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RELATION
• The way in which two or more people or things are connected
Introduction Relation
Introduction Relation
• Let A = Set of any five countries of the world
= { Nepal, India, China, America, Pakistan}
• B = Set of eight cities of the world
= {Washington D.C., New York, Islamabad, Dhaka, Beijing,
Kathmandu, New Delhi, Mumbai}
• What is one word that connect Group A to Group B ?
The word that connect set A to set B is “ capital of”.
• What are the ordered pairs connected by the word “ capital of “ ?
R = {(Nepal, Kathmandu), (India, New Delhi), (Pakistan, Islamabad),
(China, Beijing), (America, Washington D.C)}
• In the arrow diagram, the
ordered pairs formed by red
colored arrow is subset of
all the ordered pairs
A = { Nepal, India, China, America, Pakistan}
B = {Washington D.C., New York, Islamabad, Karachi, Beijing, Kathmandu, New Delhi, Mumbai}
Nepal
India
China
America
Pakistan
Washington D.C
New York
Islamabad
Dhaka
Beijing
Kathmandu
New-Delhi
Mumbai
A B
• It means the relation is
subset of A ×B
Cartesian Product of A and B
A
B
× Washingt
on D.C
New
York
Islamaba
d
Dhaka Beijing Kathman
du
New
Delhi
Mumbai
Nepal (N, W)
India
China
America
Pakistan
(N, NY) (N, IS) (N, D) (N, B) (N, K) (N, ND) (N, M)
(I, W) (I, NY) (I, IS) (I, D) (I, B) (I, K) (I, ND) (I, M)
(C, W) (C, NY) (C, IS) (C, D) (C, B) (C, K) (C, ND) (C, M)
(A, W) (A, NY) (A, IS) (A, D) (A, B) (A, K) (A, ND) (A, M)
(P, W) (P, NY) (P, IS) (P, D) (P, B) (P, K) (P, ND) (P, M)
Yes, R = {(N, K), (I, ND), (C, B), (A, W), (P, IS)} is subset of A ×B.
Does the word “ capital of “ subset of Cartesian product A ×B ?
An ordered pair represents as (INPUT, OUTPUT)
Relation shows the relationship between INPUT and OUTPUT.
• A relation from a set A to set B is a subset of A ×B. Hence a relation R
consists of order pairs (x, y), where x ∈A and y ∈B.
Relation from set A to set B is dented by R : A→B.
Let us form the following subsets or A×B:
a. R1 = {(2, 2), (3, 3), (4, 4)} ⊆ A×B
b. R2 = {((2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)}
⊆A ×B
c. R3 = {(2,3), (2, 4), (3, 4) } ⊆ A ×B
EXAMPLE:
• Let us consider the sets A = {2, 3, 4} and B = {1, 2, 3, 4}.
