Introduction to Foundations of Set Theory
Foundations of set theory are 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.Now we will see the examples of set theory.
Examples for Foundations of Union and Intersections
1)Proof the A∪(B∩C)=(A∪B) ∩(A∪C) for the following sets.
A={4,6,7,9,10} B={3,4,5,6,7} C={1,4,5,6,8}
Solution
The given sets are A={4,6,7,9,10} B={3,4,5,6,7} C={1,4,5,6,8}
A∪(B∩C)=(A∪B) ∩(A∪C)
Take the left side condition.
A∪(B∩C)
B∩C
Take the common values of the sets B and C.
B∩C={4,5,6}
A∪(B∩C)
Now joining the A set values.
A∪(B∩C)={ 4,5,6,7,9,10}
Take right side condition.
(A∪B) ∩(A∪C)
A∪B={3,4,5,6,7,9,10}
A∪C={1,4,5,6,7,8,9,10}
(A∪B) ∩(A∪C)={4,5,6,7,9,10}
So A∪(B∩C)=(A∪B) ∩(A∪C).
2) Proof the A∪(B∩C)=(A∪B) ∩(A∪C) condition for the sets A={1,2,3,4,5} B={2,4,7,9,11} C={1,5,6,8,9}
Solution
The given sets are A={1,2,3,4,5} B={2,4,7,9,11} C={1,5,6,8,9}
Take left hand side condition.
A∩(B∪C)
B∪C={1,2,4,5,6,7,8,9,11}
A∩(B∪C)={1,2,4,5}
Now take the right hand side condition.
(A∩B) ∪ (A∩C)
A∩B={2,4}
A∩C={1,5}
(A∩B) ∪ (A∩C)={1,2,4,5}
So A∩(B∪C)=(A∩B) ∪ (A∩C).
3) What is the A∪(B∪C)=(A∪B) ∪C for the following sets? A={13,17,18,20,22} B={12,13,14,15,16} C={11,12,13,15,17}
Solution
Given sets are A={13,17,18,20,22} B={12,13,14,15,16} C={11,12,13,15,17}
A∪(B∪C)=(A∪B) ∪C
Take left hand side condition
A∪(B∪C)
B∪C={11,12,13,14,15,16,17}
Grouping the set of A values with this B∪C
A∪(B∪C)={11,12,13,14,15,16,17,18,20,22}
Right hand side.
(A∪B) ∪C
A∪B={12,13,14,15,16,17,18,20,22}
Joining the values is the set C with this A∪B.
(A∪B)∪C={11,12,13,14,15,16,17,18,20,22}
So the condition is proofed.
Between, if you have problem on these topics high school math word problems, please browse expert math related websites for more help on 8th grade math word problems.
Example for Foundations of Difference Set
1)A={2,5,7,8,9} B={2,3,5,6,7} and C={3,4,5,7,9}. Find the i)A-(B∪C) ii)B-(A∪C).
Solution
The given sets are A={2,5,7,8,9} B={2,3,5,6,7} and C={3,4,5,7,9}
i) A-(B∪C)
B∪C
B∪C={2,3,5,6,7,9}
A-( B∪C) 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}
ii)B-(A∪C)
A∪C={2,3,4,5,7,8,9}
B-(A∪C)={6}
These are the examples of foundations of set theory.
Foundations of set theory are 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.Now we will see the examples of set theory.
Examples for Foundations of Union and Intersections
1)Proof the A∪(B∩C)=(A∪B) ∩(A∪C) for the following sets.
A={4,6,7,9,10} B={3,4,5,6,7} C={1,4,5,6,8}
Solution
The given sets are A={4,6,7,9,10} B={3,4,5,6,7} C={1,4,5,6,8}
A∪(B∩C)=(A∪B) ∩(A∪C)
Take the left side condition.
A∪(B∩C)
B∩C
Take the common values of the sets B and C.
B∩C={4,5,6}
A∪(B∩C)
Now joining the A set values.
A∪(B∩C)={ 4,5,6,7,9,10}
Take right side condition.
(A∪B) ∩(A∪C)
A∪B={3,4,5,6,7,9,10}
A∪C={1,4,5,6,7,8,9,10}
(A∪B) ∩(A∪C)={4,5,6,7,9,10}
So A∪(B∩C)=(A∪B) ∩(A∪C).
2) Proof the A∪(B∩C)=(A∪B) ∩(A∪C) condition for the sets A={1,2,3,4,5} B={2,4,7,9,11} C={1,5,6,8,9}
Solution
The given sets are A={1,2,3,4,5} B={2,4,7,9,11} C={1,5,6,8,9}
Take left hand side condition.
A∩(B∪C)
B∪C={1,2,4,5,6,7,8,9,11}
A∩(B∪C)={1,2,4,5}
Now take the right hand side condition.
(A∩B) ∪ (A∩C)
A∩B={2,4}
A∩C={1,5}
(A∩B) ∪ (A∩C)={1,2,4,5}
So A∩(B∪C)=(A∩B) ∪ (A∩C).
3) What is the A∪(B∪C)=(A∪B) ∪C for the following sets? A={13,17,18,20,22} B={12,13,14,15,16} C={11,12,13,15,17}
Solution
Given sets are A={13,17,18,20,22} B={12,13,14,15,16} C={11,12,13,15,17}
A∪(B∪C)=(A∪B) ∪C
Take left hand side condition
A∪(B∪C)
B∪C={11,12,13,14,15,16,17}
Grouping the set of A values with this B∪C
A∪(B∪C)={11,12,13,14,15,16,17,18,20,22}
Right hand side.
(A∪B) ∪C
A∪B={12,13,14,15,16,17,18,20,22}
Joining the values is the set C with this A∪B.
(A∪B)∪C={11,12,13,14,15,16,17,18,20,22}
So the condition is proofed.
Between, if you have problem on these topics high school math word problems, please browse expert math related websites for more help on 8th grade math word problems.
Example for Foundations of Difference Set
1)A={2,5,7,8,9} B={2,3,5,6,7} and C={3,4,5,7,9}. Find the i)A-(B∪C) ii)B-(A∪C).
Solution
The given sets are A={2,5,7,8,9} B={2,3,5,6,7} and C={3,4,5,7,9}
i) A-(B∪C)
B∪C
B∪C={2,3,5,6,7,9}
A-( B∪C) 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}
ii)B-(A∪C)
A∪C={2,3,4,5,7,8,9}
B-(A∪C)={6}
These are the examples of foundations of set theory.
