# Laws of Algebra of Sets – Commutative, Associative, Distributive, Demorgan’s | Set Operations & Laws of Set Theory

Sets are defined as a collection of well-defined elements that do not vary from person to person. It can be represented either in set-builder form or roster form. Generally, sets can be represented using curly braces {}. The different types of sets are empty set, finite set, singleton set, infinite set, equivalent set, disjoint sets, equal sets, subsets, superset, and universal sets. Get to know more about the Laws of Algebra of Sets for a better understanding of the students.

### Laws of Algebra of Sets

The operations of sets are union, intersection, and complementation. The binary operations of set union, intersection satisfy many identities. The seven fundamental laws of the algebra of sets are commutative laws, associative laws, idempotent laws, distributive laws, de morgan’s laws, and other algebra laws.

1. Commutative Laws

For any two finite sets A and B

• A U B = B U A
• A ∩ B = B ∩ A

2. Associative Laws

For any three finite sets A, B, and C

• (A U B) U C = A U (B U C)
• (A ∩ B) ∩ C = A ∩ (B ∩ C)

So, union and intersection are associative.

3. Idempotent Laws

For any finite set A

• A U A = A
• A ∩ A = A
• A ∩ A’ = ∅
• ∅’ = U
• ∅ = U’

4. Distributive Laws

For any three finite sets A, B, and C

• A U (B ∩ C) = (A U B) ∩ (A U C)
• A ∩ (B U C) = (A ∩ B) U (A ∩ C)

Thus, union and intersection are distributive over intersection and union respectively.

5. De morgan’s Laws

For any two finite sets A and B

• A – (B U C) = (A – B) ∩ (A – C)
• A – (B ∩ C) = (A – B) U (A – C)

De Morgan’s Laws can also be written as

• Law of union: (A U B)’ = A’ ∩ B’
• Law of intersection: (A ∩ B)’ = A’ U B’

6. Complement Law

For any finite set A

• A ∪ A’ = A’ ∪ A =U
• A ∩ A’ = ∅

More laws of the algebra of sets:

7. For any two finite sets A and B;

• A – B = A ∩ B’
• B – A = B ∩ A’
• A – B = A ⇔ A ∩ B = ∅
• (A – B) U B = A U B
• (A – B) ∩ B = ∅
• A ⊆ B ⇔ B’ ⊆ A’
• (A – B) U (B – A) = (A U B) – (A ∩ B)

8. For any three finite sets A, B, and C;

• A – (B ∩ C) = (A – B) U (A – C)
• A – (B U C) = (A – B) ∩ (A – C)
• A ∩ (B – C) = (A ∩ B) – (A ∩ C)
• A ∩ (B △ C) = (A ∩ B) △ (A ∩ C)

### Solved Examples on Laws of Algebra of Sets

Example 1:

If E = {1, 2, 3, 4, 5, 6, 7}, A = {1, 2, 3, 4, 5}, B = {2, 5, 7} show that

(i) (A U B)’ = A’ ∩ B’

(ii) (A U B) = B U A

(iii) A ∩ B = B ∩ A

(iv) (A ∩ B)’ = A’ U B’

Solution:

Given that

E = {1, 2, 3, 4, 5, 6, 7}, A = {1, 2, 3, 4, 5}, B = {2, 5, 7}

(i) (A U B)’ = A’ ∩ B’

A U B = {{1, 2, 3, 4, 5} U {2, 5, 7}}

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

(A U B)’ = {1, 2, 3, 4, 5, 7}’

= {6}

A’ = {6, 7}

B’ = {1, 3, 4, 6}

A’ ∩ B’ = {6, 7} ∩ {1, 3, 4, 6}

= {6}

So, L.H.S = R.H.S

(ii) (A U B) = B U A

A U B = {{1, 2, 3, 4, 5} U {2, 5, 7}}

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

B U A = {2, 5, 7} U {1, 2, 3, 4, 5}

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

So, L.H.S = R.H.S

(iii) A ∩ B = B ∩ A

L.H.S = A ∩ B

= {1, 2, 3, 4, 5} ∩ {2, 5, 7}

= {2, 5}

R.H.S = B ∩ A

= {2, 5, 7} ∩ {1, 2, 3, 4, 5}

= {2, 5}

So, L.H.S = R.H.S

(iv) (A ∩ B)’ = A’ U B’

L.H.S = (A ∩ B)’

= {{1, 2, 3, 4, 5} ∩ {2, 5, 7}}’

= {2, 5}’

= {1, 3, 4, 6, 7}

R.H.S = A’ U B’

= {1, 2, 3, 4, 5}’ U {2, 5, 7}’

= {6, 7} U {1, 3, 4, 6}

= {1, 3, 4, 6, 7}

L.H.S = R.H.S

Hence, proved.

Example 2:

If X = {a, b, c, d}, Y = {b, d, f}, Z = {a, c, e} verify that

(i) (X ∪ Y) ∪ Z = X ∪ (Y ∪ Z)

(ii) (X ∩ Y) ∩ Z = X ∩ (Y ∩ Z)

Solution:

Given that,

X = {a, b, c, d}, Y = {b, d, f}, Z = {a, c, e}

(i) (X ∪ Y) ∪ Z = X ∪ (Y ∪ Z)

L.H.S = (X ∪ Y) ∪ Z

= ({a, b, c, d} U {b, d, f}) U {a, c, e}

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

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

R.H.S = X ∪ (Y ∪ Z)

= {a, b, c, d} U ({b, d, f} U {a, c, e})

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

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

So, L.H.S = R.H.S

(ii) (X ∩ Y) ∩ Z = X ∩ (Y ∩ Z)

L.H.S = (X ∩ Y) ∩ Z

= ({a, b, c, d} ∩ {b, d, f}) ∩ {a, c, e}

= {b, d} ∩ {a, c, e}

= ∅

R.H.S = X ∩ (Y ∩ Z)

= {a, b, c, d} ∩ ({b, d, f} ∩ {a, c, e})

= {a, b, c, d} ∩ ∅

= ∅

So, L.H.S = R.H.S

Hence verified.

Example 3:

If A = {p, q, r, s}, B = {u, q, s, v} find

(i) A – B

(ii) B – A

(iii) A ∩ B

Solution:

Given that,

A = {p, q, r, s}, B = {u, q, s, v}

(i)

A – B = {p, q, r, s} – {u, q, s, v}

= {p, r}

(ii)

B – A = {u, q, s, v} – {p, q, r, s}

= {u, v}

(iii)

A ∩ B = {p, q, r, s} ∩ {u, q, s, v}

= {q, s}

### FAQs on Laws of Algebra of Sets

1. What is a set? Give an example?

A set is a collection of elements or objects or numbers represented using the curly brackets {}. The example is {1, 2, 3, 5} is a set of numbers.

2. What are the five basic properties of sets?

The five basic properties of sets are commutative property, identity property, associative property, complement property, and distributive property.

3. What are the 4 operations of sets?

The 4 set operations include set union, set intersection, set difference, the complement of a set, and cartesian product.