If x ∈ A and A ∈ B , then x ∈ B .

Let's take example A = { 1 , 2 } and B = { 1 , { 1 , 2 } , { 3 } } . Now, 2 ∈ { 1 , 2 } and { 1 , 2 } ∈ { { 3 } , 1 , { 1 , 2 } } . ∴ A ⊂ B However, 2 ∉ { { 3 } , 1 , { 1 , 2 } } Therefore, if x ∈ A , A ∈ B , then x may or may not belong to B .

Show that the following four conditions are equivalent: (I) A ⊂ B (ii) A - B = ϕ (iii) A ∪ B = B (iv) A ∩ B = A

Let us first show that i is equivalent to i i Let A ⊂ B To prove: A - B = ϕ Let us suppose A - B ≠ ϕ This means that there exists a x such that x ∈ A but x ∉ B which is not possible as A ⊂ B . ∴ A - B = ϕ ∴ A ⊂ B ⇒ A - B = ϕ Let A - B = 0 To show: A ⊂ B Let x ∈ A Clearly, x ∈ B because if x ∉ B , then A - B ≠ ϕ ∴ A - B = ϕ ⇒ A ⊂ B ∴ i ⇔ i i Let A ⊂ B To show: A ∪ B = B Clearly, B ⊂ A ∪ B Let x ∈ A ∪ B ⇒ x ∈ A or x ∈ B Case 1 : x ∈ A ⇒ x ∈ B ∵ A ⊂ B ∴ A ∪ B ⊂ B Case 2 : x ∈ B Then, A ∪ B = B Conversely, let A ∪ B = B Let x ∈ A ⇒ x ∈ A ∪ B ∵ A ⊂ A ∪ B ⇒ x ∈ B ∵ A ∪ B = B ∴ A ⊂ B Therefore, i ⇔ i i i Now, we have to show that i ⇔ i v Let A ⊂ B Clearly A ∩ B ⊂ A Let x ∈ A We have to show that x ∈ A ∩ B As A ⊂ B , x ∈ B ∴ x ∈ A ∩ B ∴ A ⊂ A ∩ B Therefore, A = A ∩ B Conversely, suppose A ∩ B = A Let x ∈ A ⇒ x ∈ A ∩ B ⇒ x ∈ A and x ∈ B ∴ A ⊂ B Therefore, i ⇔ i v Hence, i ⇔ i i ⇔ i i i ⇔ i v

Show that the following four conditions are equivalent: A ⊂ B A - B = ϕ A ∪ B = B A ∩ B = A

Let us first show that (i) is equivalent to (ii) Let A ⊂ B To prove: A - B = ϕ Let us suppose A - B ≠ ϕ This means that we have an element x such that x ∈ A but x ∉ B which is not possible as A ⊂ B . ∴ A - B = ϕ ∴ A ⊂ B ⇒ A - B = ϕ Conversely, to show (ii) ⇒ (i) Let A - B = φ To show: A ⊂ B Let x ∈ A Clearly, x ∈ B because if x ∉ B , then A - B ≠ ϕ ∴ A - B = ϕ ⇒ A ⊂ B ∴ (i) ⇔ (ii) To show (i) ⇒ (iii) Let A ⊂ B To show: A ∪ B = B Clearly, B ⊂ A ∪ B Let x ∈ A ∪ B ⇒ x ∈ A or x ∈ B Case I: x ∈ A ⇒ x ∈ B ∵ A ⊂ B ∴ A ∪ B ⊂ B Case II: x ∈ B then A ∪ B ⊂ B Thus we have A ∪ B ⊂ B and B ∪ A ⊂ A Therefore, A ∪ B = B Conversely, let A ∪ B = B To show A ⊂ B Let x ∈ A ⇒ x ∈ A ∪ B ∵ A ⊂ A ∪ B ⇒ x ∈ B ∵ A ∪ B = B So for x ∈ A we have x ∈ B ∴ A ⊂ B Therefore, (i) ⇔ (iii) Now, we have to show that (i) ⇔ (iv) Let A ⊂ B To show: A ∩ B = A Clearly A ∩ B ⊂ A We have to show that A ⊂ A ∩ B Let x ∈ A As A ⊂ B , x ∈ B ∴ x ∈ A ∩ B ∴ A ⊂ A ∩ B Therefore, A = A ∩ B Conversely, suppose A ∩ B = A Let x ∈ A ⇒ x ∈ A ∩ B ⇒ x ∈ A and x ∈ B ⇒ x ∈ B ∴ A ⊂ B Therefore, (i) ⇔ (iv).

