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Chapter 1. Relations and Functions
Exercise 1.1
1. Find A × B, A × A and B × A
(i) A = {2, -2, 3} and B = {1, -4}
(ii) A = B = {p, q}
(iii) A – {m, n} ; B = Φ
Solution:
(i) A = {2, -2, 3} and B = {1, -4}
A × B = {2, -2, 3} × {1, -4}
= {(2, 1), (2, -4), (-2, 1), (-2, -4), (3, 1), (3,-4)}
A × A = {2, -2, 3} × {2, -2, 3}
= { (2, 2), (2, -2 ), (2, 3)
(-2, 2), (-2, -2), (-2, 3)
(3, 2), (3, -2), (3, 3) }
B × A = {1,-4} × {2,-2, 3}
= {(1, 2), (1, -2), ( 1, 3) (-4, 2), (-4, -2), (-4, 3)}
(ii) A = B = {p, q}
A × B = {p, q) × {p, q}
= {(p, p), (p, q) (q, p) (q, q)}
A × A = {p, q) × (p, q)
= {(p, p) (p, q) (q ,p) (q, q)
B × A = {p, q} × {p, q}
= {(p, p) (p, q) (q, p) (q, q)
(iii) A = {m, n} × B = Φ
Note: B = Φ or {}
A × B = {m, n) × { }
= { }
A × A = {m, n) × (m, n)}
= {(m, m), (m, n), (n, m), (n, n)}
B × A = { } × {m, n}
= { }
Question 2. Let A = {1, 2, 3} and B = {x | x is a prime number less than 10}. Find A × B and B × A.
Solution:
A = {1, 2, 3}, B = {2, 3, 5, 7}
A × B = { (1, 2), (1, 3), (1, 5), (1, 7),
(2, 2), (2, 3) , (2, 5), (2, 7),
(3, 2), (3, 3), (3, 5), (3, 7)}
B × A = {(2, 1), (2, 2), (2, 3),
(3, 1), (3, 2), (3, 3) ,
(5, 1), (5, 2), (5, 3),
(7, 1), (7, 2) , (7, 3)}
Question 3. If B × A = {(-2, 3),(-2, 4),(0, 3),(0, 4), (3,3) ,(3, 4)} find A and B.
Solution:
B × A = {(-2, 3) (-2, 4) (0, 3) (0, 4) (3, 3) (3,4)}
A = {3, 4}
B = {-2, 0, 3}
Question 4. If A ={5, 6}, B = {4, 5, 6} , C = {5, 6, 7}, Show that A × A = (B × B) ∩ (C × C).
Solution:
A = {5,6}, B = {4, 5, 6}, C = {5, 6, 7}
A × A = {(5, 5), (5, 6), (6, 5), (6, 6)} ……….. (1) (LHS)
B × B = {(4, 4), (4, 5), (4, 6), (5, 4),
(5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} …............ (2)
C × C = {(5, 5), (5, 6), (5, 7), (6, 5), (6, 6),
(6, 7), (7, 5), (7, 6), (7, 7)} …........................ (3)
(B × B) ∩ (C × C) = {(5, 5), (5,6), (6, 5), (6,6)} ….... (4) RHS
- (1) = (4)
- A × A = (B × B) ∩ (C × C)
- LHS = RHS - It is proved.
Question 5. Given A = {1,2,3}, B = {2,3,5}, C = {3,4} and D = {1,3,5}, check if (A ∩ C) × (B ∩ D) = (A × B) ∩ (C × D) is true?
Solution:
A = {1,2, 3}, B = {2, 3, 5}, C = {3,4} D = {1,3,5}
LHS : (A ∩ C) × (B ∩ D)
A ∩ C = {1, 2, 3} ∩ {3, 4} = (3}
B ∩ D = {2, 3, 5} ∩ {1, 3, 5} = = {3,5}
(A ∩ C) × (B ∩ D) = {3} × {3,5}
(A ∩ C) × (B ∩ D) = {(3, 3) (3, 5)} …. (1) - LHS
RHS : (A × B) ∩ (C × D)
A × B = {1,2,3} × {2,3,5}
= {(1,2) (1,3) (1,5) (2, 2) (2, 3) (2, 5), (3, 2) (3, 3) (3, 5)}
C × D = {3,4} × {1,3,5}
= {(3,1) (3, 3) (3, 5) (4,1) (4, 3) (4, 5)}
(A × B) ∩ (C × D) = {(3, 3) (3, 5)} ….(2) - RHS
- From (1) and (2) we get
- (A ∩ C) × (B ∩ D) = (A × B) ∩ (C × D)
- {(3, 3) (3, 5)} = {(3, 3) (3, 5)}
- It is True
Question 6. Let A = {x ∈ W | x < 2}, B = {x ∈ N |1 < x ≤ 4} and C = {3, 5}. Verify that
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
(iii) (A ∪ B) × C = (A × C) ∪ (B × C)
Solution:
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
A = {x ∈ W|x < 2} = {0,1}
B = {x ∈ N |1 < x ≤ 4} = {2,3,4}
C = {3,5}
∪ is combination of two sets.
