Mathematics

                                Chapter 1 – Relations and Functions

                                                                   

Exercise 1.1

Question 1:

Determine whether each of the following relations are reflexive, symmetric and

transitive:

(i)Relation R in the set A = {1, 2, 3…13, 14} defined as

R = {(x, y): 3x− y = 0}

(ii) Relation R in the set N of natural numbers defined as

R = {(x, y): y = x + 5 and x < 4}

(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as

R = {(x, y): y is divisible by x}

(iv) Relation R in the set Z of all integers defined as

R = {(x, y): x − y is as integer}

(v) Relation R in the set A of human beings in a town at a particular time given by

(a) R = {(x, y): x and y work at the same place}

(b) R = {(x, y): x and y live in the same locality}

(c) R = {(x, y): x is exactly 7 cm taller than y}

(d) R = {(x, y): x is wife of y}

(e) R = {(x, y): x is father of y}

Solutions

(i) A = {1, 2, 3 … 13, 14}

R = {(x, y): 3x − y = 0}

∴R = {(1, 3), (2, 6), (3, 9), (4, 12)}

R is not reflexive since (1, 1), (2, 2) … (14, 14) ∉ R.

Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R. [3(3) − 1 ≠ 0]

Also, R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R.

[3(1) − 9 ≠ 0]

Hence, R is neither reflexive, nor symmetric, nor transitive.

(ii) R = {(x, y): y = x + 5 and x < 4} = {(1, 6), (2, 7), (3, 8)}

It is seen that (1, 1) ∉ R.

∴R is not reflexive.

(1, 6) ∈R But,(1, 6) ∉ R.

∴R is not symmetric.

Now, since there is no pair in R such that (x, y) and (y, z) ∈R

∴ it is not transitive

(iii) A = {1, 2, 3, 4, 5, 6}

R = {(x, y): y is divisible by x}

We know that any number (x) is divisible by itself.

(x, x) ∈R

∴R is reflexive.

Now,

(2, 4) ∈R [as 4 is divisible by 2]

But,

(4, 2) ∉ R. [as 2 is not divisible by 4]

∴R is not symmetric.

Let (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y.

∴z is divisible by x.

⇒ (x, z) ∈R

∴R is transitive.

Hence, R is reflexive and transitive but not symmetric.

(iv) R = {(x, y): x − y is an integer}

Now, for every x ∈ Z, (x, x) ∈R as x − x = 0 is an integer.

∴R is reflexive.

Now, for every x, y ∈ Z if (x, y) ∈ R, then x − y is an integer.

⇒ −(x − y) is also an integer.

⇒ (y − x) is an integer.

∴ (y, x) ∈ R

∴R is symmetric.

Now,

Let (x, y) and (y, z) ∈R, where x, y, z ∈ Z.

⇒ (x − y) and (y − z) are integers.

⇒ x − z = (x − y) + (y − z) is an integer.

∴ (x, z) ∈R

∴R is transitive.

Hence, R is reflexive, symmetric, and transitive.

(v) (a) R = {(x, y): x and y work at the same place}

(x, x) ∈ R

∴ R is reflexive.

If (x, y) ∈ R, then x and y work at the same place.

⇒ y and x work at the same place.

⇒(y, x) ∈ R.

∴R is symmetric.

Now, let (x, y), (y, z) ∈ R

⇒ x and y work at the same place and y and z work at the same place.

⇒ x and z work at the same place.

⇒ (x, z) ∈R

∴ R is transitive.

Hence, R is reflexive, symmetric, and transitive.

(b) R = {(x, y): x and y live in the same locality}

Clearly (x, x) ∈ R as x and x is the same human being.

∴ R is reflexive.

If (x, y) ∈R, then x and y live in the same locality.

⇒ y and x live in the same locality.

⇒ (y, x) ∈ R

∴R is symmetric.

Now, let (x, y) ∈ R and (y, z) ∈ R.

⇒ x and y live in the same locality and y and z live in the same locality.

⇒ x and z live in the same locality.

⇒ (x, z) ∈ R

∴ R is transitive.

Hence, R is reflexive, symmetric, and transitive.

(c) R = {(x, y): x is exactly 7 cm taller than y}

Now,

(x, x) ∉ R

Since human being x cannot be taller than himself.

∴R is not reflexive.

Now, let (x, y) ∈R.

⇒ x is exactly 7 cm taller than y.

Then, y is not taller than x.

∴ (y, x) ∉R

Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x.

∴R is not symmetric.

Now,

Let (x, y), (y, z) ∈ R.

⇒ x is exactly 7 cm taller than y and y is exactly 7 cm taller than z.

⇒ x is exactly 14 cm taller than z .

∴(x, z) ∉R

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

(d) R = {(x, y): x is the wife of y}

Now,

(x, x) ∉ R

Since x cannot be the wife of herself.

∴R is not reflexive.

Now, let (x, y) ∈ R

⇒ x is the wife of y.

Clearly y is not the wife of x.

∴(y, x) ∉ R

Indeed if x is the wife of y, then y is the husband of x.

∴ R is not transitive.

Let (x, y), (y, z) ∈ R

⇒ x is the wife of y and y is the wife of z.