Show that the following four conditions are equivalent: A ⊂ B A - B = ϕ A ∪ B = B A ∩ B = A

Let us first show that (i) is equivalent to (ii) Let A ⊂ B To prove: A - B = ϕ Let us suppose A - B ≠ ϕ This means that there exists a x such that x ∈ A but x ∉ B which is not possible as A ⊂ B . ∴ A - B = ϕ ∴ A ⊂ B ⇒ A - B = ϕ Let A - B = ϕ To show: A ⊂ B Let x ∈ A Clearly, x ∈ B because if x ∉ B , then A - B ≠ ϕ ∴ A - B = ϕ ⇒ A ⊂ B ∴ (i) ⇔ (ii) Let A ⊂ B To show: A ∪ B = B Clearly, B ⊂ A ∪ B Let x ∈ A ∪ B ⇒ x ∈ A or x ∈ B Case I: x ∈ A ⇒ x ∈ B ∵ A ⊂ B ∴ A ∪ B ⊂ B Case II: x ∈ B Then, A ∪ B = B Conversely, let A ∪ B = B Let x ∈ A ⇒ x ∈ A ∪ B ∵ A ⊂ A ∪ B ⇒ x ∈ B ∵ A ∪ B = B ∴ A ⊂ B Therefore, (i) ⇔ (iii) Now, we have to show that (i) ⇔ (iv) Let A ⊂ B Clearly A ∩ B ⊂ A Let x ∈ A We have to show that x ∈ A ∩ B As A ⊂ B , x ∈ B ∴ x ∈ A ∩ B ∴ A ⊂ A ∩ B Therefore, A = A ∩ B Conversely, suppose A ∩ B = A Let x ∈ A ⇒ x ∈ A ∩ B ⇒ x ∈ A and x ∈ B ⇒ x ∈ B ∴ A ⊂ B Therefore, (i) ⇔ (iv).

Show that for any sets A and B , A = ( A ∩ B ) ∪ ( A - B ) and A ∪ ( B - A ) = ( A ∪ B )

The union of two sets A and B is the set C which consists of all those elements which are either in A or in B . The intersection of sets A and B is the set of all elements which are common to both A a n d B .The symbol ' ∩ ' is used to represent intersection. The difference of the sets A and B in this order is the set of elements which belong to A but not to B . Let U = 1 , 2 , 3 , 4 , 5 , A = 1 , 2 , B = 2 , 3 , 4 Now, A ∩ B = 1 , 2 ∩ 2 , 3 , 4 = 2 A - B = 1 , 2 - 2 , 3 , 4 = 1 ∴ ( A ∩ B ) ∪ ( A - B ) = 2 ∪ 1 = 1 , 2 = A Hence, proved. Again, L H S = A ∪ ( B - A ) = 1 , 2 ∪ ( 2 , 3 , 4 - 1 , 2 ) = 1 , 2 ∪ 3 , 4 = 1 , 2 , 3 , 4 R H S = A ∪ B = 1 , 2 ∪ 2 , 3 , 4 = 1 , 2 , 3 , 4 L H S = R H S Hence, proved.

E M B I B E

MATHEMATICS Textbook for Class XI>Chapter 1 - Sets>Miscellaneous Exercise on Chapter 1

Mizoram Board Solutions for Chapter: Sets, Exercise 7: Miscellaneous Exercise on Chapter 1

Author:Mizoram Board

Mizoram Board Mathematics Solutions for Exercise - Mizoram Board Solutions for Chapter: Sets, Exercise 7: Miscellaneous Exercise on Chapter 1

Attempt the practice questions on Chapter 1: Sets, Exercise 7: Miscellaneous Exercise on Chapter 1 with hints and solutions to strengthen your understanding. MATHEMATICS Textbook for Class XI solutions are prepared by Experienced Embibe Experts.