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According to De Morgan's Law, the complement of the union of two sets is the intersection of their complements, and the complement of the intersection of two sets is the union of their complements.
- These are named after the renowned British mathematician Augustus De Morgan.
- De Morgan's Law is based on rules of interference.
- It is a pair of rules that is used in set theory and boolean algebra.
- The complement rules are used in set theory to link the intersection and union of sets.
- Boolean algebra determines the relationship between AND, OR, or negation of the statement.
- The laws can be represented using a Venn diagram.
- In set theory, the law expresses the intersection and union of sets by complements expression.
- This law may be written as:
( A ∪ B) ‘ = A ‘ ∩ B ‘.
(A ∩ B)’ = A’ ∪ B’
Key Terms: De Morgan’s Law, Boolen Algebra, Set Theory, Venn Diagram, Complement, Union of Sets, Universal Set, Intersection of Sets, Interference
What are De Morgan's Law
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De Morgan’s Law states that the complement of two sets' union is equal to the intersection of their complements, and the complement of two sets' intersection is equal to the union of their complements.
- De Morgan’s law can be applied to negation of disjunction and conjuction of a set.
- It can perform union and intersection of set using complement of the set.
- The concept of the law is based on the algebraization of logic gates which was proposed by George Boolean.
- De Morgan’s law is used by electronic engineer to create logic gates.
- The law is used in computer programming.
In the case of any two finite sets A and B:
- (A U B)' = A' B' (which is a union law of De Morgan).
- (A B)' = A' U B' (a De Morgan's law of intersection).
First De Morgan’s Law
First De Morgan’s Law states that complement of the union of the two given sets is equal to the intersection of the complements of each of the set taken in an experiment.
- This law is also known as De Morgan’s law of union.
- It can be represented mathematically as:
(A ∪ B)’ = A’ ∩ B’
Second De Morgan’s Law
Second De Morgan’s Law states that complement of the intersection of the two given sets is equal to the union of the complements of each of the set taken in an experiment.
- This law is also known as De Morgan’s law of intersection.
- It can be represented mathematically as:
(A ∩ B)’ = A’ ∪ B’
De Morgan’s Law ExampleExample: If U = {1,3,5,7,9,11,10} ,A = {3,5,6}, B = {5,7,9}. Solution: De Morgan's First Law is that (A∪B)=A′∩B′ A' = {1,7,9,11} B' = {1,3,11} A′∩B′={1,11} |
De Morgan’s Law Proof
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A set is a well-defined collection of objects or components. On two sets, several operations such as complement of a set, union, and intersection may be done.
- The procedures and their applications can be simplified further by using a set of laws known as De Morgan's Laws.
- These are extremely basic and straightforward laws.
- A universal set is any set that contains all of the items or elements linked to a specific context.
Consider a universal set U whose subsets are A and B. The complement of the union of two sets A and B is equal to the intersection of the complements of the sets A and B, according to De Morgan's first law.
(A∪B)’= A’∩ B’ —–(1)
Where a set's complement is defined as:
A’= {x:x ∈ U and x ∉ A}
Where A' stands for the complement.
- Venn Diagrams may be used to readily depict this law.
- The complement of the union of two sets A and B is represented by the L.H.S of equation 1.
- To begin with, the union of two sets A and B is defined as the set of all items that are either in set A or set B.
- It is possible to depict it using Venn Diagrams, as shown:
Union of Set
- AUB is denoted by the highlighted or green section. The complement of the union of A and B, i.e., (AUB), is the set of all items that are not in AUB. This may be shown as follows:
Complement of sets
- Similarly, R.H.S of equation 1 may be represented using Venn Diagrams, with the first component, A', shown as follows:
Complement of set A
- Set A is denoted by the black section, while its counterpart, A', is denoted by the blue portion.
- Similarly, B' is denoted as:
Complement of set B
- Set B is denoted by the black section, while its counterpart, B', is denoted by the yellow portion.
- When both the below figures are overlaid on one another, we get a figure that looks like the complement of sets.
Intersection of complements of sets
- As a result, L.H.S = R.H.S.
- Mathematically,
- As an example, AUB= either in A or in B.
- (AUB)'= L.H.S = neither A nor B
- A' also stands for "not in A."
- 'B' = 'Not in B'
- A'∩ B' = Not in A nor in B
- (AUB)' = A'∩ B'
Thus, the validity of De Morgan's Laws may be justified by picturing the Venn Diagrams and writing them down.
