π Miscellaneous Exercise β Chapter 1
Sets β Complete Step-by-Step Solutions
1
Decide which sets are subsets of one and another
A = {x : x β β and x satisfies xΒ² β 8x + 12 = 0}
B = {2, 4, 6}, C = {2, 4, 6, 8, β¦}, D = {6}
B = {2, 4, 6}, C = {2, 4, 6, 8, β¦}, D = {6}
π‘ Click to View Solution
1 Solve for Set A
xΒ² β 8x + 12 = 0
(x β 2)(x β 6) = 0
A = {2, 6}
(x β 2)(x β 6) = 0
A = {2, 6}
2 List All Sets
A = {2, 6}, B = {2, 4, 6}, C = {2, 4, 6, 8, β¦}, D = {6}
3 Determine Subset Relations
A β B (2, 6 both in B)
A β C (2, 6 both in C)
B β C (2, 4, 6 all in C)
D β A (6 β A)
D β B (6 β B)
D β C (6 β C)
A β C (2, 6 both in C)
B β C (2, 4, 6 all in C)
D β A (6 β A)
D β B (6 β B)
D β C (6 β C)
A β B, A β C, B β C, D β A, D β B, D β C
2
Determine whether each statement is True or False. Prove or give a counterexample.
(i) If x β A and A β B, then x β B
(ii) If A β B and B β C, then A β C
(iii) If A β B and B β C, then A β C
(iv) If A β B and B β C, then A β C
(v) If x β A and A β B, then x β B
(vi) If A β B and x β B, then x β A
(ii) If A β B and B β C, then A β C
(iii) If A β B and B β C, then A β C
(iv) If A β B and B β C, then A β C
(v) If x β A and A β B, then x β B
(vi) If A β B and x β B, then x β A
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i If x β A and A β B, then x β B
FALSE
Counterexample: Let A = {1}, B = {{1}, 2}
Clearly 1 β A and A β B, but 1 β B.
β΄ x β A and A β B does not imply x β B.
Counterexample: Let A = {1}, B = {{1}, 2}
Clearly 1 β A and A β B, but 1 β B.
β΄ x β A and A β B does not imply x β B.
ii If A β B and B β C, then A β C
FALSE
Counterexample: Let A = {1}, B = {1,2}, C = {{1,2}, 3}
A β B and B β C but A β C.
Counterexample: Let A = {1}, B = {1,2}, C = {{1,2}, 3}
A β B and B β C but A β C.
iii If A β B and B β C, then A β C
TRUE
Let x β A
β x β B [β΅ A β B]
β x β C [β΅ B β C]
β΄ x β A β x β C for all x β A, therefore A β C.
Let x β A
β x β B [β΅ A β B]
β x β C [β΅ B β C]
β΄ x β A β x β C for all x β A, therefore A β C.
iv If A β B and B β C, then A β C
FALSE
Counterexample: A = {1,2}, B = {2,3}, C = {1,2,5}
A β B (since 1 β A, 1 β B) and B β C (since 3 β B, 3 β C), but A β C.
Counterexample: A = {1,2}, B = {2,3}, C = {1,2,5}
A β B (since 1 β A, 1 β B) and B β C (since 3 β B, 3 β C), but A β C.
v If x β A and A β B, then x β B
FALSE
Counterexample: A = {1,2}, B = {2,3}
1 β A and A β B, but 1 β B.
Counterexample: A = {1,2}, B = {2,3}
1 β A and A β B, but 1 β B.
vi If A β B and x β B, then x β A
TRUE
Let A β B and x β B. Suppose x β A.
Then x β A and A β B β x β B β contradiction!
β΄ Our supposition is wrong. Hence x β A.
Let A β B and x β B. Suppose x β A.
Then x β A and A β B β x β B β contradiction!
β΄ Our supposition is wrong. Hence x β A.
3
Let A βͺ B = A βͺ C and A β© B = A β© C. Show that B = C.
