Chapter 2
RELATIONS AND FUNCTIONS
Mathematics is the indispensable instrument of
all physical research. – BERTHELOT
2.1 Introduction
Much of mathematics is about finding a pattern – a
recognisable link between quantities that change. In our
daily life, we come across many patterns that characterise
relations such as brother and sister, father and son, teacher
and student. In mathematics also, we come across many
relations such as number m is less than number n, line l is
parallel to line m, set A is a subset of set B. In all these, we
notice that a relation involves pairs of objects in certain
order. In this Chapter, we will learn how to link pairs of
objects from two sets and then introduce relations between
the two objects in the pair. Finally, we will learn about G . W. Leibnitz
special relations which will qualify to be functions. The (1646–1716)
concept of function is very important in mathematics since it captures the idea of a
mathematically precise correspondence between one quantity with the other.
2.2 Cartesian Products of Sets
Suppose A is a set of 2 colours and B is a set of 3 objects, i.e.,
A = {red, blue}and B = {b, c, s},
where b, c and s represent a particular bag, coat and shirt, respectively.
How many pairs of coloured objects can be made from these two sets?
Proceeding in a very orderly manner, we can see that there will be 6
distinct pairs as given below:
(red, b), (red, c), (red, s), (blue, b), (blue, c), (blue, s).
Thus, we get 6 distinct objects (Fig 2.1).
Let us recall from our earlier classes that an ordered pair of elements Fig 2.1
taken from any two sets P and Q is a pair of elements written in small
2022-23
, RELATIONS AND FUNCTIONS 31
brackets and grouped together in a particular order, i.e., (p,q), p ∈ P and q ∈ Q . This
leads to the following definition:
Definition 1 Given two non-empty sets P and Q. The cartesian product P × Q is the
set of all ordered pairs of elements from P and Q, i.e.,
P × Q = { (p,q) : p ∈ P, q ∈ Q }
If either P or Q is the null set, then P × Q will also be empty set, i.e., P × Q = φ
From the illustration given above we note that
A × B = {(red,b), (red,c), (red,s), (blue,b), (blue,c), (blue,s)}.
Again, consider the two sets:
A = {DL, MP, KA}, where DL, MP, KA represent Delhi,
Madhya Pradesh and Karnataka, respectively and B = {01,02, 03
03}representing codes for the licence plates of vehicles issued 02
by DL, MP and KA . 01
If the three states, Delhi, Madhya Pradesh and Karnataka
were making codes for the licence plates of vehicles, with the DL MP KA
restriction that the code begins with an element from set A,
Fig 2.2
which are the pairs available from these sets and how many such
pairs will there be (Fig 2.2)?
The available pairs are:(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03),
(KA,01), (KA,02), (KA,03) and the product of set A and set B is given by
A × B = {(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02),
(KA,03)}.
It can easily be seen that there will be 9 such pairs in the Cartesian product, since
there are 3 elements in each of the sets A and B. This gives us 9 possible codes. Also
note that the order in which these elements are paired is crucial. For example, the code
(DL, 01) will not be the same as the code (01, DL).
As a final illustration, consider the two sets A= {a1, a2} and
B = {b1, b2, b3, b4} (Fig 2.3).
A × B = {( a1, b1), (a1, b2), (a1, b3), (a1, b4), (a2, b1), (a2, b2),
(a2, b3), (a2, b4)}.
The 8 ordered pairs thus formed can represent the position of points in
the plane if A and B are subsets of the set of real numbers and it is
obvious that the point in the position (a1, b2) will be distinct from the point
Fig 2.3
in the position (b2, a1).
Remarks
(i) Two ordered pairs are equal, if and only if the corresponding first elements
are equal and the second elements are also equal.
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, 32 MATHEMATICS
(ii) If there are p elements in A and q elements in B, then there will be pq
elements in A × B, i.e., if n(A) = p and n(B) = q, then n(A × B) = pq.
(iii) If A and B are non-empty sets and either A or B is an infinite set, then so is
A × B.
(iv) A × A × A = {(a, b, c) : a, b, c ∈ A}. Here (a, b, c) is called an ordered
triplet.
Example 1 If (x + 1, y – 2) = (3,1), find the values of x and y.
Solution Since the ordered pairs are equal, the corresponding elements are equal.
Therefore x + 1 = 3 and y – 2 = 1.
Solving we get x = 2 and y = 3.
Example 2 If P = {a, b, c} and Q = {r}, form the sets P × Q and Q × P.
Are these two products equal?
Solution By the definition of the cartesian product,
P × Q = {(a, r), (b, r), (c, r)} and Q × P = {(r, a), (r, b), (r, c)}
Since, by the definition of equality of ordered pairs, the pair (a, r) is not equal to the pair
(r, a), we conclude that P × Q ≠ Q × P.
However, the number of elements in each set will be the same.
Example 3 Let A = {1,2,3}, B = {3,4} and C = {4,5,6}. Find
(i) A × (B ∩ C) (ii) (A × B) ∩ (A × C)
(iii) A × (B ∪ C) (iv) (A × B) ∪ (A × C)
Solution (i) By the definition of the intersection of two sets, (B ∩ C) = {4}.
Therefore, A × (B ∩ C) = {(1,4), (2,4), (3,4)}.
(ii) Now (A × B) = {(1,3), (1,4), (2,3), (2,4), (3,3), (3,4)}
and (A × C) = {(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)}
Therefore, (A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}.
(iii) Since, (B ∪ C) = {3, 4, 5, 6}, we have
A × (B ∪ C) = {(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,3),
(3,4), (3,5), (3,6)}.
(iv) Using the sets A × B and A × C from part (ii) above, we obtain
(A × B) ∪ (A × C) = {(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6),
(3,3), (3,4), (3,5), (3,6)}.
