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Chapter
1
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RELATIONS AND FUNCTIONS
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1
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It may
be very hard to define mathematical beauty but that is just as true of
beauty of any kind, we may not know quite what we mean by a
beautiful poem, but that does not prevent us from recognising
one when we read it
...
H
...
The concept of the term ‘relation’ in
mathematics has been drawn from the meaning of relation
in English language, according to which two objects or
quantities are related if there is a recognisable connection
or link between the two objects or quantities
...
Then some
of the examples of relations from A to B are
(i) {(a, b) ∈ A × B: a is brother of b},
Lejeune Dirichlet
(1805-1859)
(ii) {(a, b) ∈ A × B: a is sister of b},
(iii) {(a, b) ∈ A × B: age of a is greater than age of b},
(iv) {(a, b) ∈ A × B: total marks obtained by a in the final examination is less than
the total marks obtained by b in the final examination},
(v) {(a, b) ∈ A × B: a lives in the same locality as b}
...
If (a, b) ∈ R, we say that a is related to b under the relation R and we write as
a R b
...
As seen in Class XI, functions are special kind of
relations
...
2
MATHEMATICS
1
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We know that a
relation in a set A is a subset of A × A
...
For illustration, consider a relation R in the set A = {1, 2, 3, 4} given by
R = {(a, b): a – b = 10}
...
Similarly, R′ = {(a, b) : | a – b | ≥ 0} is the whole set A × A, as all pairs
(a, b) in A × A satisfy | a – b | ≥ 0
...
is
Definition 1 A relation R in a set A is called empty relation, if no element of A is
related to any element of A, i
...
, R = φ ⊂ A × A
...
e
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Both the empty relation and the universal relation are some times called trivial
relations
...
Show that the relation R
in A given by R = {(a, b) : a is sister of b} is the empty relation and R′ = {(a, b) : the
difference between heights of a and b is less than 3 meters} is the universal relation
...
Hence, R = φ, showing that R is the empty relation
...
This shows that R′ = A × A is the universal relation
...
However, a relation R in the set {1, 2, 3, 4} defined by R
= {(a, b) : b = a + 1} is also expressed as a R b if and only if
b = a + 1 by many authors
...
If (a, b) ∈ R, we say that a is related to b and we denote it as a R b
...
To study equivalence relation, we first consider three
types of relations, namely reflexive, symmetric and transitive
...
(iii) transitive, if (a1, a2) ∈ R and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2,
a3 ∈ A
...
Example 2 Let T be the set of all triangles in a plane with R a relation in T given by
R = {(T1, T2) : T1 is congruent to T2}
...
is
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Solution R is reflexive, since every triangle is congruent to itself
...
Hence,
R is symmetric
...
Therefore, R is an equivalence
relation
...
Show that R is symmetric but neither
reflexive nor transitive
...
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R is symmetric as (L1, L2) ∈ R
⇒
L1 is perpendicular to L2
⇒
L2 is perpendicular to L1
⇒
(L2, L1) ∈ R
...
Indeed, if L1 is perpendicular to L2 and
Fig 1
...
In fact, L1 is parallel to L3, i
...
, (L1, L2) ∈ R, (L2, L3) ∈ R but (L1, L3) ∉ R
...
Solution R is reflexive, since (1, 1), (2, 2) and (3, 3) lie in R
...
Similarly, R is not transitive, as (1, 2) ∈ R and (2, 3) ∈ R
but (1, 3) ∉ R
...
Solution R is reflexive, as 2 divides (a – a) for all a ∈ Z
...
Therefore, 2 divides b – a
...
Similarly, if (a, b) ∈ R and (b, c) ∈ R, then a – b and b – c are divisible by
2
...
So, (a – c) is divisible by 2
...
Thus, R is an equivalence relation in Z
...
, lie in R and no odd integer is related to 0, as (0, ± 1), (0, ± 3) etc
...
Similarly, all odd integers are related to one and no even integer is related to one
...
(ii) No element of E is related to any element of O and vice-versa
...
The subset E is called the equivalence class containing zero and is denoted by
[0]
...
Note that
[0] ≠ [1], [0] = [2r] and [1] = [2r + 1], r ∈ Z
...
Given an arbitrary equivalence
relation R in an arbitrary set X, R divides X into mutually disjoint subsets Ai called
partitions or subdivisions of X satisfying:
(i) all elements of Ai are related to each other, for all i
...
(iii) ∪ Aj = X and Ai ∩ Aj = φ, i ≠ j
...
The interesting part of the situation
is that we can go reverse also
...
, – 6, – 3, 0, 3, 6,
...
, – 5, – 2, 1, 4, 7,
...
, – 4, – 1, 2, 5, 8,
...