• Cartesian product A×B is
A
B
× 1 2 3 4
2
3
4
(2, 1)
(4, 4)
(2, 2) (2, 3) (2, 4)
(3, 1) (3, 2) (3, 3) (3, 4)
(4, 1) (4, 2) (4, 3)
Subsets of A×B Property RelationSubsets of A×B Property Relation
R1
S P R
x component = y component
S P R
“is equal to”
S P R
R2
X component > y component “is greater than”
S P R
R2
R3
X component < y component “is less than”
If A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2),
(3,3),(3,4)}, find the relations
a. “is less than”
b. “is equal to”
c. “is square of”
d. “is greater than”
EXAMPLE:
a. “is less than “ relation
Here, A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)}
“is less than” relation = {(x, y): x < y}
= {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}
b. “is equal to “ relation
Here, A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)}
“is equal to” relation = {(x, y): x = y}
= {(2, 2),(3, 3)}
c. “is square of “ relation
Here, A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)}
“is square of” relation = {(x, y): x = y2}
= ∅
d. “is greater than “ relation
Here, A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)}
“is greater than” relation = {(x, y): x > y}
= {(3, 2)
Representation of Relation
a. Tabulation method
b. Mapping/Arrow diagram
c. Set of order pair
d. Description or formula method
e. Graphical Method
Let A = {1,2, 3} and B ={2, 3, 4}. Represent the
relation “is less than” by all the methods
a. Tabulation method
b. Mapping/Arrow diagram
c. Set of order pair
d. Description or formula method
e. Graphical Method
a. Tabulation Method
Here, A = {1, 2, 3} and B = {2, 3, 4}
∴A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)
A relation “is less than” from set A to set B in the table is
A 1 1 1 2 2 3
B 2 3 4 3 4 4
b. Mapping/Arrow diagram
Here, A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)}
A relation “is less than” from set A to set B in the arrow diagram is
1
2
3
A
2
3
4
B
R
c. Set of ordered pair
Here, A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)}
A relation “is less than” from set A to set B in set of ordered form :
The relation “is less than” = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}
d. Description or formula form
Here, A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)}
A relation “is less than” from set A to set B in description form :
The relation “is less than” = {(x, y): x∈A, y ∈B and x < y}
d. Graphical method
Here, A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)}
A relation “is less than” from set A to set B in in graphical form :
1. If A = {0, 1, 3} then A× 𝐀 and the set of ordered
pairs in the following relation from A to A
a) Is less than
b) Is greater than
c) Is equal to
a) “is less than” relation
Here, A = {0, 1, 3}.
A×A = {0, 1, 3} ×{0, 1, 3}
= {(0, 0), (0, 1), (0, 3), (1, 0),(1, 1),(1, 3), (3, 0),(3, 1),(3, 3)}
∴ “is less than” relation = {x : x<y)
= {(0, 1),(0, 3),(1, 3)}
b) “is greater than” relation
Here, A = {0, 1, 3}.
A×A = {0, 1, 3} ×{0, 1, 3}
= {(0, 0), (0, 1), (0, 3), (1, 0),(1, 1),(1, 3), (3, 0),(3, 1),(3, 3)}
∴ “is greater than” relation = {x : x > y)
= {(1, 0),(3, 0),(3,1)}
c) “is equal to” relation
Here, A = {0, 1, 3}.
A×A = {0, 1, 3} ×{0, 1, 3}
= {(0, 0), (0, 1), (0, 3), (1, 0),(1, 1),(1, 3), (3, 0),(3, 1),(3, 3)}
∴ “is less than” relation = {x : x = y)
= {(0, 0),(1, 1),(3, 3)}
2. If A = {4, 8, 12, 16} and B = {0, 1, 2, 3, 5} and R is a relation from set A to
set B defined by ‘is four time of , find all the elements of R.
• Here, A = {4, 8, 12, 16} and B = {0, 1, 2, 3, 5}
A
B
× 0 1 2 3 5
4 (4, 0) (4, 1) (4, 2) (4, 3) (4, 5)
8 (8, 0) (8, 1) (8, 2) (8, 3) (8, 5)
12 (12, 0) (12, 1) (12, 2) (12, 3) (12, 5)
16 (16, 0) (16, 1) (16, 2) (16, 3) (16, 5)
A ×B is
R = ‘ is four times of’ ={(4, 1), (8, 2), (12, 3)}
3. If A ×B = {(1, 4), (1, 5), (1, 6), (2 ,4),(2 ,5),(2 ,6),(3 ,4,),
(3, 5),(3, 6)}, find the following relations and show in
arrow diagram.
