,33} z={1,2,6,8,10,11}. What is the A-(B∩C}?