Algebra of SetsSets under the operations of union, intersection, and complement satisfy various laws (identities) which are listed in Table 1. Table: Law of Algebra of Sets
Table 1 shows the law of algebra of sets. Example 1: Prove Idempotent Laws:Solution: Since, B ⊂ A ∪ B, therefore A ⊂ A ∪ A Let x ∈ A ∪ A ⇒ x ∈ A or x ∈ A ⇒ x ∈ A ∴ A ∪ A ⊂ A As A ∪ A ⊂ A and A ⊂ A ∪ A ⇒ A =A ∪ A. Hence Proved. Solution: Since, A ∩ B ⊂ B, therefore A ∩ A ⊂ A Let x ∈ A ⇒ x ∈ A and x ∈ A ⇒ x ∈ A ∩ A ∴ A ⊂ A ∩ A As A ∩ A ⊂ A and A ⊂ A ∩ A ⇒ A = A ∩ A. Hence Proved. Example 2: Prove Associative Laws:Solution: Let some x ∈ (A'∪ B) ∪ C ⇒ (x ∈ A or x ∈ B) or x ∈ C ⇒ x ∈ A or x ∈ B or x ∈ C ⇒ x ∈ A or (x ∈ B or x ∈ C) ⇒ x ∈ A or x ∈ B ∪ C ⇒ x ∈ A ∪ (B ∪ C). Similarly, if some x ∈ A ∪ (B ∪ C), then x ∈ (A ∪ B) ∪ C. Thus, any x ∈ A ∪ (B ∪ C) ⇔ x ∈ (A ∪ B) ∪ C. Hence Proved. Solution: Let some x ∈ A ∩ (B ∩ C) ⇒ x ∈ A and x ∈ B ∩ C ⇒ x ∈ A and (x ∈ B and x ∈ C) ⇒ x ∈ A and x ∈ B and x ∈ C ⇒ (x ∈ A and x ∈ B) and x ∈ C) ⇒ x ∈ A ∩ B and x ∈ C ⇒ x ∈ (A ∩ B) ∩ C. Similarly, if some x ∈ A ∩ (B ∩ C), then x ∈ (A ∩ B) ∩ C Thus, any x ∈ (A ∩ B) ∩ C ⇔ x ∈ A ∩ (B ∩ C). Hence Proved. Example3: Prove Commutative LawsSolution: To Prove A ∪ B = B ∪ A A ∪ B = {x: x ∈ A or x ∈ B} = {x: x ∈ B or x ∈ A} (∵ Order is not preserved in case of sets) A ∪ B = B ∪ A. Hence Proved. Solution: To Prove A ∩ B = B ∩ A A ∩ B = {x: x ∈ A and x ∈ B} = {x: x ∈ B and x ∈ A} (∵ Order is not preserved in case of sets) A ∩ B = B ∩ A. Hence Proved. Example 4: Prove Distributive LawsSolution: To Prove Let x ∈ A ∪ (B ∩ C) ⇒ x ∈ A or x ∈ B ∩ C ⇒ (x ∈ A or x ∈ A) or (x ∈ B and x ∈ C) ⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C) ⇒ x ∈ A ∪ B and x ∈ A ∪ C ⇒ x ∈ (A ∪ B) ∩ (A ∪ C) Therefore, A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C)............(i) Again, Let y ∈ (A ∪ B) ∩ (A ∪ C) ⇒ y ∈ A ∪ B and y ∈ A ∪ C ⇒ (y ∈ A or y ∈ B) and (y ∈ A or y ∈ C) ⇒ (y ∈ A and y ∈ A) or (y ∈ B and y ∈ C) ⇒ y ∈ A or y ∈ B ∩ C ⇒ y ∈ A ∪ (B ∩ C) Therefore, (A ∪ B) ∩ (A ∪ C) ⊂ A ∪ (B ∩ C)............(ii) Combining (i) and (ii), we get A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). Hence Proved Solution: To Prove Let x ∈ A ∩ (B ∪ C) ⇒ x ∈ A and x ∈ B ∪ C ⇒ (x ∈ A and x ∈ A) and (x ∈ B or x ∈ C) ⇒ (x ∈ A and x ∈ B) or (x ∈ A and x ∈ C) ⇒ x ∈ A ∩ B or x ∈ A ∩ C ⇒ x ∈ (A ∩ B) ∪ (A ∪ C) Therefore, A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∪ C)............ (i) Again, Let y ∈ (A ∩ B) ∪ (A ∪ C) ⇒ y ∈ A ∩ B or y ∈ A ∩ C ⇒ (y ∈ A and y ∈ B) or (y ∈ A and y ∈ C) ⇒ (y ∈ A or y ∈ A) and (y ∈ B or y ∈ C) ⇒ y ∈ A and y ∈ B ∪ C ⇒ y ∈ A ∩ (B ∪ C) Therefore, (A ∩ B) ∪ (A ∪ C) ⊂ A ∩ (B ∪ C)............ (ii) Combining (i) and (ii), we get A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∪ C). Hence Proved Example 5: Prove De Morgan's Laws(a) (A ∪B)^{c}=A^{c}∩ B^{c} Solution: To Prove (A ∪B)^{c}=A^{c}∩ B^{c} Let x ∈ (A ∪B)^{c} ⇒ x ∉ A ∪ B (∵ a ∈ A ⇔ a ∉ A^{c}) ⇒ x ∉ A and x ∉ B ⇒ x ∉ A^{c} and x ∉ B^{c} ⇒ x ∉ A^{c}∩ B^{c} Therefore, (A ∪B)^{c} ⊂ A^{c}∩ B^{c}............. (i) Again, let x ∈ A^{c}∩ B^{c} ⇒ x ∈ A^{c} and x ∈ B^{c} ⇒ x ∉ A and x ∉ B ⇒ x ∉ A ∪ B ⇒ x ∈ (A ∪B)^{c} Therefore, A^{c}∩ B^{c} ⊂ (A ∪B)^{c}............. (ii) Combining (i) and (ii), we get A^{c}∩ B^{c} =(A ∪B)^{c}. Hence Proved. (b) (A ∩B)^{c} = A^{c}∪ B^{c} Solution: Let x ∈ (A ∩B)^{c} ⇒ x ∉ A ∩ B (∵ a ∈ A ⇔ a ∉ A^{c}) ⇒ x ∉ A or x ∉ B ⇒ x ∈ A^{c} and x ∈ B^{c} ⇒ x ∈ A^{c}∪ B^{c} ∴ (A ∩B)^{c}⊂ (A ∪B)^{c}.................. (i) Again, Let x ∈ A^{c}∪ B^{c} ⇒ x ∈ A^{c} or x ∈ B^{c} ⇒ x ∉ A or x ∉ B ⇒ x ∉ A ∩ B ⇒ x ∈ (A ∩B)^{c} ∴ A^{c}∪ B^{c}⊂ (A ∩B)^{c}.................... (ii) Combining (i) and (ii), we get(A ∩B)^{c}=A^{c}∪ B^{c}. Hence Proved. Example 6: Prove Identity Laws.Solution: To Prove A ∪ ∅ = A Let x ∈ A ∪ ∅ ⇒ x ∈ A or x ∈ ∅ ⇒ x ∈ A (∵x ∈ ∅, as ∅ is the null set ) Therefore, x ∈ A ∪ ∅ ⇒ x ∈ A Hence, A ∪ ∅ ⊂ A. We know that A ⊂ A ∪ B for any set B. But for B = ∅, we have A ⊂ A ∪ ∅ From above, A ⊂ A ∪ ∅ , A ∪ ∅ ⊂ A ⇒ A = A ∪ ∅. Hence Proved. Solution: To Prove A ∩ ∅ = ∅ If x ∈ A, then x ∉ ∅ (∵∅ is a null set) Therefore, x ∈ A, x ∉ ∅ ⇒ A ∩ ∅ = ∅. Hence Proved. Solution: To Prove A ∪ U = U Every set is a subset of a universal set. ∴ A ∪ U ⊆ U Also, U ⊆ A ∪ U Therefore, A ∪ U = U. Hence Proved. Solution: To Prove A ∩ U = A We know A ∩ U ⊂ A................. (i) So we have to show that A ⊂ A ∩ U Let x ∈ A ⇒ x ∈ A and x ∈ U (∵ A ⊂ U so x ∈ A ⇒ x ∈ U ) ∴ x ∈ A ⇒ x ∈ A ∩ U ∴ A ⊂ A ∩ U................. (ii) From (i) and (ii), we get A ∩ U = A. Hence Proved. Example7: Prove Complement Laws(a) A ∪ A^{c}= U Solution: To Prove A ∪ A^{c}= U Every set is a subset of U ∴ A ∪ A^{c} ⊂ U.................. (i) We have to show that U ⊆ A ∪ A^{c} Let x ∈ U ⇒ x ∈ A or x ∉ A ⇒ x ∈ A or x ∈ A^{c} ⇒ x ∈ A ∪ A^{c} ∴ U ⊆ A ∪ A^{c}................... (ii) From (i) and (ii), we get A ∪ A^{c}= U. Hence Proved. (b) A ∩ A^{c}=∅ Solution: As ∅ is the subset of every set ∴ ∅ ⊆ A ∩ A^{c}..................... (i) We have to show that A ∩ A^{c} ⊆ ∅ Let x ∈ A ∩ A^{c} ⇒ x ∈ A and x ∈ A^{c} ⇒ x ∈ A and x ∉ A ⇒ x ∈ ∅ ∴ A ∩ A^{c} ⊂∅..................... (ii) From (i) and (ii), we get A∩ A^{c}=∅. Hence Proved. (c) U^{c}= ∅ Solution: Let x ∈ U^{c} ⇔ x ∉ U ⇔ x ∈ ∅ ∴ U^{c}= ∅. Hence Proved. (As U is the Universal Set). (d) ∅^{c} = U Solution: Let x ∈ ∅^{c} ⇔ x ∉ ∅ ⇔ x ∈ U (As ∅ is an empty set) ∴ ∅^{c} = U. Hence Proved. Example8: Prove Involution Law(a) (A^{c} )^{c} A. Solution: Let x ∈ (A^{c} )^{c} ⇔ x ∉ A^{c}⇔ x ∈ a ∴ (A^{c} )^{c} =A. Hence Proved. Duality:The dual E∗ of E is the equation obtained by replacing every occurrence of ∪, ∩, U and ∅ in E by ∩, ∪, ∅, and U, respectively. For example, the dual of It is noted as the principle of duality, that if any equation E is an identity, then its dual E∗ is also an identity. Principle of Extension:According to the Principle of Extension two sets, A and B are the same if and only if they have the same members. We denote equal sets by A=B. Cartesian product of two sets:The Cartesian Product of two sets P and Q in that order is the set of all ordered pairs whose first member belongs to the set P and second member belong to set Q and is denoted by P x Q, i.e., Example: Let P = {a, b, c} and Q = {k, l, m, n}. Determine the Cartesian product of P and Q. Solution: The Cartesian product of P and Q is
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