a) R1 = {(x, y): x + y = 6}
b) R2 = {(x, y) : x < y}
c) R3 = {(x, y): y = x2}
a) R1 = {(x, y): x + y = 6}
• Here, A ×B = {(1, 4), (1, 5), (1, 6), (2 ,4),(2 ,5),(2 ,6),(3 ,4,), (3, 5),(3, 6)},
• R1 = {(x, y) : x + y = 6}
= {(1, 5), (2, 4)}
4
5
6
1
2
3
A B
R1
b) R2 = {(x, y) : x < y}
• Here, A ×B = {(1, 4), (1, 5), (1, 6), (2 ,4),(2 ,5),(2 ,6),(3 ,4,), (3, 5),(3, 6)},
• R2 = {(x, y) : x < y }
= {(1, 4), (1, 5), (1, 6), (2, 4),(2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
4
5
6
1
2
3
A B
R2
c) R3 = {(x, y): y = x2}
• Here, A ×B = {(1, 4), (1, 5), (1, 6), (2 ,4),(2 ,5),(2 ,6),(3 ,4,), (3, 5),(3, 6)},
• R3 = {(x, y) : y = x2}
= {(2, 4) }
4
5
6
1
2
3
A B
R3
4. If A = {6, 7, 8, 10} and B{2, 4, 6}, then represent the
relation R = {(x, y):x + y < 12, x ∈A, x ∈B} by
following methods:
a) Ordered pair form
b) Mapping diagram
c) Table
d) Graph
a) Here, A = {6, 7, 8, 10} and B{2, 4, 6},
R = {(x, y):x + y < 12, x ∈A, x ∈B}
A
B
× 2 4 6
6 (6, 2) (6, 4) (6, 6)
7 (7, 2) (7, 4) (7, 6)
8 (8, 2) (8, 4) (8, 6)
10 (10, 2) (10, 4) (10, 6)
A ×B =
R = {(6, 2), (6, 4), (7, 2), (7, 4), (8, 2)}
In ordered pair form
b) Here, A = {6, 7, 8, 10} and B{2, 4, 6},
R = {(x, y):x + y < 12, x ∈A, x ∈B}
A
B
× 2 4 6
6 (6, 2) (6, 4) (6, 6)
7 (7, 2) (7, 4) (7, 6)
8 (8, 2) (8, 4) (8, 6)
10 (10, 2) (10, 4) (10, 6)
A ×B =
2
4
6
6
7
8
10
A B
R
R in arrow diagram
c) Here, A = {6, 7, 8, 10} and B{2, 4, 6},
R = {(x, y):x + y < 12, x ∈A, x ∈B}
A
B
× 2 4 6
6 (6, 2) (6, 4) (6, 6)
7 (7, 2) (7, 4) (7, 6)
8 (8, 2) (8, 4) (8, 6)
10 (10, 2) (10, 4) (10, 6)
A ×B = In table form
A 6 6 7 7 8
B 2 4 2 4 2
c) Here, A = {6, 7, 8, 10} and B{2, 4, 6},
R = {(x, y):x + y < 12, x ∈A, x ∈B}
A
B
× 2 4 6
6 (6, 2) (6, 4) (6, 6)
7 (7, 2) (7, 4) (7, 6)
8 (8, 2) (8, 4) (8, 6)
10 (10, 2) (10, 4) (10, 6)
A ×B =
Representing R in graph
Practice questions
1. Let A = {1, 2} and B = {1, 2, 3}. If a relation from A to B is defined by
R = {(x, y): x + y = 3}, find R
2. If A = {(1, 2, 3 }, B = {(4, 5}. Find A×B and B×A. Check whether
a) R1 = {(1, 4), (2, 5), (3, 4) is a relation from A to B
b) R2 = {(1, 4), (1, 5), (2, 5) (3, 5), (4, 3)} is a relation from A to B.
c) R3= {(5, 1), (5, 2), (4, 3)} is a relation from B to A.
3. If A = {1, 2} and B = {4, 5}, find a relation from set A to B such that y = x + 4.
4. If A = {3, 4, 5, 6} and B = {6, 7, 8, 10} and R is the relation from set A to set B
defined by “is half of”, then find the elements of R.
5. If A = {1, 4, 9} and B = {1, 2, 3} and a relation R : A→B is defined by “is square of”
represent the relation by carious ways.