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|---|---|---|
| Bayes Theorem | Ordered Pairs | Real Valued Functions |
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De Morgan’s Law in Boolean Algebra
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De Morgan’s law can be used in the field of boolean algebra. It defines relationship between the AND, OR, and the complements of variables. The law is used for complement of AND and OR on two set of values.
- There are two De Morgan’s Law in boolean algebra which are as follows:
First De Morgan’s Law in Boolean Algebra
First De Morgan’s Law in Boolean Algebra states that the complement of the OR of two values is equal to the AND of the complement of each of the value.
- It can be represented mathematically as:
(A + B)’ = A’ . B’
- where + represents the OR operator
- . represents AND operator
- ‘ represents complement operator
First De Morgan’s Law in Boolean Algebra Truth Table
The truth table for first De Morgan’s Law in Boolean Algebra is as follows:
| A | B | A + B | (A + B)’ | A’ | B’ | A’. B’ |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 | 0 | 0 |
Second De Morgan’s Law in Boolean Algebra
Second De Morgan’s Law in Boolean Algebra states that the complement of the AND of two values is equal to the OR of the complement of each of the value.
- It can be represented mathematically as:
(A . B)’ = A’ + B’
- where + represents the OR operator
- . represents AND operator
- ‘ represents complement operator
Second De Morgan’s Law in Boolean Algebra Truth Table
The truth table for second De Morgan’s Law in Boolean Algebra is as follows:
| A | B | AB | (AB)’ | A’ | B’ | A’ + B’ |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 | 0 | 0 |
De Morgan‘s Law Formulas
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The important de morgan’s law formulas for set theory and boolen algebra is as follows:
Formula for Set Theory
- (A ∪ B)’ = A’ ∩ B’
- (A ∩ B)’ = A’ ∪ B’
Formula for Boolean Algebra
- (A + B)’ = A’ . B’
- (A . B)’ = A’ + B’
Formula for Logic
- ∼ (a ∧ b) ≡ ∼ a ∨ ∼ b
- ∼ (a ∨ b) ≡ ∼ a ∧ ∼ b
Applications of De Morgan’s Theorem
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The important applications of De Morgan’s theorem is as follows:
- It is used in the logic gates of the circuit as it is cheaper hardware.
- An individual can easily implement NAND, NOT and NOR gates.
- De Morgan’s law is used to simplify logical expressions written in codes.
- The law can be used for the verification of SAS codes.
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|---|---|---|
| Group Theory | Set Operations | Degree of polynomial |
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Things to Remember
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- De Morgan’s is used for set theory and boolean algebra.
- It determines the relation between the complement of the union of two sets, which is the intersection of their complements.
- Sets can perform several operations, such as complement of a set, union, and intersection may be done.
- Commutative, associative and distributive are three properties of Boolean algebra.
- It is a powerful tool for the design of digital equipment.
Sample Questions
Ques: If U = j, k, l, m, n, then X = j, k, m, and Y = k, m, n. (X Y)' = X' U Y' is proof of De Morgan's law. (4 Marks)
Ans: We know that U = j, k, l, m, n.
X = j, k, and m
Y = k, m, and n
(X Y) = j, k, m (X Y) = j, k, m (X Y) = j, k, m (X Y) = j, k, m (X Y) = j, k
= k, m
As a result, (X Y)' = j, l, n .................... (i)
Because X = j, k, m, X' = l, n
and Y = k, m, n, thus Y' = j, l
X' and Y' = l, n, j, l
As a result, X' Y' = j, l, n .................... (ii)
When we combine I with (ii), we obtain;
(X Y)' = X' U Y ' Proved
Ques: Let U equal 1, 2, 3, 4, 5, 6, 7, 8, P equal 4, 5, 6, and Q equal 5, 6, 8. Demonstrate that (P Q)' = P' Q'. (3 Marks)
Ans: We know that U = 1, 2, 3, 4, 5, 6, 7, 8, and 9.