π‘ Click to View Solution
1 Prove B β C
Let x β B
β x β A βͺ B [β΅ B β A βͺ B always]
β x β A βͺ C [β΅ A βͺ B = A βͺ C]
β x β A or x β C
Case I: x β A
Then x β A and x β B β x β A β© B β x β A β© C β x β C
Case II: x β C
In each case x β C. β΄ B β C ...(i)
β x β A βͺ B [β΅ B β A βͺ B always]
β x β A βͺ C [β΅ A βͺ B = A βͺ C]
β x β A or x β C
Case I: x β A
Then x β A and x β B β x β A β© B β x β A β© C β x β C
Case II: x β C
In each case x β C. β΄ B β C ...(i)
2 Prove C β B (Similarly)
By symmetry (interchanging B and C in the argument above),
C β B ...(ii)
C β B ...(ii)
From (i) and (ii): B = C
4
Show that the following four conditions are equivalent: (i) A β B (ii) A β B = Ο (iii) A βͺ B = B (iv) A β© B = A
π‘ Click to View Solution
1 (i) βΉ (ii): A β B βΉ A β B = Ο
Suppose A β B β Ο. Let x β A β B β x β A and x β B.
But A β B β x β B β contradiction!
β΄ A β B = Ο
But A β B β x β B β contradiction!
β΄ A β B = Ο
2 (i) βΉ (iii): A β B βΉ A βͺ B = B
Let x β A βͺ B β x β A or x β B β x β B or x β B (since A β B)
β x β B β A βͺ B β B
Also B β A βͺ B always. β΄ A βͺ B = B
β x β B β A βͺ B β B
Also B β A βͺ B always. β΄ A βͺ B = B
3 (i) βΉ (iv): A β B βΉ A β© B = A
We know A β© B β A always.
Now let x β A β x β B [β΅ A β B] β x β A β© B
β΄ A β A β© B. Therefore A β© B = A
Now let x β A β x β B [β΅ A β B] β x β A β© B
β΄ A β A β© B. Therefore A β© B = A
π‘ Key Insight: All four conditions are logically equivalent β proving any one of them from A β B establishes all four.
5
Show that if A β B, then C β B β C β A.
π‘ Click to View Solution
1 Start with an element of C β B
Given: A β B
Let x β (C β B)
β x β C and x β B
Let x β (C β B)
β x β C and x β B
2 Use A β B to conclude
Since A β B and x β B β x β A
β΄ x β C and x β A β x β (C β A)
β΄ x β C and x β A β x β (C β A)
β΄ (C β B) β (C β A) β
6
Assume that P(A) = P(B). Show that A = B.
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1 Prove A β B
Let x β A
β {x} β A β {x} β P(A)
β {x} β P(B) [β΅ P(A) = P(B)]
β {x} β B β x β B
β΄ A β B
β {x} β A β {x} β P(A)
β {x} β P(B) [β΅ P(A) = P(B)]
β {x} β B β x β B
β΄ A β B
2 Prove B β A (Similarly)
By identical argument (interchanging A and B):
B β A
B β A
Since A β B and B β A, we have A = B β
7
Is it true that for any sets A and B, P(A) βͺ P(B) = P(A βͺ B)? Justify your answer.