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RELATION

  • 2. • The way in which two or more people or things are connected Introduction Relation
  • 3. Introduction Relation • Let A = Set of any five countries of the world = { Nepal, India, China, America, Pakistan} • B = Set of eight cities of the world = {Washington D.C., New York, Islamabad, Dhaka, Beijing, Kathmandu, New Delhi, Mumbai} • What is one word that connect Group A to Group B ? The word that connect set A to set B is “ capital of”. • What are the ordered pairs connected by the word “ capital of “ ? R = {(Nepal, Kathmandu), (India, New Delhi), (Pakistan, Islamabad), (China, Beijing), (America, Washington D.C)}
  • 4. • In the arrow diagram, the ordered pairs formed by red colored arrow is subset of all the ordered pairs A = { Nepal, India, China, America, Pakistan} B = {Washington D.C., New York, Islamabad, Karachi, Beijing, Kathmandu, New Delhi, Mumbai} Nepal India China America Pakistan Washington D.C New York Islamabad Dhaka Beijing Kathmandu New-Delhi Mumbai A B • It means the relation is subset of A ×B
  • 5. Cartesian Product of A and B A B × Washingt on D.C New York Islamaba d Dhaka Beijing Kathman du New Delhi Mumbai Nepal (N, W) India China America Pakistan (N, NY) (N, IS) (N, D) (N, B) (N, K) (N, ND) (N, M) (I, W) (I, NY) (I, IS) (I, D) (I, B) (I, K) (I, ND) (I, M) (C, W) (C, NY) (C, IS) (C, D) (C, B) (C, K) (C, ND) (C, M) (A, W) (A, NY) (A, IS) (A, D) (A, B) (A, K) (A, ND) (A, M) (P, W) (P, NY) (P, IS) (P, D) (P, B) (P, K) (P, ND) (P, M) Yes, R = {(N, K), (I, ND), (C, B), (A, W), (P, IS)} is subset of A ×B. Does the word “ capital of “ subset of Cartesian product A ×B ? An ordered pair represents as (INPUT, OUTPUT) Relation shows the relationship between INPUT and OUTPUT.
  • 6. • A relation from a set A to set B is a subset of A ×B. Hence a relation R consists of order pairs (x, y), where x ∈A and y ∈B. Relation from set A to set B is dented by R : A→B.
  • 7. Let us form the following subsets or A×B: a. R1 = {(2, 2), (3, 3), (4, 4)} ⊆ A×B b. R2 = {((2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)} ⊆A ×B c. R3 = {(2,3), (2, 4), (3, 4) } ⊆ A ×B EXAMPLE: • Let us consider the sets A = {2, 3, 4} and B = {1, 2, 3, 4}. • Cartesian product A×B is A B × 1 2 3 4 2 3 4 (2, 1) (4, 4) (2, 2) (2, 3) (2, 4) (3, 1) (3, 2) (3, 3) (3, 4) (4, 1) (4, 2) (4, 3) Subsets of A×B Property RelationSubsets of A×B Property Relation R1 S P R x component = y component S P R “is equal to” S P R R2 X component > y component “is greater than” S P R R2 R3 X component < y component “is less than”
  • 8. If A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)}, find the relations a. “is less than” b. “is equal to” c. “is square of” d. “is greater than” EXAMPLE:
  • 9. a. “is less than “ relation Here, A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)} “is less than” relation = {(x, y): x < y} = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}
  • 10. b. “is equal to “ relation Here, A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)} “is equal to” relation = {(x, y): x = y} = {(2, 2),(3, 3)}
  • 11. c. “is square of “ relation Here, A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)} “is square of” relation = {(x, y): x = y2} = ∅
  • 12. d. “is greater than “ relation Here, A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)} “is greater than” relation = {(x, y): x > y} = {(3, 2)
  • 13. Representation of Relation a. Tabulation method b. Mapping/Arrow diagram c. Set of order pair d. Description or formula method e. Graphical Method
  • 14.
