BICONDITIONAL:UNION, VENN DIAGRAM FOR UNION

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MTH001 Elementary Mathematics

LECTURE # 7

UNION:

Let A and B be subsets of a universal set U. The union of sets A and B is the set of

all elements in U that belong to A or to B or to both, and is denoted A ∪ B.

Symbolically:

A ∪ B = {x ∈U | x ∈A or x ∈ B}

EMAMPLE:

Let U = {a, b, c, d, e, f, g}

A = {a, c, e, g},

B = {d, e, f, g}

Then A ∪ B = {x ∈U | x ∈A or x ∈ B}

={a, c, d, e, f, g}

VENN DIAGRAM FOR UNION:

A ∪ B is shaded

REMARK:

A ∪ B = B ∪ A that is union is commutative you can

prove this very easily only by using definition.

A⊆A∪B

B⊆A∪B

and

The above remark of subset is easily seen by the definition of union.

MEMBERSHIP TABLE FOR UNION:

A∪B

REMARK:

This membership table is similar to the truth table for logical

connective, disjunction (∨).

INTERSECTION:

Let A and B subsets of a universal set U. The intersection of sets

A and B is the set of all elements in U that belong to both A and B and is denoted

A ∩ B.

Symbolically:

A ∩ B = {x ∈U | x ∈ A and x ∈B}

EXMAPLE:

Let

U = {a, b, c, d, e, f, g}

A = {a, c, e, g},

B = {d, e, f, g}

Then A ∩ B = {e, g}

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MTH001 Elementary Mathematics

A ∩ B is shaded

VENN DIAGRAM FOR INTERSECTION:

REMARK:

1. A ∩ B = B ∩ A

2. A ∩ B ⊆ A

A∩B⊆B

and

3. If A ∩ B = φ, then A & B are called disjoint sets.

MEMBERSHIP TABLE FOR INTERSECTION:

A∩B

REMARK:

This membership table is similar to the truth table for logical

connective, conjunction (∧).

DIFFERENCE:

Let A and B be subsets of a universal set U. The difference of "A and B" (or

relative complement of B in A) is the set of all elements in U that belong to A but not

to B, and is denoted A B or A \ B.

Symbolically:

A B = {x ∈U | x ∈ A and x ∈B}

EXAMPLE:

Let

U = {a, b, c, d, e, f, g}

A = {a, c, e, g},

B = {d, e, f, g}

Then A B = {a, c}

VENN DIAGRAM FOR SET DIFFERENCE:

A-B is shaded

REMARK:

1. A B ≠ B A that is Set difference is not commutative.

2. A B ⊆ A

3. A B, A ∩ B and

B A are mutually disjoint sets.

MEMBERSHIP TABLE FOR SET DIFFERENCE:

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MTH001 Elementary Mathematics

REMARK:

The membership table is similar to the truth table for ~ (p →q).

COMPLEMENT:

Let A be a subset of universal set U. The complement of A is the set of all

element in U that do not belong to A, and is denoted AΝ, A or Ac

Symbolically:

Ac = {x ∈U | x ∉A}

EXAMPLE:

Let

U = {a, b, c, d, e, f, g]

A = {a, c, e, g}

Then Ac = {b, d, f}

VENN DIAGRAM FOR COMPLEMENT:

Ac is shaded

REMARK :

1. Ac = U A

2. A ∩ Ac = φ

3. A ∪ Ac = U

MEMBERSHIP TABLE FOR COMPLEMENT:

REMARK

This membership table is similar to the truth table for logical connective

negation (~)

EXERCISE:

Let

U = {1, 2, 3, ..., 10},

X = {1, 2, 3, 4, 5}

Y = {y | y = 2 x, x ∈X}, Z = {z | z2 9 z + 14 = 0}

Enumerate:

(1)X ∩ Y

(2) Y ∪ Z

(3) X Z

(6) (X Z) c

(4)Y

(5) X Z

Firstly we enumerate the given sets.

