05 tat math class xi_r&f-1_for website

Relations & Functions-1
Lecture - 5

1

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)

2

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.


and B ≠ ,

3

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)}


4

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


b

5

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))}.


6

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)}.

7

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)

8

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.


9

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)


B}.

R for some a

A}.
10

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)}.


11

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.

12

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

13

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


14

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}.


15

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