  • 15. Let A = {1,2, 3} and B ={2, 3, 4}. Represent the relation “is less than” by all the methods a. Tabulation method b. Mapping/Arrow diagram c. Set of order pair d. Description or formula method e. Graphical Method
  • 16. a. Tabulation Method Here, A = {1, 2, 3} and B = {2, 3, 4} ∴A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4) A relation “is less than” from set A to set B in the table is A 1 1 1 2 2 3 B 2 3 4 3 4 4
  • 17. b. Mapping/Arrow diagram Here, A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)} A relation “is less than” from set A to set B in the arrow diagram is 1 2 3 A 2 3 4 B R
  • 18. c. Set of ordered pair Here, A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)} A relation “is less than” from set A to set B in set of ordered form : The relation “is less than” = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}
  • 19. d. Description or formula form Here, A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)} A relation “is less than” from set A to set B in description form : The relation “is less than” = {(x, y): x∈A, y ∈B and x < y}
  • 20. d. Graphical method Here, A×B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3,2), (3,3),(3,4)} A relation “is less than” from set A to set B in in graphical form :
  • 21. 1. If A = {0, 1, 3} then A× 𝐀 and the set of ordered pairs in the following relation from A to A a) Is less than b) Is greater than c) Is equal to
  • 22. a) “is less than” relation Here, A = {0, 1, 3}. A×A = {0, 1, 3} ×{0, 1, 3} = {(0, 0), (0, 1), (0, 3), (1, 0),(1, 1),(1, 3), (3, 0),(3, 1),(3, 3)} ∴ “is less than” relation = {x : x<y) = {(0, 1),(0, 3),(1, 3)}
  • 23. b) “is greater than” relation Here, A = {0, 1, 3}. A×A = {0, 1, 3} ×{0, 1, 3} = {(0, 0), (0, 1), (0, 3), (1, 0),(1, 1),(1, 3), (3, 0),(3, 1),(3, 3)} ∴ “is greater than” relation = {x : x > y) = {(1, 0),(3, 0),(3,1)}
  • 24. c) “is equal to” relation Here, A = {0, 1, 3}. A×A = {0, 1, 3} ×{0, 1, 3} = {(0, 0), (0, 1), (0, 3), (1, 0),(1, 1),(1, 3), (3, 0),(3, 1),(3, 3)} ∴ “is less than” relation = {x : x = y) = {(0, 0),(1, 1),(3, 3)}
  • 25. 2. If A = {4, 8, 12, 16} and B = {0, 1, 2, 3, 5} and R is a relation from set A to set B defined by ‘is four time of , find all the elements of R. • Here, A = {4, 8, 12, 16} and B = {0, 1, 2, 3, 5} A B × 0 1 2 3 5 4 (4, 0) (4, 1) (4, 2) (4, 3) (4, 5) 8 (8, 0) (8, 1) (8, 2) (8, 3) (8, 5) 12 (12, 0) (12, 1) (12, 2) (12, 3) (12, 5) 16 (16, 0) (16, 1) (16, 2) (16, 3) (16, 5) A ×B is R = ‘ is four times of’ ={(4, 1), (8, 2), (12, 3)}