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MTH001 Elementary Mathematics

Given

U = {1, 2, 3, ..., 10},

X = {1, 2, 3, 4, 5}

Y = {y | y = 2 x, x ∈X} = {2, 4, 6, 8, 10}

Z = {z | z2 9 z + 14 = 0} = {2, 7}

X ∩ Y = {1, 2, 3, 4, 5} ∩ {2, 4, 6, 8, 10}

(1)

= {2, 4}

Y ∪ Z = {2, 4, 6, 8, 10} ∪ {2, 7}

(2)

= {2, 4, 6, 7, 8, 10}

(3)

X Z = {1, 2, 3, 4, 5} {2, 7}

= {1, 3, 4, 5}

(4)

Y = U Y = {1, 2, 3, ..., 10} {2, 4, 6, 8, 10}

= {1, 3, 5, 7, 9

(5)

X Z = {6, 7, 8, 9, 10} {1, 3, 4, 5, 6, 8, 9, 10}

= {7}

(6)

(X Z)c = U (X Z)

= {1, 2, 3, ..., 10} {1, 3, 4, 5}

= {2, 6, 7, 8, 9, 10}

(X Z)c ≠ Xc - Zc

NOTE

EXERCISE:

Given the following universal set U and its two subsets P and Q, where

U = {x | x ∈ Z,0 ≤ x ≤ 10}

P = {x | x is a prime number}

Q = {x | x2 < 70}

(i)

Draw a Venn diagram for the above

List the elements in Pc ∩ Q

(ii)

SOLUTION:

First we write the sets in Tabular form.

U = {x | x ∈Z, 0 ≤ x ≤ 10}

Since it is the set of integers that are greater then or equal 0 and less or equal to 10.

So we have

U= {0, 1, 2, 3, ..., 10}

P = {x | x is a prime number}

It is the set of prime numbers between 0 and 10. Remember Prime numbers are

those numbers which have only two distinct divisors.

P = {2, 3, 5, 7}

Q = {x | x2 < 70}

The set Q contains the elements between 0 and 10 which has their square less or

equal to 70.

Q= {0, 1, 2, 3, 4, 5, 6, 7, 8}

Thus we write the sets in Tabular form.

VENN DIAGRAM:

2,3,5,7P

0,1,4,6,8

9,10

Pc ∩ Q = ?

(i)

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MTH001 Elementary Mathematics

Pc = U P = {0, 1, 2, 3, ..., 10}- {2, 3, 5, 7}

= {0, 1, 4, 6, 8, 9, 10}

and

Pc ∩ Q = {0, 1, 4, 6, 8, 9, 10} ∩ {0, 1, 2, 3, 4, 5, 6, 7, 8}

= {0, 1, 4, 6, 8}

EXERCISE:

Let

U = {1, 2, 3, 4, 5},

C = {1, 3}

and A and B are non empty sets. Find A in each of the following:

A ∪ B = U,

A∩B=φ

(i)

and

B = {1}

A ⊂ B and

A ∪ B = {4, 5}

(ii)

A ∩ B = {3},

A ∪ B = {2, 3, 4}

B ∪ C = {1,2,3}

(iii)

and

(iv)

A and B are disjoint, B and C are disjoint, and the union of A and B is

the set {1, 2}.

