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

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: \( (A \cup B)’ = A’ \cap B’ \).