P = 4, 5, and 6
Q = 5, 6, 8
PUQ = 4, 5, 6, 5, 6, 8
(4), (5), (6), (8)
As a result, (PUQ)' = 1, 2, 3, 7,.................... (i)
Now P = {4, 5, 6} so, P' = {1, 2, 3, 7, 8}
and Q = {5, 6, 8} so, Q' = {1, 2, 3, 4, 7}
P' ∩ Q' = {1, 2, 3, 7, 8} ∩ {1, 2, 3, 4, 7}
Therefore, P' ∩ Q' = {1, 2, 3, 7} ……………….. (ii)
Combining (i)and (ii) we get;
(P ∪ Q)' = P' ∩ Q'. Proved
As a result, P' Q' = 1, 2, 3, 7,.................... (ii)
When we combine I with (ii), we obtain;
(P Q)' = P' Q' Proved
Ques: Write a description of each shaded area. Use the symbols
(i) U,A,B,∪,∩
(ii) U,A,B,∪,∩
and as necessary. (5 Marks)
Ans: We begin by writing definitions for each phrase in the question.
Universal set (U): A universal set is one that contains all other sets.
Union of sets (): The union of two or more sets is the set that contains items from both sets.
Intersection of sets (): The set that contains the common items of both sets is denoted by the intersection of two or more sets.
Subtraction of sets (): Subtraction of a set from another set signifies the elements in one set but not in the other.
We will now solve each section independently.
We denote the two sets in the shape of circles as 'A' and 'B,' respectively, and the rectangle as 'U.' The diagram is then transformed.
The shaded region here contains the whole set 'B,' the entire universal set 'U,' and only excludes the area of set 'A.'
Shaded region is U\(A\(A∩B))
(ii) The three sets presented in the shape of circles are denoted by the letters 'A,' 'B,' and 'C,' and the rectangle is denoted by the letter 'U.' The diagram is then transformed.
The shaded region here contains the intersection of sets 'A' and 'B,' as well as the intersection of sets 'A' and 'C.'
Shaded region is (A∩B)∪(A∩C)
(iii) The three sets presented in the shape of circles are denoted by the letters 'A,' 'B,' and 'C,' while the rectangle is denoted by the letter 'U.' The diagram is then transformed.
(iv) The three sets presented in the shape of circles are denoted by the letters 'A,' 'B,' and 'C,' and the rectangle is denoted by the letter 'U.' The diagram is then transformed.
Ques: Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}.
Show that (A ∪B) ' = A ' ∩ B'. (3 Marks)
Ans: U = {1, 2, 3, 4, 5, 6}
A = {2, 3}
B = {3, 4, 5}
A ∪ B = {2, 3} ∪ {3, 4, 5}
= {2, 3, 4, 5}
∴ (A ∪ B) ' = {1, 6}
Also A ' = {1, 4, 5, 6}
B ' = {1, 2, 6}
∴ A' ∩ B' = {1, 4, 5, 6} ∩ {1, 2, 6}
= {1, 6}
Hence (A ∪ B)' = A ' ∩ B
Ques: Let A = { a, b, c, d, e, f, g, x, y, z }, B = { 1, 2, c, d, e } and C = { d, e, f, g, 2, y }. Verify De Morgan’s laws of set difference. (3 Marks)
Ans: First, we shall verify A \ (B u C) = (A \ B) n (A \ C)
To do this, we consider
B u C = { 1, 2, c, d, e } u { d, e, f, g, 2, y }
B u C = { 1, 2, c, d, e, f, g, y }
We know that
A/(B u C) = {a, b, c, d, e, f, g, x, y, z}\{1, 2, c, d, e, f, g, y}
A / (B u C) = { a, b, x, z } ---------(1)
A \ B = { a, b, f, g, x, y, z }
A \ C = { a, b, c, x, z }
(A \ B) n (A \ C) = { a, b, x, z } ---------(2)
From (1) and (2), it is clear that A\(B u C) = (A\B)n(A\C)
Ques: Prove the expression: ((A+BC')'+D(E+F')')'. (5 Marks)
Ans: For applying the DeMorgan's theorem on this expression, we have to follow the following expressions:
1) In complete expression, first, we find those terms on which we can apply DeMorgan's theorem and treat each term as a single variable.
2) Next, we apply DeMorgan's first theorem. So,
3) Next, we use rule number 9, i.e., (A=(A')') for canceling the double bars.
4) Next, we apply DeMorgan's second theorem. So,
5) Again apply rule number 9 to cancel the double bar
Now, this expression has no term in which we can apply any rule or theorem. So, this is the final expression.
Ques: The table below displays the proportion of a school's pupils that participated in elections and drawing contests.
| Competition | Election | Drawing | Both |
| Percentage of students | 55 | 45 | 20 |
Draw a Venn diagram to illustrate this information and use it to calculate the proportion of students who I participated only in the election, (ii) participated only in the drawing, and (iii) did not participate in any of the contests. (5 Marks)
Ans: We can see from the table that the following information is available.
