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Misc 9 - Chapter 1 Class 11 Sets
Last updated at April 16, 2024 by Teachoo
Chapter 1 Class 11 Sets
Concept wise
- Depiction and Definition
- Depicition of sets - Roster form
- Depicition of sets - Set builder form
- Intervals
- Null Set
- Finite/Infinite
- Equal sets
- Subset
- Power Set
- Universal Set
- Venn Diagram and Union of Set
- Intersection of Sets
- Difference of sets
- Complement of set
- Number of elements in set - 2 sets (Direct)
- Number of elements in set - 2 sets - (Using properties)
- Number of elements in set - 3 sets
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Proof - Using properties of sets
- Proof - where properties of sets cant be applied,using element
Transcript
Misc 9 Let A and B be sets. If A ∩ X = B ∩ X = ∅ and A ∪ X = B ∪ X for some set X, show that A = B. (Hints: A = A ∩ (A ∪ X), B = B ∩ (B ∪ X) and use distributive law) Given: Let A and B be two sets such that A ∩ X = B ∩ X = ∅ and A ∪ X = B ∪ X for some set X. To prove: A = B Proof: Let A = A ∩ (A ∪ X) A = A ∩ (B ∪ X) Let A = A ∩ (A ∪ X) Given A ∪ X = B ∪ X A = A ∩ (B ∪ X) Using distributive law : A ∩ (B ∪ C)= (A ∩ B) ∪ (A ∩ C) = (A ∩ B) ∪ (A ∩ X) As A ∩ X = ∅ given = (A ∩ B) ∪ ∅ A = A ∩ B Let B = B ∩ (B ∪ X) Given A ∪ X = B ∪ X B = B ∩ (A ∪ X) Using distributive law: A ∪ (B ∩ C)= (A ∩ B) ∪ (A ∩ C) = (B ∩ A) ∪ (B ∩ X) As B ∩ X = Φ = (B ∩ A) ∪ Φ B = B ∩ A B = A ∩ B From (1) and (2), A = A ∩ B & B = A ∩ B ∴ A = B Hence proved