Following the
arguments similar to those used in Example 5, we can show that R is an equivalence
relation
...
Thus, A1 = [0], A2 = [1] and A3 = [2]
...
Example 6 Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7} by
R = {(a, b) : both a and b are either odd or even}
...
Further, show that all the elements of the subset {1, 3, 5, 7} are related to each
other and all the elements of the subset {2, 4, 6} are related to each other, but no
element of the subset {1, 3, 5, 7} is related to any element of the subset {2, 4, 6}
...
1
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Solution Given any element a in A, both a and a must be either odd or even, so
that (a, a) ∈ R
...
Similarly, (a, b) ∈ R and (b, c) ∈ R ⇒ all elements a, b, c, must be
either even or odd simultaneously ⇒ (a, c) ∈ R
...
Further, all the elements of {1, 3, 5, 7} are related to each other, as all the elements
of this subset are odd
...
Also, no element of the subset {1, 3, 5, 7} can be
related to any element of {2, 4, 6}, as elements of {1, 3, 5, 7} are odd, while elements
of {2, 4, 6} are even
...
Determine whether each of the following relations are reflexive, symmetric and
transitive:
(i) Relation R in the set A = {1, 2, 3,
...
Show that the relation R in the set R of real numbers, defined as
R = {(a, b) : a ≤ b2} is neither reflexive nor symmetric nor transitive
...
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive
...
Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and
transitive but not symmetric
...
Check whether the relation R in R defined by R = {(a, b) : a ≤ b3} is reflexive,
symmetric or transitive
...
Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is
symmetric but neither reflexive nor transitive
...
Show that the relation R in the set A = {1, 2, 3, 4, 5} given by
he
7
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R = {(a, b) : |a – b| is even}, is an equivalence relation
...
But no element of {1, 3, 5} is related to any element of {2, 4}
...
Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by
(ii) R = {(a, b) : a = b}
bl
(i) R = {(a, b) : |a – b| is a multiple of 4}
is an equivalence relation
...
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Which is
(i) Symmetric but neither reflexive nor transitive
...
(iii) Reflexive and symmetric but not transitive
...
(v) Symmetric and transitive but not reflexive
...
Show that the relation R in the set A of points in a plane given by
R = {(P, Q) : distance of the point P from the origin is same as the distance of the
point Q from the origin}, is an equivalence relation
...
12
...
Consider three right angle triangles T1
with sides 3, 4, 5, T2 with sides 5, 12, 13 and T3 with sides 6, 8, 10
...
Show that the relation R defined in the set A of all polygons as R = {(P1, P2) :
P1 and P2 have same number of sides}, is an equivalence relation
...
Let L be the set of all lines in XY plane and R be the relation in L defined as
R = {(L1, L2) : L1 is parallel to L2}
...
Find
the set of all lines related to the line y = 2x + 4
...
Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4,4),
(1, 3), (3, 3), (3, 2)}
...
(A) R is reflexive and symmetric but not transitive
...
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(C) R is symmetric and transitive but not reflexive
...
16
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Choose
the correct answer
...
3 Types of Functions
(C) (6, 8) ∈ R
(D) (8, 7) ∈ R
is
(A) (2, 4) ∈ R
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The notion of a function along with some special functions like identity function, constant
function, polynomial function, rational function, modulus function, signum function etc
...
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Addition, subtraction, multiplication and division of two functions have also been
studied
...
In this section, we would like to study different types of
functions
...
In Fig 1
...
Further, there are some elements like e and f in X2 which are not images of
any element of X1 under f1, while all elements of X3 are images of some elements of X1
under f3
...
e
...
Otherwise, f is called many-one
...
2 (i) and (iv) are one-one and the function f2 and f3
in Fig 1
...
Definition 6 A function f : X → Y is said to be onto (or surjective), if every element
of Y is the image of some element of X under f, i
...
, for every y ∈ Y, there exists an
element x in X such that f (x) = y
...
2 (iii), (iv) are onto and the function f1 in Fig 1
...
MATHEMATICS
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8
Fig 1
...
Definition 7 A function f : X → Y is said to be one-one and onto (or bijective), if f is
both one-one and onto
...
2 (iv) is one-one and onto
...
Let f : A → N be
function defined by f (x) = roll number of the student x
...
Solution No two different students of the class can have same roll number
...
We can assume without any loss of generality that roll numbers of
students are from 1 to 50
...
Hence, f is not onto
...
Solution The function f is one-one, for f (x1) = f (x2) ⇒ 2x1 = 2x2 ⇒ x1 = x2
...
RELATIONS AND FUNCTIONS
9
Example 9 Prove that the function f : R → R, given by f (x) = 2x, is one-one and onto
...
Also, given any real
y
y
y
in R such that f ( ) = 2
...
Hence, f is onto
...