The percentage of students that voted = 55...........................................
(A) The percentage of pupils that participated in drawing competitions is 45.........................................
(B) The percentage of pupils that participated in both elections and drawing contests = 20.................................................
(C) We can get the percentage of students who voted in elections by subtracting equation (3) from equation (2). (1).
The percentage of students who only voted in elections = 55-20=35....................................
(D) Similarly, we can get the proportion of students that participated in drawing contests by subtracting equation (3) from equation (2)
The percentage of pupils who solely participated in drawing contests = 45-20=25.................................... (5)
Now, using the information provided, create a Venn diagram.
From the Venn diagram, we have
(a) The percentage of students who participated in elections = 35%.
(b) The percentage of students who participated in drawing competitions = 25%.
The percentage of students who participated in both elections and drawing competitions = 20%.
(c) The percentage of students who participated in neither elections nor drawing competitions = 20%.
Ques: Represents the shaded part of the Venn diagram in terms of union and intersection of sets: P and Q their complements. (5 Marks)
Ans: Step 1: From the diagram below, the universal set U is the complete set along with sets
P and Q
Step 2: From the below diagram, P∪Q is the set contains all the shaded area of both the sets as shown below:
Step 3: Similarly, the intersection is that area which is common in two or more sets. So, P∩Q
is that shaded area which is common in both sets as shown below:
Step 4: Now, for finding the shaded area from the below diagram, we need to subtract P∪Q from the universal set and after that, we will add the shaded area which is common in the set P and Q as shown below:
∴ The Shaded area = U−P∪Q+P∩Q.
Ques: Given that U = {2, 3, 7, 8, 9,10}, A = {2, 7,3} and B = {2, 3, 9,10}. Prove De Morgan’s Second Law. (3 Marks)
Ans: Since it is given that U = {2, 3, 7, 8, 9,10}, A = {2, 7,3} and B = {2, 3, 9,10}
To Prove: (A ∩ B)’ = A’ ∪ B’
- (A ∩ B) = {2,3}
- (A ∩ B)’ = U – (A ∩ B) = {2, 3, 7, 8, 9,10} – {2,3}
- (A ∩ B)’ = {7, 8, 9,10}
- A’ = U – A = {2, 3, 7, 8, 9,10} – {2, 7,3}
- A’ = {8, 9,10}
- B’ = U – B = {2, 3, 7, 8, 9,10} – {2, 3, 9,10}
- B’ = {7, 8}
- A’ ∪ B’ = {10, 8, 9} ∪ {7, 8}
- A’ ∪ B’ = {10, 7, 8, 9}
- (A ∩ B)’ = A’ ∪ B’
Ques: Given that U = {1, 4, 6, 8, 9,2}, A = {1, 9,2} and B = {4, 6, 9}. Prove De Morgan’s First Law. (3 Marks)
Ans: Given that U = {1, 4, 6, 8, 9,2}, A = {1, 9,2} and B = {4, 6, 9}
- To Prove: (A ∪ B)’ = A’ ∩ B’
- (A ∪ B) = {1, 4, 6, 9,2}
- (A ∪ B)’ = U – (A ∪ B) = {1, 4, 6, 8, 9,2} – {1, 4, 6, 9,2}
- (A ∪ B)’ = {8}
- A’ = U – A = {1, 4, 6, 8, 9,2} – {1, 9,2}
- A’ = {4, 6, 8}
- B’ = U – B = {1, 4, 6, 8, 9,2} – {4, 6, 9}
- B’ = {1, 8,2}
- A’ ∩ B’ = {4, 6, 8} ∩ {1, 8,2}
- A’ ∩ B’ = {8}
- (A ∪ B)’ = A’ ∩ B’
- Hence Proved
Ques: If U = {1, 2, 3, 5, 7, 9, 11}, A = {3, 5, 4} and B = {5, 7, 9}, then prove De Morgan's first law. (2 Marks)
Ans: According to De Morgan's First law, (A ∪ B)’ = A’ ∩ B’
(A ∪ B) = {3, 5, 7, 9, 4}, (A ∪ B)’ = {1, 2, 11}
A’ = {1, 7, 9, 11, 2} and B' = {1, 3, 11, 2,4}
A’ ∩ B’ = {1, 11, 2}.
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