IIT JEE Main Maths -Unit 1- Algebraic properties of sets- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 1- Algebraic properties of sets – Study Notes – New syllabus
IIT JEE Main Maths -Unit 1- Algebraic properties of sets – Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Algebraic Properties of Sets
- De Morgan’s Laws
- Number of Elements in Sets
Algebraic Properties of Sets
| Property | Symbolic Form | Meaning |
|---|---|---|
| Commutative Laws | \( A \cup B = B \cup A \), \( A \cap B = B \cap A \) | Order of sets does not matter in union or intersection |
| Associative Laws | \( A \cup (B \cup C) = (A \cup B) \cup C \) \( A \cap (B \cap C) = (A \cap B) \cap C \) | Grouping of sets does not affect result |
| Distributive Laws | \( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \) \( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \) | Union and intersection distribute over each other |
| Identity Laws | \( A \cup \varnothing = A \), \( A \cap U = A \) | Union with empty set or intersection with universal set gives same set |
| Domination Laws | \( A \cup U = U \), \( A \cap \varnothing = \varnothing \) | Union with universal set or intersection with empty set dominates the result |
| Idempotent Laws | \( A \cup A = A \), \( A \cap A = A \) | Combining a set with itself gives same set |
| Complement Laws | \( A \cup A’ = U \), \( A \cap A’ = \varnothing \) | Union of a set with its complement gives universal set; intersection gives empty set |
| Double Complement Law | \( (A’)’ = A \) | Complement of complement is original set |
| De Morgan’s Laws | \( (A \cup B)’ = A’ \cap B’ \) \( (A \cap B)’ = A’ \cup B’ \) | Complement of union is intersection of complements and vice versa |
Example
Verify the commutative law of union and intersection for \( A = \{1, 2, 3\} \) and \( B = \{3, 4, 5\} \).
▶️ Answer / Explanation
\( A \cup B = \{1, 2, 3, 4, 5\} \)
\( B \cup A = \{3, 4, 5, 1, 2\} = \{1, 2, 3, 4, 5\} \)
So, \( A \cup B = B \cup A \)
Similarly, \( A \cap B = \{3\} \) and \( B \cap A = \{3\} \)
Hence, commutative law is verified.
Example
Prove the distributive law \( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \) for the sets \( A = \{1, 2\}, B = \{2, 3\}, C = \{3, 4\} \).
▶️ Answer / Explanation
Step 1: \( B \cap C = \{3\} \)
\( A \cup (B \cap C) = \{1, 2, 3\} \)
Step 2: \( A \cup B = \{1, 2, 3\} \), \( A \cup C = \{1, 2, 3, 4\} \)
\( (A \cup B) \cap (A \cup C) = \{1, 2, 3\} \)
Conclusion: Both sides are equal, so the distributive law holds.
Example
Using De Morgan’s Law, find \( (A \cup B)’ \) when \( U = \{1, 2, 3, 4, 5, 6\} \), \( A = \{1, 2, 3\} \), \( B = \{3, 4, 5\} \).
▶️ Answer / Explanation
\( A \cup B = \{1, 2, 3, 4, 5\} \)
\( (A \cup B)’ = U – (A \cup B) = \{6\} \)
Using De Morgan’s Law: \( (A \cup B)’ = A’ \cap B’ \)
\( A’ = \{4, 5, 6\}, B’ = \{1, 2, 6\} \)
\( A’ \cap B’ = \{6\} \)
Hence verified: \( (A \cup B)’ = A’ \cap B’ \).
De Morgan’s Laws
De Morgan’s Laws are important identities in set theory that relate the complement of unions and intersections of sets.
They are widely used in simplifying set expressions and solving problems involving complements.
First Law:
\( (A \cup B)’ = A’ \cap B’ \)
Meaning: The complement of the union of two sets is equal to the intersection of their complements.
Explanation: An element belongs to \( (A \cup B)’ \) if it is not in \( A \) and not in \( B \).
This is exactly the condition for being in both \( A’ \) and \( B’ \).
Second Law:
\( (A \cap B)’ = A’ \cup B’ \)
Meaning: The complement of the intersection of two sets is equal to the union of their complements.