3
Example 10 Show that the function f : N → N, given by f (1) = f (2) = 1 and f (x) = x – 1,
for every x > 2, is onto but not one-one
...
But f is onto, as given any y ∈ N, y ≠ 1,
we can choose x as y + 1 such that f (y + 1) = y + 1 – 1 = y
...
Example 11 Show that the function f : R → R,
defined as f (x) = x2, is neither one-one nor onto
...
Also, the element – 2 in the co-domain R is
not image of any element x in the domain R
(Why?)
...
Example 12 Show that f : N → N, given by
f ( x)
x 1,if x is odd,
x 1,if x is even
is both one-one and onto
...
4
10
MATHEMATICS
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Solution Suppose f (x1) = f (x2)
...
e
...
Similarly, the possibility of x1 being
even and x2 being odd can also be ruled out, using the similar argument
...
Suppose both x1 and x2 are odd
...
Similarly, if both x1 and x2 are even, then also
f (x1) = f (x2) ⇒ x1 – 1 = x2 – 1 ⇒ x1 = x2
...
Also, any odd number
2r + 1 in the co-domain N is the image of 2r + 2 in the domain N and any even number
2r in the co-domain N is the image of 2r – 1 in the domain N
...
Example 13 Show that an onto function f : {1, 2, 3} → {1, 2, 3} is always one-one
...
Then there exists two elements, say 1 and 2 in the
domain whose image in the co-domain is same
...
Therefore, the range set can have at the most two elements of the
co-domain {1, 2, 3}, showing that f is not onto, a contradiction
...
Example 14 Show that a one-one function f : {1, 2, 3} → {1, 2, 3} must be onto
...
Hence, f has to be onto
...
e
...
In contrast to this, Examples 8
and 10 show that for an infinite set, this may not be true
...
EXERCISE 1
...
Is the result true, if the domain
R∗ is replaced by N with co-domain being same as R∗?
2
...
Prove that the Greatest Integer Function f : R → R, given by f (x) = [x], is neither
one-one nor onto, where [x] denotes the greatest integer less than or equal to x
...
Show that the function f : R∗ → R∗ defined by f (x) =
RELATIONS AND FUNCTIONS
11
4
...
5
...
is
6
...
Show that f is one-one
...
In each of the following cases, state whether the function is one-one, onto or
bijective
...
(ii) f : R → R defined by f (x) = 1 + x2
8
...
Show that f : A × B → B × A such that f (a, b) = (b, a) is
bijective function
...
Let f : N → N be defined by f (n) =
n 1
, if n is odd
2
for all n ∈ N
...
Justify your answer
...
Let A = R – {3} and B = R – {1}
...
Is f one-one and onto? Justify your answer
...
Let f : R → R be defined as f(x) = x4
...
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto
...
Let f : R → R be defined as f (x) = 3x
...
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto
...
4 Composition of Functions and Invertible Function
This leads to the following definition:
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In this section, we will study composition of functions and the inverse of a bijective
function
...
Each student appearing in the Board Examination is assigned a
roll number by the Board which is written by the students in the answer script at the
time of examination
...
Let B ⊂ N be the set of all roll numbers and C ⊂ N be the set of all code
numbers
...
In this process each student is assigned a roll number through the function f
and each roll number is assigned a code number through the function g
...
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Definition 8 Let f : A → B and g : B → C be two functions
...
Fig 1
...
Find gof
...
Example 16 Find gof and fog, if f : R → R and g : R → R are given by f (x) = cos x
and g (x) = 3x2
...
Solution We have gof (x) = g (f (x)) = g (cos x) = 3 (cos x)2 = 3 cos2 x
...
Note that 3cos2 x ≠ cos 3x2, for x = 0
...
RELATIONS AND FUNCTIONS
13
3x + 4
⎧7 ⎫
⎧3⎫
Example 17 Show that if f : R − ⎨ ⎬ → R − ⎨ ⎬ is defined by f ( x) =
and
5x − 7
⎩5⎭
⎩5⎭
he
7x + 4
⎧3⎫
⎧7 ⎫
g : R − ⎨ ⎬ → R − ⎨ ⎬ is defined by g ( x) =
, then fog = IA and gof = IB, where,
5x − 3
⎩5 ⎭
⎩5 ⎭
⎧3⎫
⎧7 ⎫
A = R – ⎨ ⎬ , B = R – ⎨ ⎬ ; IA (x) = x, ∀ x ∈ A, IB (x) = x, ∀ x ∈ B are called identity
5⎭
⎩
⎩5⎭
functions on sets A and B, respectively
...
Example 18 Show that if f : A → B and g : B → C are one-one, then gof : A → C is
also one-one
...
Example 19 Show that if f : A → B and g : B → C are onto, then gof : A → C is
also onto
...
