Operations on Sets – Union, Intersection, Commutative, Associative & Distributive Laws Explained | Math With Raabi
📘 Operations on Sets — Union, Intersection, Distributive & Associative Laws
📚 In set theory, the main operations are union, intersection and complement. These operations help us combine and compare sets. Knowing the formal laws — commutative, associative and distributive — makes it easier to simplify set expressions and verify equalities.
🧭 Before studying operations, make sure you have read the previous topic: Understanding Sets in Mathematics . This provides a clear introduction to sets and notation. 🔗
Union, Intersection and Complement — Quick Reminder
Intersection (A ∩ B): The set of elements common to both A and B.
Union (A ∪ B): The set of elements that are in A or in B (or in both).
Complement (A′) (with respect to universal set U): All elements of U that are not in A.
When finding A ∩ B, it’s easier to write the first set (A) first and check which of its elements are also in B. This reduces unnecessary comparison and helps avoid mistakes. ✅
Commutative Laws
The commutative laws say that changing the order of sets does not change the result:
- A ∪ B = B ∪ A
- A ∩ B = B ∩ A
Let A = {a, b, c, d} and B = {b, d, f}.
Union:
A ∪ B = {a, b, c, d, f} and B ∪ A = {a, b, c, d, f}.
Intersection:
A ∩ B = {b, d} and B ∩ A = {b, d}.
Both results match, so commutative laws are verified.
Associative Laws
Associative laws allow grouping of sets without changing the result when three sets are involved:
- A ∪ (B ∪ C) = (A ∪ B) ∪ C
- A ∩ (B ∩ C) = (A ∩ B) ∩ C
Let A = {1,2,3,4}, B = {0,2,4,6} and C = {0,4,5}.
B ∪ C = {0,2,4,5,6}. Then A ∪ (B ∪ C) = {0,1,2,3,4,5,6}.
A ∪ B = {0,1,2,3,4,6}. Then (A ∪ B) ∪ C = {0,1,2,3,4,5,6}.
Both give the same set, so the associative law of union holds.
Let A = {0,2,4,6}, B = {1,2,3,4} and C = {0,4,5}.
Step 1️⃣: Find B ∩ C = {4}.
Step 2️⃣: Now find A ∩ (B ∩ C) = A ∩ {4} = {4}.
Step 3️⃣: Find A ∩ B = {2,4}.
Step 4️⃣: Then (A ∩ B) ∩ C = {2,4} ∩ {0,4,5} = {4}.
✅ Both results match, so the associative law of intersection is verified.
Distributive Laws
The distributive laws relate union and intersection. They are important because they show how one operation distributes over the other:
- A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) — Union distributes over intersection.
- A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) — Intersection distributes over union.
Let A = {1,3,5}, B = {0,5,7}, C = {3,4,5}.
LHS: A ∪ (B ∩ C) = {1,3,5} ∪ ({0,5,7} ∩ {3,4,5}) = {1,3,5} ∪ {5} = {1,3,5}.
RHS: (A ∪ B) ∩ (A ∪ C) = ({1,3,5} ∪ {0,5,7}) ∩ ({1,3,5} ∪ {3,4,5})
= {0,1,3,5,7} ∩ {1,3,4,5} = {1,3,5}.
LHS = RHS, so union distributes over intersection.
Let A = {3,9,12}, B = {2,6,12} and C = {6,12,18}.
LHS: A ∩ (B ∪ C) = {3,9,12} ∩ ({2,6,12} ∪ {6,12,18}) = {3,9,12} ∩ {2,6,12,18} = {12}.
RHS: (A ∩ B) ∪ (A ∩ C) = ({3,9,12} ∩ {2,6,12}) ∪ ({3,9,12} ∩ {6,12,18})
= {12} ∪ {12} = {12}.
LHS = RHS, so intersection distributes over union.
Exercises — Practice
- Verify the commutative law of union and intersection for sets A = {1,2,3} and B = {2,3,4}.
- For A = {0,2,4}, B = {1,2,3}, C = {0,4,5}, verify A ∪ (B ∪ C) = (A ∪ B) ∪ C.
- For A = {0,2,4,6}, B = {1,2,3,4}, C = {0,4,5}, verify A ∩ (B ∩ C) = (A ∩ B) ∩ C.
- Verify A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) for A = {1,3,5}, B = {0,5,7}, C = {3,4,5}.
- Verify A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) for A = {3,9,12}, B = {2,6,12}, C = {6,12,18}.
🎯 Operations on Sets — Quick Quiz (Class 8)
Check your understanding of union, intersection, and distributive laws. 🌸
Q1: State the commutative laws of union and intersection.
Q2: Write one example to verify the associative law of union.
Q3: What do the distributive laws of sets show?
Q4: If A = {1,2}, B = {2,3}, find A ∪ B and A ∩ B.
Q5: True or False: (A ∪ B)′ = A′ ∪ B′
If A = {a, b} and B = {b, c}, find A ∪ B and A ∩ B.
Answer: A ∪ B = {a, b, c}, A ∩ B = {b}.
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© Math with Raabi — Class 8 Mathematics Notes | Operations on Sets explained with examples.
MashaAllah 😊
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