Explanation: An element belongs to \( (A \cap B)’ \) if it is not common to both sets.
This means it lies in at least one of \( A’ \) or \( B’ \).
Important Notes:
- These laws are always defined with respect to a universal set \( U \).
- They are very useful in simplifying complex set expressions.
- They can be extended to three or more sets.
Extension to Three Sets:
- \( (A \cup B \cup C)’ = A’ \cap B’ \cap C’ \)
- \( (A \cap B \cap C)’ = A’ \cup B’ \cup C’ \)
Example 1
If \( U = \{1,2,3,4,5,6\} \), \( A = \{1,2,3\} \), \( B = \{3,4,5\} \), verify that \( (A \cup B)’ = A’ \cap B’ \).
▶️ Answer / Explanation
\( A \cup B = \{1,2,3,4,5\} \)
\( (A \cup B)’ = \{6\} \)
\( A’ = \{4,5,6\}, \quad B’ = \{1,2,6\} \)
\( A’ \cap B’ = \{6\} \)
Hence, LHS = RHS
Example 2
Verify \( (A \cap B)’ = A’ \cup B’ \) for the same sets.
▶️ Answer / Explanation
\( A \cap B = \{3\} \)
\( (A \cap B)’ = \{1,2,4,5,6\} \)
\( A’ = \{4,5,6\}, \quad B’ = \{1,2,6\} \)
\( A’ \cup B’ = \{1,2,4,5,6\} \)
Hence, verified
Example 3
Simplify \( (A \cup B)’ \cap A \).
▶️ Answer / Explanation
Using De Morgan’s Law:
\( (A \cup B)’ = A’ \cap B’ \)
So expression becomes:
\( (A’ \cap B’) \cap A \)
\( = (A’ \cap A) \cap B’ \)
\( A’ \cap A = \varnothing \)
Therefore, result \( = \varnothing \)
Number of Elements in Sets
This topic deals with finding the number of elements in unions and intersections of sets. It is one of the most important and frequently asked concepts in JEE Main.
Basic Idea:
When combining two or more sets, some elements may be counted more than once. Therefore, we subtract the common elements to avoid repetition.
Formula for Two Sets:
\( n(A \cup B) = n(A) + n(B) – n(A \cap B) \)
Explanation:
- Elements of \( A \) and \( B \) are added.
- Common elements \( (A \cap B) \) are counted twice, so we subtract them once.
Formula for Three Sets:
\( n(A \cup B \cup C) = n(A) + n(B) + n(C) \)
\( – n(A \cap B) – n(B \cap C) – n(C \cap A) + n(A \cap B \cap C) \)
Explanation:
- Add individual sets.
- Subtract pairwise intersections (counted twice).
- Add triple intersection (subtracted too many times).
Special Cases:
- If \( A \) and \( B \) are disjoint: \( n(A \cap B) = 0 \)
- Then \( n(A \cup B) = n(A) + n(B) \)
Important Notes:
- Always identify overlapping elements carefully.
- Use Venn diagrams for better understanding.
- This concept is based on the inclusion–exclusion principle.
Example 1
In a class of 40 students, 25 like Mathematics, 20 like Physics, and 10 like both. Find how many students like at least one subject.
▶️ Answer / Explanation
Use formula:
\( n(A \cup B) = n(A) + n(B) – n(A \cap B) \)
\( = 25 + 20 – 10 = 35 \)
So, 35 students like at least one subject.
Example 2
In a group of 100 people, 60 like tea, 50 like coffee, and 30 like both. Find how many like neither.
▶️ Answer / Explanation
\( n(A \cup B) = 60 + 50 – 30 = 80 \)
People who like neither:
\( 100 – 80 = 20 \)
Example 3
In a survey of 100 students, 50 study Maths, 60 study Physics, 40 study Chemistry, 20 study Maths & Physics, 25 study Physics & Chemistry, 15 study Maths & Chemistry, and 10 study all three. Find how many study at least one subject.
▶️ Answer / Explanation
Use three-set formula:
\( n(A \cup B \cup C) = 50 + 60 + 40 \)
\( – 20 – 25 – 15 + 10 \)
\( = 150 – 60 + 10 = 100 \)
So, all 100 students study at least one subject.