Further, for y ∈ B, there exists an element x in A
14
MATHEMATICS
with f (x) = y, since f is onto
...
Example 20 Consider functions f and g such that composite gof is defined and is oneone
...
Example 21 Are f and g both necessarily onto, if gof is onto?
he
Solution Consider f : {1, 2, 3, 4} → {1, 2, 3, 4, 5, 6} defined as f (x) = x, ∀ x and
g : {1, 2, 3, 4, 5, 6} → {1, 2, 3, 4, 5, 6} as g (x) = x, for x = 1, 2, 3, 4 and g (5) = g (6) = 5
...
But g is clearly not one-one
...
It can be
seen that gof is onto but f is not onto
...
Similarly, gof is onto implies that g is onto
...
Each student appearing
in Class X Examination of the Board is assigned a roll number under the function f and
each roll number is assigned a code number under g
...
The Board officials decode by assigning roll number
back to each code number through a process reverse to g and thus mark gets attached
to roll number rather than code number
...
This helps in assigning mark to the
student scoring that mark
...
Example 22 Let f : {1, 2, 3} → {a, b, c} be one-one and onto function given by
f (1) = a, f (2) = b and f (3) = c
...
Solution Consider g : {a, b, c} → {1, 2, 3} as g (a) = 1, g (b) = 2 and g (c) = 3
...
Remark The interesting fact is that the result mentioned in the above example is true
for an arbitrary one-one and onto function f : X → Y
...
e
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The above discussion, Example 22 and Remark lead to the following definition:
RELATIONS AND FUNCTIONS
15
Definition 9 A function f : X → Y is defined to be invertible, if there exists a function
g : Y → X such that gof = IX and fog = IY
...
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Thus, if f is invertible, then f must be one-one and onto and conversely, if f is
one-one and onto, then f must be invertible
...
is
Example 23 Let f : N → Y be a function defined as f (x) = 4x + 3, where,
Y = {y ∈ N : y = 4x + 3 for some x ∈ N }
...
Find the inverse
...
By the definition of Y, y = 4x + 3,
(4 x + 3 − 3)
( y − 3)
= x and
...
Define g : Y → N by
4
bl
for some x in the domain N
...
This shows that gof = IN
fog (y) = f (g (y)) = f ⎜
⎟=
4
⎝ 4 ⎠
and fog = IY, which implies that f is invertible and g is the inverse of f
...
Consider f : N → Y as f (n) = n2
...
Find the inverse of f
...
This
implies that n =
gof (n) = g (n2) =
y
...
Now,
= y , which shows that
gof = IN and fog = IY
...
Example 25 Let f : N → R be a function defined as f (x) = 4x2 + 12x + 15
...
Find the inverse of f
...
Then y = 4x2 + 12x + 15, for some
x in N, which implies that y = (2x + 3)2 + 6
...
MATHEMATICS
y 6
Let us define g : S → N by g (y) =
...
2
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16
gof = IN and fog =IS
...
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Example 26 Consider f : N → N, g : N → N and h : N → R defined as f (x) = 2x,
g (y) = 3y + 4 and h (z) = sin z, ∀ x, y and z in N
...
Solution We have
ho(gof) (x) = h(gof (x)) = h(g (f (x))) = h (g (2x))
= h(3(2x) + 4) = h(6x + 4) = sin (6x + 4)
Also,
x N
...
This shows that ho(gof) = (hog) o f
...
Theorem 1 If f : X → Y, g : Y → Z and h : Z → S are functions, then
ho(gof ) = (hog) o f
...
ho(gof) = (hog) o f
...
Show that f, g and gof are invertible
...
RELATIONS AND FUNCTIONS
17
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Solution Note that by definition, f and g are bijective functions
...
It is easy to verify that f –1 o f = I{1, 2, 3}, f o f –1 = I{a, b, c}, g –1og = I{a, b, c} and g o g–1 = ID,
where, D = {apple, ball, cat}
...
We can define
(gof)–1 : {apple, ball, cat} → {1, 2, 3} by (gof)–1 (apple) = 1, (gof)–1 (ball) = 2 and
(g o f)–1 (cat) = 3
...
Thus, we have seen that f, g and gof are invertible
...
Hence
(gof)–1 = f –1og –1
...
Theorem 2 Let f : X → Y and g : Y → Z be two invertible functions
...
Proof To show that gof is invertible with (gof)–1 = f –1og–1, it is enough to show that
( f –1og–1)o(gof) = IX and (gof)o( f –1og–1) = IZ
...
Similarly, it can be shown that (gof ) o (f –1 o g –1) = IZ
...
Determine whether the functions f : S → S defined as
below have inverses
...