π‘ Click to View Solution
1 Take a Counterexample
Let A = {1}, B = {2}
P(A) = {Ο, {1}} and P(B) = {Ο, {2}}
P(A) βͺ P(B) = {Ο, {1}, {2}}
P(A) = {Ο, {1}} and P(B) = {Ο, {2}}
P(A) βͺ P(B) = {Ο, {1}, {2}}
2 Compute P(A βͺ B)
A βͺ B = {1, 2}
P(A βͺ B) = {Ο, {1}, {2}, {1,2}}
P(A βͺ B) = {Ο, {1}, {2}, {1,2}}
3 Compare
{1, 2} β P(A βͺ B) but {1, 2} β P(A) βͺ P(B)
β΄ P(A) βͺ P(B) β P(A βͺ B)
β΄ P(A) βͺ P(B) β P(A βͺ B)
No, it is NOT true in general. P(A) βͺ P(B) β P(A βͺ B)
8
Show that for any sets A and B: A = (A β© B) βͺ (A β B) and A βͺ (B β A) = A βͺ B
π‘ Click to View Solution
1 Prove A = (A β© B) βͺ (A β B)
(A β© B) βͺ (A β B) = (A β© B) βͺ (A β© Bβ²)
= A β© (B βͺ Bβ²) [Distributive Law]
= A β© U = A
= A β© (B βͺ Bβ²) [Distributive Law]
= A β© U = A
2 Prove A βͺ (B β A) = A βͺ B
A βͺ (B β A) = A βͺ (B β© Aβ²)
= (A βͺ B) β© (A βͺ Aβ²) [Distributive Law]
= (A βͺ B) β© U = A βͺ B
= (A βͺ B) β© (A βͺ Aβ²) [Distributive Law]
= (A βͺ B) β© U = A βͺ B
Both identities are proved using Distributive Law β
9
Using properties of sets, show that: (i) A βͺ (A β© B) = A (ii) A β© (A βͺ B) = A
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1 Prove A βͺ (A β© B) = A
We know: X β Y β X βͺ Y = Y (the superset)
Here A β© B β A
β΄ A βͺ (A β© B) = A
Here A β© B β A
β΄ A βͺ (A β© B) = A
2 Prove A β© (A βͺ B) = A
We know: X β Y β X β© Y = X (the subset)
Here A β A βͺ B
β΄ A β© (A βͺ B) = A
Here A β A βͺ B
β΄ A β© (A βͺ B) = A
π‘ Key Insight: These are called Absorption Laws β A "absorbs" any set obtained by intersecting or unioning with itself.
10
Show that A β© B = A β© C need not imply B = C.
π‘ Click to View Solution
1 Counterexample
Let A = {1, 2, 3}, B = {2, 4}, C = {2, 5}
A β© B = {2} and A β© C = {2}
β΄ A β© B = A β© C = {2}
But B = {2, 4} β {2, 5} = C
A β© B = {2} and A β© C = {2}
β΄ A β© B = A β© C = {2}
But B = {2, 4} β {2, 5} = C
A β© B = A β© C does NOT necessarily imply B = C β
11
If A β© X = B β© X = Ο and A βͺ X = B βͺ X for some set X, show that A = B.
π‘ Click to View Solution
1 Express A using given conditions
A β© (A βͺ X) = A [β΅ A β A βͺ X always]
= A β© (B βͺ X) [β΅ A βͺ X = B βͺ X]
= (A β© B) βͺ (A β© X) [Distributive Law]
= (A β© B) βͺ Ο [β΅ A β© X = Ο]
= A β© B
β΄ A = A β© B ...(i)
= A β© (B βͺ X) [β΅ A βͺ X = B βͺ X]
= (A β© B) βͺ (A β© X) [Distributive Law]
= (A β© B) βͺ Ο [β΅ A β© X = Ο]
= A β© B
β΄ A = A β© B ...(i)
2 Interchange A and B
By the same argument (interchanging A and B):
B = B β© A = A β© B ...(ii) [Commutative Law]
B = B β© A = A β© B ...(ii) [Commutative Law]
From (i) and (ii): A = A β© B = B β΄ A = B β
12
Find sets A, B and C such that A β© B, B β© C and A β© C are non-empty but A β© B β© C = Ο.
π‘ Click to View Solution
1 Choose the Sets
Let A = {x, y}, B = {x, z}, C = {y, z}
2 Verify Pairwise Intersections
A β© B = {x} β Ο β
B β© C = {z} β Ο β
A β© C = {y} β Ο β
B β© C = {z} β Ο β
A β© C = {y} β Ο β
3 Verify Triple Intersection
A β© B β© C = {x,y} β© {x,z} β© {y,z} = Ο β
A = {x, y}, B = {x, z}, C = {y, z} satisfies all conditions β
13
In a survey of 600 students, 150 drink tea, 225 drink coffee, 100 drink both. Find how many drink neither.