B∪C= {2,3,4,5}
Now, L.H.S=A × (B∪C)
={0, 1} × (2, 3, 4, 5}
={(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)} ---- ( 1 )
R.H.S=(A×B)∪(A×C)
={0,1}×{2,3,4}∪{0,1}×{3,5}
={(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)}∪{(0,3),(0,5),(1,3),(1,5)}
={(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)} --- ( 2 )
- From ( 1 ) and ( 2 )
- ⇒ L.H.S=R.H.S
- ∴ A×(B∪C)=(A×B)∪(A×C) Hence It is Verified
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
LHS = A × (B ∩ C)
(B ∩ C) = {3}
A × (B ∩ C) = {(0, 3), (1, 3)} …(1)
RHS = (A × B) ∩ (A × C)
(A × B) = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
(A × C) = {(0, 3), (0, 5), (1, 3), (1, 5)}
(A × B) ∩ (A × C) = {(0, 3), (1, 3)} ……….. (2)
- (1) = (2) ⇒ LHS = RHS.
- A × (B ∩ C) = (A × B) ∩ (A × C)
- Hence it is verified.
(iii) (A ∪ B) × C = (A × C) ∪ (B × C)
LHS = (A ∪ B) × C
A ∪ B = {0, 1, 2, 3, 4}
(A ∪ B) × C = {(0, 3), (0, 5), (1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)} …………. (1)
RHS = (A × C) ∪ (B × C)
(A × C) = {(0, 3), (0, 5), (1, 3), (1, 5)}
(B × C) = {(2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)}
(A × C) ∪ (B × C) = {(0, 3), (0, 5), (1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)} ………… (2)
- (1) = (2)
- ∴ LHS = RHS. Hence it is verified.
Question 7.
Let A = The set of all natural numbers less than 8,
B = The set of all prime numbers less than 8,
C = The set of even prime number. Verify that
(i) (A ∩ B) × C = (A × C) ∩ (B × C)
(ii) A × (B – C) = (A × B) – (A × C)
Solution:
A = {1,2, 3, 4, 5,6, 7}
B = {2, 3, 5,7}
C = {2}
(i) (A ∩ B) × C = (A × C) ∩ (B × C)
A ∩ B = {1, 2, 3, 4, 5, 6, 7} ∩ {2, 3, 5, 7}
= {2, 3, 5, 7}
(A ∩ B) × C = {2, 3, 5, 7} × {2}
= {(2, 2) (3, 2) (5, 2) (7, 2)} ….(1) (LHS)
A × C = {1,2, 3, 4, 5, 6, 7} × {2}
= {(1,2) (2, 2) (3, 2) (4, 2)
(5.2) (6, 2) (7, 2)}
B × C = {2, 3, 5, 7} × {2}
= {(2, 2) (3, 2) (5, 2) (7, 2)}
(A × C) ∩ (B × C) = {(2, 2) (3, 2) (5, 2) (7, 2)} ….(2) (RHS)
- From (1) and (2) we get
- (A ∩ B) × C = (A × C) ∩ (B × C)
(ii) A × (B – C) = (A × B) – (A × C)
B – C = {2, 3, 5, 7} – {2}
= {3,5,7}
A × (B – C) = {1,2, 3, 4, 5, 6,7} × {3,5,7}
= {(1, 3) (1, 5) (1, 7) (2, 3) (2, 5)
(2, 7) (3, 3) (3, 5) (3, 7) (4, 3)
(4, 5) (4, 7) (5, 3) (5, 5) (5, 7)
(6, 3) (6, 5) (6, 7) (7, 3) (7, 5) (7, 7)} ………….(1)
A × B = {1,2, 3, 4, 5, 6, 7} × {2, 3, 5,7}
= {(1, 2) (1, 3) (1, 5) (1, 7) (2, 2) (2, 3)
(2, 5) (2, 7) (3, 2) (3, 3) (3, 5) (3, 7)
(4, 2) (4, 3) (4, 5) (4, 7) (5, 2) (5, 3) (5, 5)
(5, 7) (6, 2) (6, 3) (6, 5) (6, 7)
(7, 2) (7, 3) (7, 5) (7, 7)}
A × C = {1,2, 3,4, 5, 6, 7} × {2}
= {(1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6.2) (7,2)}
(A × B) – (A × C) = {(1, 3) (1, 5) (1, 7)
(2, 3) (2, 5) (2, 7) (3, 3) (3, 5)
(3, 7) (4, 3) (4, 5) (4, 7) (5, 3) (5, 5)
(5, 7) (6, 3) (6, 5) (6, 7) (7, 3) (7, 5) (7, 7)} ….(2)
- From (1) and (2) we get
- A × (B – C) = (A × B) – (A × C)
Let A and B be any two non-empty sets. A “relation” R from A to B is a subset of A × B satisfying some specified conditions.
Note:
The domain of the relations R = {x ∈ A/xRy, for some y ∈ B}
The co-domain of the relation R is B
The range of the ralation
R = (y ∈ B/xRy for some x ∈ A}