(a) f = {(1, 1), (2, 2), (3, 3)}
(b) f = {(1, 2), (2, 1), (3, 1)}
(c) f = {(1, 3), (3, 2), (2, 1)}
Solution
(a) It is easy to see that f is one-one and onto, so that f is invertible with the inverse
f –1 of f given by f –1 = {(1, 1), (2, 2), (3, 3)} = f
...
(c) It is easy to see that f is one-one and onto, so that f is invertible with
f –1 = {(3, 1), (2, 3), (1, 2)}
...
3
1
...
Write down gof
...
Let f, g and h be functions from R to R
...
g) o h = (foh)
...
Find gof and fog, if
(i) f (x) = | x | and g(x) = | 5x – 2 |
1
(ii) f (x) = 8x and g(x) = x 3
...
What is the
(6 x − 4)
3
3
inverse of f ?
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State with reason whether following functions have inverse
(i) f : {1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
6
...
Find the inverse
( x + 2)
of the function f : [–1, 1] → Range f
...
e
...
Consider f : R → R given by f (x) = 4x + 3
...
Find the
inverse of f
...
Consider f : R+ → [4, ∞) given by f (x) = x2 + 4
...
y − 4 , where R+ is the set of all non-negative
RELATIONS AND FUNCTIONS
19
9
...
Show that f is invertible
1
(B) x 3
, then fof (x) is
(C) x
(D) (3 – x3)
...
If f : R → R be given by f (x) = (3 −
1
3 3
x )
is
he
⎛ ( y + 6 ) −1⎞
⎟
...
Let f : X → Y be an invertible function
...
(Hint: suppose g1 and g2 are two inverses of f
...
Use one-one ness of f)
...
Consider f : {1, 2, 3} → {a, b, c} given by f (1) = a, f (2) = b and f (3) = c
...
12
...
Show that the inverse of f –1 is f, i
...
,
(f –1)–1 = f
...
Let f : R – ⎨− ⎬ → R be a function defined as f (x) =
...
5 Binary Operations
Right from the school days, you must have come across four fundamental operations
namely addition, subtraction, multiplication and division
...
It is to be noted that only two numbers can be added or
b
multiplied at a time
...
Thus, addition, multiplication, subtraction
or a – b or ab or
20
MATHEMATICS
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and division are examples of binary operation, as ‘binary’ means two
...
This gives rise to a general definition as follows:
Definition 10 A binary operation ∗ on a set A is a function ∗ : A × A → A
...
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Example 29 Show that addition, subtraction and multiplication are binary operations
on R, but division is not a binary operation on R
...
+ : R × R → R is given by
(a, b) → a + b
– : R × R → R is given by
(a, b) → a – b
× : R × R → R is given by
(a, b) → ab
Since ‘+’, ‘–’ and ‘×’ are functions, they are binary operations on R
...
b
However, ÷ : R∗ × R∗ → R∗, given by (a, b) →
a
is a function and hence a
b
binary operation on R∗
...
Solution – : N × N → N, given by (a, b) → a – b, is not binary operation, as the image
of (3, 5) under ‘–’ is 3 – 5 = – 2 ∉ N
...
5
Example 31 Show that ∗ : R × R → R given by (a, b) → a + 4b2 is a binary
operation
...
RELATIONS AND FUNCTIONS
21
Example 32 Let P be the set of all subsets of a given set X
...
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Solution Since union operation ∪ carries each pair (A, B) in P × P to a unique element
A ∪ B in P, ∪ is binary operation on P
...
Example 33 Show that the ∨ : R × R → R given by (a, b) → max {a, b} and the
∧ : R × R → R given by (a, b) → min {a, b} are binary operations
...
Using the similar argument,
one can say that ∧ is also a binary operation
...
When number of elements in a set A is small, we can express a binary operation ∗ on
the set A through a table called the operation table for the operation ∗
...
Then, the operation ∨ on A defined in Example 33 can be expressed
by the following operation table (Table 1
...
Here, ∨ (1, 3) = 3, ∨ (2, 3) = 3, ∨ (1, 2) = 2
...
1
Here, we are having 3 rows and 3 columns in the operation table with (i, j) the
entry of the table being maximum of ith and jth elements of the set A
...
If A = {a1, a2,
...
Then the
operation table will be having n rows and n columns with (i, j)th entry being ai ∗ aj
...
, an}, we can define a binary operation ∗ : A × A → A
given by ai ∗ aj = the entry in the ith row and jth column of the operation table
...
e
...
e
...
Similarly, in case of multiplication of 3 and 4, order is immaterial, but
division of 3 and 4 in different order give different results
...
For subtraction and division we have to write ‘subtract 3 from 4’, ‘subtract
4 from 3’, ‘divide 3 by 4’ or ‘divide 4 by 3’
...