2022-23
RELATIONS AND FUNCTIONS
Mathematics is the indispensable instrument of
all physical research. – BERTHELOT
2.1 Introduction
Much of mathematics is about finding a pattern – a
recognisable link between quantities that change. In our
daily life, we come across many patterns that characterise
relations such as brother and sister, father and son, teacher
and student. In mathematics also, we come across many
relations such as number m is less than number n, line l is
parallel to line m, set A is a subset of set B. In all these, we
notice that a relation involves pairs of objects in certain
order. In this Chapter, we will learn how to link pairs of
objects from two sets and then introduce relations between
the two objects in the pair. Finally, we will learn about G . W. Leibnitz
special relations which will qualify to be functions. The (1646–1716)
concept of function is very important in mathematics since it captures the idea of a
mathematically precise correspondence between one quantity with the other.
2.2 Cartesian Products of Sets
Suppose A is a set of 2 colours and B is a set of 3 objects, i.e.,
A = {red, blue}and B = {b, c, s},
where b, c and s represent a particular bag, coat and shirt, respectively.
How many pairs of coloured objects can be made from these two sets?
Proceeding in a very orderly manner, we can see that there will be 6
distinct pairs as given below:
(red, b), (red, c), (red, s), (blue, b), (blue, c), (blue, s).
Thus, we get 6 distinct objects (Fig 2.1).
Let us recall from our earlier classes that an ordered pair of elements Fig 2.1
taken from any two sets P and Q is a pair of elements written in small
2022-23
, RELATIONS AND FUNCTIONS 31
brackets and grouped together in a particular order, i.e., (p,q), p ∈ P and q ∈ Q . This
leads to the following definition:
Definition 1 Given two non-empty sets P and Q. The cartesian product P × Q is the
set of all ordered pairs of elements from P and Q, i.e.,
P × Q = { (p,q) : p ∈ P, q ∈ Q }
If either P or Q is the null set, then P × Q will also be empty set, i.e., P × Q = φ
From the illustration given above we note that
A × B = {(red,b), (red,c), (red,s), (blue,b), (blue,c), (blue,s)}.
Again, consider the two sets:
A = {DL, MP, KA}, where DL, MP, KA represent Delhi,
Madhya Pradesh and Karnataka, respectively and B = {01,02, 03
03}representing codes for the licence plates of vehicles issued 02
by DL, MP and KA . 01
If the three states, Delhi, Madhya Pradesh and Karnataka
were making codes for the licence plates of vehicles, with the DL MP KA
restriction that the code begins with an element from set A,
Fig 2.2
which are the pairs available from these sets and how many such
pairs will there be (Fig 2.2)?
The available pairs are:(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03),
(KA,01), (KA,02), (KA,03) and the product of set A and set B is given by
A × B = {(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02),
(KA,03)}.
It can easily be seen that there will be 9 such pairs in the Cartesian product, since
there are 3 elements in each of the sets A and B. This gives us 9 possible codes. Also
note that the order in which these elements are paired is crucial. For example, the code
(DL, 01) will not be the same as the code (01, DL).
As a final illustration, consider the two sets A= {a1, a2} and
B = {b1, b2, b3, b4} (Fig 2.3).
A × B = {( a1, b1), (a1, b2), (a1, b3), (a1, b4), (a2, b1), (a2, b2),
(a2, b3), (a2, b4)}.
The 8 ordered pairs thus formed can represent the position of points in
the plane if A and B are subsets of the set of real numbers and it is
obvious that the point in the position (a1, b2) will be distinct from the point
Fig 2.3
in the position (b2, a1).
Remarks
(i) Two ordered pairs are equal, if and only if the corresponding first elements
are equal and the second elements are also equal.
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, 32 MATHEMATICS
(ii) If there are p elements in A and q elements in B, then there will be pq
elements in A × B, i.e., if n(A) = p and n(B) = q, then n(A × B) = pq.
(iii) If A and B are non-empty sets and either A or B is an infinite set, then so is
A × B.
(iv) A × A × A = {(a, b, c) : a, b, c ∈ A}. Here (a, b, c) is called an ordered
triplet.
Example 1 If (x + 1, y – 2) = (3,1), find the values of x and y.
Solution Since the ordered pairs are equal, the corresponding elements are equal.
Therefore x + 1 = 3 and y – 2 = 1.
Solving we get x = 2 and y = 3.
Example 2 If P = {a, b, c} and Q = {r}, form the sets P × Q and Q × P.
Are these two products equal?
Solution By the definition of the cartesian product,
P × Q = {(a, r), (b, r), (c, r)} and Q × P = {(r, a), (r, b), (r, c)}
Since, by the definition of equality of ordered pairs, the pair (a, r) is not equal to the pair
(r, a), we conclude that P × Q ≠ Q × P.
However, the number of elements in each set will be the same.
Example 3 Let A = {1,2,3}, B = {3,4} and C = {4,5,6}. Find
(i) A × (B ∩ C) (ii) (A × B) ∩ (A × C)
(iii) A × (B ∪ C) (iv) (A × B) ∪ (A × C)
Solution (i) By the definition of the intersection of two sets, (B ∩ C) = {4}.
Therefore, A × (B ∩ C) = {(1,4), (2,4), (3,4)}.
(ii) Now (A × B) = {(1,3), (1,4), (2,3), (2,4), (3,3), (3,4)}
and (A × C) = {(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)}
Therefore, (A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}.
(iii) Since, (B ∪ C) = {3, 4, 5, 6}, we have
A × (B ∪ C) = {(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,3),
(3,4), (3,5), (3,6)}.
(iv) Using the sets A × B and A × C from part (ii) above, we obtain
(A × B) ∪ (A × C) = {(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6),
(3,3), (3,4), (3,5), (3,6)}.
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