A ∪ B = U,

A∩B=φ

(i)

and B = {1}

SOLUTION

Since A ∪ B = U = {1, 2, 3, 4, 5}

and A ∩ B = φ,

A = Bc = {1}c = {2, 3, 4, 5}

Therefore

A ⊂ B and

A ∪ B = {4, 5}

(i)

also

C = {1, 3}

SOLUTION

When A ⊂ B, then

A ∪ B = B = {4, 5}

Also A being a proper subset of B implies

A = {4} or

A = {5}

A ∩ B = {3}, A ∪ B = {2, 3, 4}and B ∪ C = {1,2,3}

(iii)

Also C = {1, 3}

SOLUTION

3 B

Since we have 3 in the intersection of A and B as well as in C so we place 3

common part shared by the three sets in the Venn diagram. Now since 1 is

in the

union of B and C it means that 1 may be in C or may be in B, but 1cannot be in B because if

be in A ∪ B but 1 is not there, thus we place 1 in the part of

1 is in the B then it must

C which is not shared by any other set. Same is the

reason for 4 and we place it in

the set which is not shared by any other set.

Now 2 will be in B, 2 cannot be in A

because A ∩ B = {3}, and is not in C.

So A = {3, 4} and B = {2, 3}

A ∩ B = φ,

B ∩ C = φ,

A ∪ B = {1, 2}.

(i)

Also C = {1, 3}

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MTH001 Elementary Mathematics

SOLUTION

4, 5

A = {1}

EXERCISE:

Use a Venn diagram to represent the following:

(A ∩ B) ∩ Cc

(i)

Ac ∪ (B ∪ C)

(ii)

(A B) ∩ C

(iii)

(A ∩ Bc) ∪ Cc

(iv)

3 B

A 2

1 1

(A ∩ B) ∩ Cc

(1)

1 4

(A ∩ B) ∩ Cc is shaded

Ac ∪ (B ∪ C) is shaded.

(ii)

7 C

(A B) ∩ C

(iii)

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MTH001 Elementary Mathematics

(A B) ∩ C is shaded

(iii)

(A ∩ Bc) ∪ Cc is shaded.

PROVING SET IDENTITIES BY VENN DIAGRAMS:

Prove the following using Venn Diagrams:

A (A B) = A ∩ B

(i)

(A ∩ B)c = A c ∪ B c

(ii)

A B = A ∩ Bc

(iii)

SOLUTION (i)

A - (A B) = A ∩ B

(a)

A = { 1, 2 }

B = { 2, 3 }

A B ={ 1 }

A B is shaded

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MTH001 Elementary Mathematics

(b)

A = { 1, 2 }

AB={1}

A (A B) = { 2 }

A (A B) is shaded

(c)

A ∩ B is shaded

A= { 1, 2 }

B = { 2, 3 }

A ∩ B = {2}

RESULT: A (A B) = A ∩ B

SOLUTION (ii)

(A ∩ B)c = A c ∪ B c

(a)

A∩B

(b)

123

(A ∩ B)c

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MTH001 Elementary Mathematics

(c)

1 A2

Ac is shaded.

(d)

Bc is shaded.

(e)

Ac ∪ Bc is shaded.

Now diagrams (b) and (e) are same hence

(A ∩ B)c = A c ∪ B c

RESULT:

SOLUTION (iii)

A B = A ∩ Bc

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MTH001 Elementary Mathematics

(a)

A B is shaded.

(b)

Bc is shaded.

(c)

A A

A ∩ Bc is shaded

From diagrams (a) and (b) we can say

A B = A ∩ Bc

RESULT:

PROVING SET IDENTITIES BY MEMBERSHIP TABLE:

Prove the following using Membership Table:

A (A B) = A ∩ B

(i)

(A ∩ B)c = A c ∪ B c

(ii)

A B = A ∩ Bc

(iii)

SOLUTION (i)

A (A B) = A ∩ B

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MTH001 Elementary Mathematics

A∩B

A-B

A-(A-B)

Since the last two columns of the above table are same hence the

corresponding set expressions are same. That is

A (A B) = A ∩ B

SOLUTION (ii)

(A ∩ B)c = A c ∪ B c

(A∩B)c A B A c ∪ B c

A∩B

Since the fourth and last columns of the above table are same hence the

corresponding set expressions are same. That is

(A ∩ B)c = A c ∪ B c

SOLUTION (iii)

A ∩ Bc

B AB

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