π‘ Click to View Solution
1 Define the Sets
Total students = 600
n(T) = 150, n(C) = 225, n(T β© C) = 100
n(T) = 150, n(C) = 225, n(T β© C) = 100
2 Apply Union Formula
n(T βͺ C) = n(T) + n(C) β n(T β© C)
= 150 + 225 β 100 = 275
= 150 + 225 β 100 = 275
3 Find Neither
Neither = Total β n(T βͺ C)
= 600 β 275 = 325
= 600 β 275 = 325
325 students drink neither tea nor coffee
14
In a group, 100 know Hindi, 50 know English, 25 know both. Each knows at least one. How many in the group?
π‘ Click to View Solution
1 Define and Apply Formula
n(H) = 100, n(E) = 50, n(H β© E) = 25
n(H βͺ E) = n(H) + n(E) β n(H β© E)
= 100 + 50 β 25 = 125
n(H βͺ E) = n(H) + n(E) β n(H β© E)
= 100 + 50 β 25 = 125
There are 125 students in the group
15
Survey of 60 people: H=25, T=26, I=26, Hβ©I=9, Hβ©T=11, Tβ©I=8, Hβ©Tβ©I=3. Find (i) at least one newspaper (ii) exactly one newspaper.
π‘ Click to View Solution
1 Given Data
n(H)=25, n(T)=26, n(I)=26
n(Hβ©T)=11, n(Hβ©I)=9, n(Tβ©I)=8
n(Hβ©Tβ©I)=3
n(Hβ©T)=11, n(Hβ©I)=9, n(Tβ©I)=8
n(Hβ©Tβ©I)=3
2 (i) At Least One Newspaper
n(HβͺTβͺI) = n(H)+n(T)+n(I) β n(Hβ©T) β n(Hβ©I) β n(Tβ©I) + n(Hβ©Tβ©I)
= 25+26+26 β 11 β 9 β 8 + 3 = 52
= 25+26+26 β 11 β 9 β 8 + 3 = 52
3 (ii) Exactly One Newspaper
Only H = n(H) β n(Hβ©T) β n(Hβ©I) + n(Hβ©Tβ©I) = 25β11β9+3 = 8
Only T = n(T) β n(Tβ©H) β n(Tβ©I) + n(Hβ©Tβ©I) = 26β11β8+3 = 10
Only I = n(I) β n(Iβ©H) β n(Iβ©T) + n(Hβ©Tβ©I) = 26β9β8+3 = 12
Total = 8 + 10 + 12 = 30
Only T = n(T) β n(Tβ©H) β n(Tβ©I) + n(Hβ©Tβ©I) = 26β11β8+3 = 10
Only I = n(I) β n(Iβ©H) β n(Iβ©T) + n(Hβ©Tβ©I) = 26β9β8+3 = 12
Total = 8 + 10 + 12 = 30
(i) 52 people read at least one newspaper
(ii) 30 people read exactly one newspaper
(ii) 30 people read exactly one newspaper
16
21 liked A, 26 liked B, 29 liked C. 14 liked A&B, 12 liked C&A, 14 liked B&C, 8 liked all three. Find how many liked C only.
π‘ Click to View Solution
1 Define Variables
Let d = people who liked all three = 8
b = people who liked B and C only (not A)
c = people who liked C and A only (not B)
b = people who liked B and C only (not A)
c = people who liked C and A only (not B)
2 Solve for b and c
b + d = 14 β b = 14 β 8 = 6
c + d = 12 β c = 12 β 8 = 4
c + d = 12 β c = 12 β 8 = 4
3 Find C Only
n(Z) = 29 (total who liked C)
z + b + c + d = 29
z + 6 + 4 + 8 = 29
z = 29 β 18 = 11
z + b + c + d = 29
z + 6 + 4 + 8 = 29
z = 29 β 18 = 11
11 people liked product C only
