The document defines key concepts related to relations and functions: 1. Ordered pairs, Cartesian products, and domains and ranges of relations are introduced. An ordered pair lists two objects in a particular order, while a Cartesian product is the set of all ordered pairs with the first element from one set and the second from another. 2. The domain of a relation is the set of first elements in its ordered pairs, while the range is the set of second elements. Inverse relations are also defined. 3. Several examples illustrate calculating Cartesian products of sets, determining the domains and ranges of given relations, and finding inverse relations.

1. Relations & Functions-1
Lecture - 5
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2. Lecture-5
Ordered Pair:
A pair of objects written in a particular order is called an
Ordered Pair.
An ordered pair is written by listing its two members in a
particular order, separated by a comma and enclosing the pair
in parentheses, like (a, b).
In the ordered pair (a, b), a is called the first component or
the first element/member/coordinate and b is called the
second component or the second element/member/
coordinate.
Also
(1, 3) = (1, 3)
But
(1, 3) ≠ (3, 1) and (1, 3) ≠ (1, 2)
Class - XI Relations & Functions
(equal ordered pairs)
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3. Lecture-5
Cartesian Product:
The Cartesian product of two sets A and B is the set of all
those ordered pairs whose first co-ordinate is an element of A
and the second co-ordinate is an element of B.
This set is denoted by A × B and is read as ‘A cross B’ or
‘product set of A and B’.
Cartesian Product – Few Points:

A×B≠
A≠

A × B may or may not be equal to B × A.