Example 34 Show that + : R × R → R and × : R × R → R are commutative binary
operations, but – : R × R → R and ÷ : R∗ × R∗ → R∗ are not commutative
...
However, ‘–’ is not commutative, since 3 – 4 ≠ 4 – 3
...
is
Example 35 Show that ∗ : R × R → R defined by a ∗ b = a + 2b is not commutative
...
If we want to associate three elements of a set X through a binary operation on X,
we encounter a natural problem
...
For example,
(8 – 5) – 2 ≠ 8 – (5 – 2)
...
But in case
of addition, 8 + 5 + 2 has the same value whether we look at it as ( 8 + 5) + 2 or as
8 + (5 + 2)
...
This leads to the following:
Definition 12 A binary operation ∗ : A × A → A is said to be associative if
(a ∗ b) ∗ c = a ∗ (b ∗ c), ∀ a, b, c, ∈ A
...
But subtraction is not associative on R
...
Solution Addition and multiplication are associative, since (a + b) + c = a + (b + c) and
(a × b) × c = a × (b × c) ∀ a, b, c ∈ R
...
Example 37 Show that ∗ : R × R → R given by a ∗ b → a + 2b is not associative
...
Remark Associative property of a binary operation is very important in the sense that
with this property of a binary operation, we can write a1 ∗ a2 ∗
...
But in absence of this property, the expression a1 ∗ a2 ∗
...
Recall that in the earlier classes brackets were used whenever
subtraction or division operations or more than one operation occurred
...
e
...
But in case of
multiplication, the number 1 plays this role, as a × 1 = a = 1 × a, ∀ a in R
...
Example 38 Show that zero is the identity for addition on R and 1 is the identity for
multiplication on R
...
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Solution a + 0 = 0 + a = a and a × 1 = a = 1 × a, ∀ a ∈ R implies that 0 and 1 are
identity elements for the operations ‘+’ and ‘×’ respectively
...
Similarly, we can not find any element e in R∗ such that
a ÷ e = e ÷ a, ∀ a in R∗
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Remark Zero is identity for the addition operation on R but it is not identity for the
addition operation on N, as 0 ∉ N
...
One further notices that for the addition operation + : R × R → R, given any
a ∈ R, there exists – a in R such that a + (– a) = 0 (identity for ‘+’) = (– a) + a
...
This leads to the following definition:
a
a
Definition 14 Given a binary operation ∗ : A × A → A with the identity element e in A,
an element a ∈ A is said to be invertible with respect to the operation ∗, if there exists
an element b in A such that a ∗ b = e = b ∗ a and b is called the inverse of a and is
denoted by a–1
...
a
Solution As a + (– a) = a – a = 0 and (– a) + a = 0, – a is the inverse of a for addition
...
a
a
a
24
MATHEMATICS
Example 40 Show that – a is not the inverse of a ∈ N for the addition operation + on
N and
1
is not the inverse of a ∈ N for multiplication operation × on N, for a ≠ 1
...
1
∉ N, which implies that other than 1 no element of N
a
has inverse for multiplication operation on N
...
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Examples 34, 36, 38 and 39 show that addition on R is a commutative and associative
binary operation with 0 as the identity element and – a as the inverse of a in R ∀ a
...
Determine whether or not each of the definition of ∗ given below gives a binary
operation
...
(i) On Z+, define ∗ by a ∗ b = a – b
(ii) On Z+, define ∗ by a ∗ b = ab
(iii) On R, define ∗ by a ∗ b = ab2
(iv) On Z+, define ∗ by a ∗ b = | a – b |
(v) On Z+, define ∗ by a ∗ b = a
2
...
(i) On Z, define a ∗ b = a – b
(ii) On Q, define a ∗ b = ab + 1
(iii) On Q, define a ∗ b =
ab
2
(iv) On Z+, define a ∗ b = 2ab
(v) On Z+, define a ∗ b = ab
(vi) On R – {– 1}, define a ∗ b =
a
b +1
3
...
Write the operation table of the operation ∧
...
Consider a binary operation ∗ on the set {1, 2, 3, 4, 5} given by the following
multiplication table (Table 1
...
(i) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4)
(ii) Is ∗ commutative?
(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5)
...
2
5
...
C
...
of a and b
...
6
...
C
...
of a and b
...
Is ∗ defined on the set {1, 2, 3, 4, 5} by a ∗ b = L
...
M
...
8
...
C
...
of a and b
...
Let ∗ be a binary operation on the set Q of rational numbers as follows:
(i) a ∗ b = a – b
(ii) a ∗ b = a2 + b2
(iii) a ∗ b = a + ab
(iv) a ∗ b = (a – b)2
ab
(v) a ∗ b =
(vi) a ∗ b = ab2
4
Find which of the binary operations are commutative and which are associative
...