  • 26. 3. If A ×B = {(1, 4), (1, 5), (1, 6), (2 ,4),(2 ,5),(2 ,6),(3 ,4,), (3, 5),(3, 6)}, find the following relations and show in arrow diagram. a) R1 = {(x, y): x + y = 6} b) R2 = {(x, y) : x < y} c) R3 = {(x, y): y = x2}
  • 27. a) R1 = {(x, y): x + y = 6} • Here, A ×B = {(1, 4), (1, 5), (1, 6), (2 ,4),(2 ,5),(2 ,6),(3 ,4,), (3, 5),(3, 6)}, • R1 = {(x, y) : x + y = 6} = {(1, 5), (2, 4)} 4 5 6 1 2 3 A B R1
  • 28. b) R2 = {(x, y) : x < y} • Here, A ×B = {(1, 4), (1, 5), (1, 6), (2 ,4),(2 ,5),(2 ,6),(3 ,4,), (3, 5),(3, 6)}, • R2 = {(x, y) : x < y } = {(1, 4), (1, 5), (1, 6), (2, 4),(2, 5), (2, 6), (3, 4), (3, 5), (3, 6)} 4 5 6 1 2 3 A B R2
  • 29. c) R3 = {(x, y): y = x2} • Here, A ×B = {(1, 4), (1, 5), (1, 6), (2 ,4),(2 ,5),(2 ,6),(3 ,4,), (3, 5),(3, 6)}, • R3 = {(x, y) : y = x2} = {(2, 4) } 4 5 6 1 2 3 A B R3
  • 30. 4. If A = {6, 7, 8, 10} and B{2, 4, 6}, then represent the relation R = {(x, y):x + y < 12, x ∈A, x ∈B} by following methods: a) Ordered pair form b) Mapping diagram c) Table d) Graph
  • 31. a) Here, A = {6, 7, 8, 10} and B{2, 4, 6}, R = {(x, y):x + y < 12, x ∈A, x ∈B} A B × 2 4 6 6 (6, 2) (6, 4) (6, 6) 7 (7, 2) (7, 4) (7, 6) 8 (8, 2) (8, 4) (8, 6) 10 (10, 2) (10, 4) (10, 6) A ×B = R = {(6, 2), (6, 4), (7, 2), (7, 4), (8, 2)} In ordered pair form
  • 32. b) Here, A = {6, 7, 8, 10} and B{2, 4, 6}, R = {(x, y):x + y < 12, x ∈A, x ∈B} A B × 2 4 6 6 (6, 2) (6, 4) (6, 6) 7 (7, 2) (7, 4) (7, 6) 8 (8, 2) (8, 4) (8, 6) 10 (10, 2) (10, 4) (10, 6) A ×B = 2 4 6 6 7 8 10 A B R R in arrow diagram
  • 33. c) Here, A = {6, 7, 8, 10} and B{2, 4, 6}, R = {(x, y):x + y < 12, x ∈A, x ∈B} A B × 2 4 6 6 (6, 2) (6, 4) (6, 6) 7 (7, 2) (7, 4) (7, 6) 8 (8, 2) (8, 4) (8, 6) 10 (10, 2) (10, 4) (10, 6) A ×B = In table form A 6 6 7 7 8 B 2 4 2 4 2
  • 34. c) Here, A = {6, 7, 8, 10} and B{2, 4, 6}, R = {(x, y):x + y < 12, x ∈A, x ∈B} A B × 2 4 6 6 (6, 2) (6, 4) (6, 6) 7 (7, 2) (7, 4) (7, 6) 8 (8, 2) (8, 4) (8, 6) 10 (10, 2) (10, 4) (10, 6) A ×B = Representing R in graph
  • 35. Practice questions 1. Let A = {1, 2} and B = {1, 2, 3}. If a relation from A to B is defined by R = {(x, y): x + y = 3}, find R 2. If A = {(1, 2, 3 }, B = {(4, 5}. Find A×B and B×A. Check whether a) R1 = {(1, 4), (2, 5), (3, 4) is a relation from A to B b) R2 = {(1, 4), (1, 5), (2, 5) (3, 5), (4, 3)} is a relation from A to B. c) R3= {(5, 1), (5, 2), (4, 3)} is a relation from B to A. 3. If A = {1, 2} and B = {4, 5}, find a relation from set A to B such that y = x + 4. 4. If A = {3, 4, 5, 6} and B = {6, 7, 8, 10} and R is the relation from set A to set B defined by “is half of”, then find the elements of R. 5. If A = {1, 4, 9} and B = {1, 2, 3} and a relation R : A→B is defined by “is square of” represent the relation by carious ways.