A × A is also denoted by A2.
Class - XI Relations & Functions
and B ≠ ,
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4. Lecture-5
Ex.1.
Let A = (1, 2), B = {4, 5}, C = {6, 8}, find A × B × C.
Sol:
A × B × C = A × (B × C)
= {1, 2} × [{4, 5} × {6, 8}]
= {1, 2} × {(4, 6), (4, 8), (5, 6), (5, 8)}
= {(1, 4, 6), (1, 4, 8), (1, 5, 6), (1, 5, 8), (2, 4, 6),
(2, 4,8), (2, 5, 6), (2, 5, 8)}
Class - XI Relations & Functions
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5. Lecture-5
Number of Elements in the Cartesian Products
If A and B are two finite sets, then n(A × B) = n (A) × n(B)
Representing the Cartesian Product of two sets by
arrow Diagrams
Let A = {1, 2, 3}, B = {a, b}
Cartesian product of A and B, i.e., A × B can be represented
by arrow diagram as shown in the given figure.
A
B
1
a
2
3
Class - XI Relations & Functions
b
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6. Lecture-5
Ex.2.
Let A = {1, 2} and B = {3, 4, 5}. Find
(i)
(ii)
B×A
(iii)
Sol:
A×B
B×B
(iv)
A×A×A
(i)
A × B = {1, 2} × {3, 4, 5}
= {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)}.
(ii)
B × A = {3, 4, 5} × {1, 2}
= {(3, 1), (3, 2), (4, 1), (4, 2), (5, 1), (5, 2)}.
(iii)
B × B = {3, 4, 5} × {3, 4, 5}
= {(3, 3), (3, 4), (3, 5), (4, 3), (4, 4), (4, 5),
(5, 3), (5, 4), (5, 5))}.
Class - XI Relations & Functions
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7. Lecture-5
(iv)
A × A × A = {1, 2} × {1, 2} × {1, 2}
= {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2),
(2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}.
Ex.3.
Express R = {(x, y) : y + 2x = 5, x, y
set of ordered pairs i.e. roster form.
Sol:
Given, y + 2x = 5
W} as the
y = 5 – 2x
…(i)
x, y W, i.e., x and y are whole numbers
(no negative integers)
from (1),
x=0
y=5
x=1
y=3
x=2
y=1
R = {(0, 5), (1, 3), (2, 1)}.
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8. Lecture-5
Ex.4:
If A = {1, 2, 3}, B = {3, 4}, C = {4, 5}. Then verify if A
× (B C) = (A × B) (A × C).
Sol:
A × (B
C) = {2, 4, 5}
A × (B
C) = {1, 2, 3} × {3, 4, 5}
= {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4),
(2, 5), (3, 3), (3, 4), (3, 5),
A × B = {1, 2, 3} × {3, 4}
= {(1, 3), (1, 4), (2, 3), (2, 4),(3, 3),(3, 4)
A × C = {1, 2, 3} × {4, 5}
= {(1, 4), (1, 5), (2, 4), (2, 5),(3, 4),(3, 5)
Class - XI Relations & Functions
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9. Lecture-5
A × B)
(A × C) = {(1, 3),(1, 4), (2, 3), (2, 4), (3, 3)}
(3, 4), (1, 5), (2, 5), (3, 5)}
Thus , A × (B
C) = (A × B)
(A × C).
Questions for Discussion.
1.
If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G
and the number of elements in both of these.
2.
If A × B = {(a, x), (a, y), (b, x), (b, y)}, find A and B.
3.
Let A = {1, 2} and B = {3, 4}. Write A x B. How
many subsets will A x B have ? List them.
4.
If A = {–1, 1}, find A × A × A.
Class - XI Relations & Functions
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10. Lecture-5
Definition: A relation R, from a non-empty set A to another
non-empty set B, is a subset of A × B.
In other words any subset of A × B is a relation
from A to B.
Thus, R is a relation from A to B
R
R
{(a, b): a
A × B.
A, b
B}.
Domain and Range of a relation :
Let R be a relation from A to B. the domain of relation R
is the set of all those elements a A such that (a, b)
R for some b B.
Dom(R) = {a
A: (a, b)
Range of R = {b
R for some b
B : (a, b)
Class - XI Relations & Functions
B}.
R for some a
A}.
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11. Lecture-5
Co-domain of a relation:
If R be a relation from A to B. Then B is called the
co-domain of relation R.
Inverse Relation
Let R
A × B be a relation from A to B. Then, the
Inverse relation of R, to be denoted by R–1, is a relation
from B to A defined by :
R–1 = {(b, a) : (a, b) R}.
Ex.5.
Let A = {a, b, c} and B = {1, 2}
Let R = {(a, 1), (c, 1)}, find R–1.
Sol:
Given, R = {(a, 1), (c, 1)}
R–1 = {(1, a), (1, c)}.
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12. Lecture-5
Ex.6:
Let A = {1, 2, 3, 4} and B = {x, y, z}. Let R be a
relation from A into B defined by
R = {(1, x), (1, z), (3, x), (4, y)}
Find the domain and range of R.
Sol:
Given R = {(1, x), (1, 3), (3, x), (4, y)}.
Domain of R = set of first components of all elements of R
= {1, 3, 4}.
Range of R = set of second components of all elements of R
= {x, y, z} = B.
Class - XI Relations & Functions
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13. Lecture-5
Ex.7.
Let A = {1, 2, 3, 4, 6}. Let R be a relation on A
defined by R = {a, b) : a A, b A, a divides b}
Find (i) R (ii) domain of R (iii) Range of R.
Sol:
Given A = {(1, 2, 3, 4, 6)
R = {(a, b) : a
A, b
A and a divides b }
R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2),
(2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}
Domain of R = {1, 2, 3, 4, 6} = A
Range of R = {1, 2, 3, 4, 6} = A
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14. Lecture-5
Ex.8.
Determine the domain and range of the relation R
defined by
R = { (x + 1, x + 5) : x
Sol:
x
{0, 1, 2, 3, 5} }.
{0, 1, 2, 4, 5}.
x=0
x + 1 = 0 + 1 = 1 and x + 5 = 0 + 5 = 5
x=1
x + 1 = 1 + 1 = 2 and x + 5 = 1 + 5 = 6
x=2
x + 1 = 2 + 1 = 3 and x + 5 = 2 + 5 = 7
x=3
x + 1 = 3 + 1 = 4 and x + 5 = 3 + 5 = 8
x=4
x + 1 = 4 + 1 = 5 and x + 5 = 4 + 5 = 9
x=5
x + 1 = 5 + 1 = 6 and x + 5 = 5 + 5 = 10
Class - XI Relations & Functions
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15. Lecture-5
Hence,
R = {(1, 5), (2, 6), (3, 7), (4, 8), (5, 9), (6, 10)}.
Domain of R = {a : (a, b) R} = set of first components
of all ordered pairs belonging to R.
= {(1, 2, 3, 4, 5, 6)}.
Range of R = {b : (a, b)
R}
= Set of second components of all ordered pairs
belonging to R
= {5, 6, 7, 8, 9, 10}.
Class - XI Relations & Functions
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16. Thank You