Find which of the operations given above has identity
...
Let A = N × N and ∗ be the binary operation on A defined by
(a, b) ∗ (c, d) = (a + c, b + d)
26
MATHEMATICS
Show that ∗ is commutative and associative
...
12
...
Justify
...
(ii) If ∗ is a commutative binary operation on N, then a ∗ (b ∗ c) = (c ∗ b) ∗ a
13
...
Choose the
correct answer
...
Solution Since R1 and R2 are equivalence relations, (a, a) ∈ R1, and (a, a) ∈ R2 ∀ a ∈ A
...
Further,
(a, b) ∈ R1 ∩ R2 ⇒ (a, b) ∈ R1 and (a, b) ∈ R2 ⇒ (b, a) ∈ R1 and (b, a) ∈ R2 ⇒
(b, a) ∈ R1 ∩ R2, hence, R1 ∩ R2 is symmetric
...
This shows that
R1 ∩ R2 is transitive
...
Example 42 Let R be a relation on the set A of ordered pairs of positive integers
defined by (x, y) R (u, v) if and only if xv = yu
...
Solution Clearly, (x, y) R (x, y), ∀ (x, y) ∈ A, since xy = yx
...
Further, (x, y) R (u, v) ⇒ xv = yu ⇒ uy = vx and hence (u, v) R (x, y)
...
Similarly, (x, y) R (u, v) and (u, v) R (a, b) ⇒ xv = yu and
b
a
a
a
= yu ⇒ xv = yu ⇒ xb = ya and hence (x, y) R (a, b)
...
Thus, R is an equivalence relation
...
Let R1 be a relation in X given
by R1 = {(x, y) : x – y is divisible by 3} and R2 be another relation on X given by
R2 = {(x, y): {x, y} ⊂ {1, 4, 7}} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9}}
...
RELATIONS AND FUNCTIONS
27
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Solution Note that the characteristic of sets {1, 4, 7}, {2, 5, 8} and {3, 6, 9} is
that difference between any two elements of these sets is a multiple of 3
...
Hence, R1 ⊂ R2
...
This shows that R2 ⊂ R1
...
Example 44 Let f : X → Y be a function
...
Examine whether R is an equivalence relation or not
...
Similarly, (a, b) ∈ R ⇒ f (a) = f (b) ⇒ f (b) = f (a) ⇒ (b, a) ∈ R
...
Further, (a, b) ∈ R and (b, c) ∈ R ⇒ f (a) = f (b) and f (b) = f (c) ⇒ f (a)
= f (c) ⇒ (a, c) ∈ R, which implies that R is transitive
...
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Example 45 Determine which of the following binary operations on the set R are
associative and which are commutative
...
Also
(a ∗ b) ∗ c = (1 ∗ c) =1 and a ∗ (b ∗ c) = a ∗ (1) = 1, ∀ a, b, c ∈ R
...
(b) a ∗ b =
a+b b+a
=
= b ∗ a, shows that ∗ is commutative
...
⎝ 2 ⎠
⎛a+b⎞
⎜
⎟ + c a + b + 2c
⎝ 2 ⎠
=
...
b+c
2 = 2a + b + c ≠ a + b + 2c in general
...
Solution One-one function from {1, 2, 3} to itself is simply a permutation on three
symbols 1, 2, 3
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Example 47 Let A = {1, 2, 3}
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Solution The smallest relation R1 containing (1, 2) and (2, 3) which is reflexive and
transitive but not symmetric is {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}
...
Similarly, we can obtain R3 by adding (3, 2) to R1 to get the desired relation
...
Thus,
the total number of desired relations is three
...
Solution The smallest equivalence relation R1 containing (1, 2) and (2, 1) is {(1, 1),
(2, 2), (3, 3), (1, 2), (2, 1)}
...
If we add any one, say (2, 3) to R1, then for symmetry we must add
(3, 2) also and now for transitivity we are forced to add (1, 3) and (3, 1)
...
This shows that the total
number of equivalence relations containing (1, 2) and (2, 1) is two
...
Solution A binary operation ∗ on {1, 2} is a function from {1, 2} × {1, 2} to {1, 2}, i
...
,
a function from {(1, 1), (1, 2), (2, 1), (2, 2)} → {1, 2}
...
Since 2 is the inverse of 2, i
...
, ∗ (2, 2) must be equal to 1
...
Example 50 Consider the identity function IN : N → N defined as IN (x) = x ∀ x ∈ N
...
Solution Clearly IN is onto
...
RELATIONS AND FUNCTIONS
29
⎡ π⎤
Example 51 Consider a function f : ⎢0, ⎥ → R given by f (x) = sin x and
⎣ 2⎦
he
π
g : ⎡0, ⎤ → R given by g(x) = cos x
...
is
⎡ π⎤
Solution Since for any two distinct elements x1 and x2 in ⎢0, ⎥ , sin x1 ≠ sin x2 and
⎣ 2⎦
cos x1 ≠ cos x2, both f and g must be one-one
...
Therefore, f + g is not one-one
...
Let f : R → R be defined as f (x) = 10x + 7
...
2
...
Show that f is invertible
...
Here, W is the set of all
whole numbers
...
If f : R → R is defined by f(x) = x2 – 3x + 2, find f (f (x))
...
Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by f ( x) =
x
,
1+ | x |
x ∈ R is one one and onto function
...
Show that the function f : R → R given by f (x) = x3 is injective
...
Give examples of two functions f : N → Z and g : Z → Z such that g o f is
injective but g is not injective
...
7
...
⎧ x − 1 if x > 1
(Hint : Consider f (x) = x + 1 and g ( x ) = ⎨
⎩ 1 if x = 1
8
...
10
...
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12
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Is R an equivalence relation
on P(X)? Justify your answer
...
Show that X is the identity element for this operation and X is the only invertible
element in P(X) with respect to the operation ∗
...
, n} to itself
...
Find F–1 of the following functions F from S
to T, if it exists
...
Show that ∗ is commutative but not
associative, o is associative but not commutative
...
[If it is so, we say that the operation ∗ distributes
over the operation o]
...
Given a non-empty set X, let ∗ : P(X) × P(X) → P(X) be defined as
A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X)
...
(Hint : (A – φ) ∪ (φ – A) = A and (A – A) ∪ (A – A) = A ∗ A = φ)
...
MATHEMATICS
is
30
13
...
if a + b < 6
⎧ a + b,
a ∗b = ⎨
⎩ a + b − 6 if a + b ≥ 6
Show that zero is the identity for this operation and each element a ≠ 0 of the set
is invertible with 6 – a being the inverse of a
...
Let A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2} and f, g : A → B be functions defined
1
− 1, x ∈ A
...
(Hint: One may note that two functions f : A → B and
g : A → B such that f (a) = g (a) ∀ a ∈ A, are called equal functions)
...
Let A = {1, 2, 3}
...
Let A = {1, 2, 3}
...
Let f : R → R be the Signum Function defined as
⎧ 1, x > 0
⎪
f ( x) = ⎨ 0, x = 0
⎪−1, x < 0
⎩
Summary
is
he
and g : R → R be the Greatest Integer Function given by g (x) = [x], where [x] is
greatest integer less than or equal to x
...
Number of binary operations on the set {a, b} are
(A) 10
(B) 16
(C) 20
(D ) 8
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In this chapter, we studied different types of relations and equivalence relation,
composition of functions, invertible functions and binary operations
...
Universal relation is the relation R in X given by R = X × X
...
Symmetric relation R in X is a relation satisfying (a, b) ∈ R implies (b, a) ∈ R
...
Equivalence relation R in X is a relation which is reflexive, symmetric and
transitive
...
A function f : X → Y is one-one (or injective) if
f (x1) = f (x2) ⇒ x1 = x2 ∀ x1, x2 ∈ X
...
A function f : X → Y is one-one and onto (or bijective), if f is both one-one
and onto
...
A function f : X → Y is invertible if ∃ g : Y → X such that gof = IX and
fog = IY
...
32
MATHEMATICS
Historical Note
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is
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Given a finite set X, a function f : X → X is one-one (respectively onto) if and
only if f is onto (respectively one-one)
...
This is not true for infinite set
A binary operation ∗ on a set A is a function ∗ from A × A to A
...
An element a ∈ X is invertible for binary operation ∗ : X × X → X, if
there exists b ∈ X such that a ∗ b = e = b ∗ a where, e is the identity for the
binary operation ∗
...
An operation ∗ on X is commutative if a ∗ b = b ∗ a ∀ a, b in X
...
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The concept of function has evolved over a long period of time starting from
R
...
James
Gregory (1636-1675) in his work “ Vera Circuli et Hyperbolae Quadratura”
(1667) considered function as a quantity obtained from other quantities by
successive use of algebraic operations or by any other operations
...
W
...
However,
in his manuscript “Historia” (1714), Leibnitz used the word ‘function’ to mean
quantities that depend on a variable
...
John Bernoulli (1667-1748) used the notation φx for the first time in 1718 to
indicate a function of x
...
to
represent functions was made by Leonhard Euler (1707-1783) in 1734 in the first
part of his manuscript “Analysis Infinitorium”
...
for different function of x
...
The set theoretic definition of function
known to us presently is simply an abstraction of the definition given by Dirichlet
in a rigorous manner