This case is not possible. Also, this does not imply that x is the wife of z.

∴(x, z) ∉ R

∴R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

(e) R = {(x, y): x is the father of y}

(x, x) ∉ R

As x cannot be the father of himself.

∴R is not reflexive.

Now, let (x, y) ∈R.

⇒ x is the father of y.

⇒ y cannot be the father of y.

Indeed, y is the son or the daughter of y.

∴(y, x) ∉ R

∴ R is not symmetric.

Now, let (x, y) ∈ R and (y, z) ∈ R.

⇒ x is the father of y and y is the father of z.

⇒ x is not the father of z.

Indeed x is the grandfather of z.

∴ (x, z) ∉ R

∴R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

Question 2:

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.

Solution

R = {(a, b): a ≤ b²}

It can be observed that ()R, since

∴R is not reflexive.

Now, (1, 4) ∈ R as 1 <

But, 4 is not less than.

∴(4, 1) ∉ R

∴R is not symmetric.

Now,(3, 2), (2, 1.5) ∈ R

(as 3 < 22 = 4 and 2 < (1.5)² = 2.25)

But, 3 > (1.5)² = 2.25

∴(3, 1.5) ∉ R

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

Question 3:

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.

Solution

Let A = {1, 2, 3, 4, 5, 6}.

A relation R is defined on set A as:

R = {(a, b): b = a + 1}

∴R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

We can find (a, a) ∉ R, where a ∈ A.

For instance,

(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R

∴R is not reflexive.

It can be observed that (1, 2) ∈ R, but (2, 1) ∉ R.

∴R is not symmetric.

Now, (1, 2), (2, 3) ∈ R

But,

(1, 3) ∉ R

∴R is not transitive

Hence, R is neither reflexive, nor symmetric, nor transitive.

Question 4:

Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive

but not symmetric.

Solution

R = {(a, b); a ≤ b}

Clearly (a, a) ∈ R as a = a.

∴R is reflexive.

Now,

(2, 4) ∈ R (as 2 < 4)

But, (4, 2) ∉ R as 4 is greater than 2.

∴ R is not symmetric.

Now, let (a, b), (b, c) ∈ R.

Then,

a ≤ b and b ≤ c

⇒ a ≤ c

⇒ (a, c) ∈ R

∴R is transitive.

Hence, R is reflexive and transitive but not symmetric.

Question 5:

Check whether the relation R in R defined as R = {(a, b): a ≤ b³} is reflexive, symmetric

or transitive.

Solution

R = {(a, b): a ≤ b³}

It is observed that

∴ R is not reflexive.

Now,

(1, 2) ∈ R (as 1 < 23 = 8)

But,

(2, 1) ∉ R (as 23 > 1)

∴ R is not symmetric.

We have

But

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive

Question 6:

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.

Solution

Let A = {1, 2, 3}.

A relation R on A is defined as R = {(1, 2), (2, 1)}.

It is seen that (1, 1), (2, 2), (3, 3) ∉R.

∴ R is not reflexive.

Now, as (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric.

Now, (1, 2) and (2, 1) ∈ R

However,

(1, 1) ∉ R

∴ R is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

Question 7:

Show that the relation R in the set A of all the books in a library of a college, given by R

= {(x, y): x and y have same number of pages} is an equivalence relation.

Solution

Set A is the set of all books in the library of a college.

R = {x, y): x and y have the same number of pages}

Now, R is reflexive since (x, x) ∈ R as x and x has the same number of pages.

Let (x, y) ∈ R ⇒ x and y have the same number of pages.

⇒ y and x have the same number of pages.

⇒ (y, x) ∈ R

∴R is symmetric.

Now, let (x, y) ∈R and (y, z) ∈ R.

⇒ x and y and have the same number of pages and y and z have the same number ofpages.

⇒ x and z have the same number of pages.

⇒ (x, z) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

Question 8:

Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a,b):|a – b| is even} , is an equivalence relation. Show that all the elements of {1,3, 5} are related to each other and all the elements of {2, 4} are related to each other.

But no element of {1, 3, 5} is related to any element of 2, 4}.

Solution

A = {1, 2, 3, 4, 5} R = {(a,b):|a–b| is even} It is clear that for any element a ∈A, we have |a–b| = 0 (which is even).

∴R is reflexive.

Let (a, b) ∈ R. => |a–b| is even.

ð  |-(a – b)| = |b–a| is also even

ð  (b,a) ∈ R

∴R is symmetric.

Now, let (a, b) ∈ R and (b, c) ∈ R.

ð  |a–b| is even and |b-a| is even

ð  (a–b) is even and (b-a) is even

ð  (a-c) = (a-b)+(b-c) is even

ð  |a-c| is even

⇒ (a, c) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

Now, all elements of the set {1, 2, 3} are related to each other as all the elements of

this subset are odd. Thus, the modulus of the difference between any two elements will be even.

Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even. Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.

 Question 9:

Show that each of the relation R in the set A = {x∈Z:0≤x≤12}, given by

(i)R = {(a,b):|a-b| is a multiple of 4}

(ii) R = {(a,b):a = b}

is an equivalence relation. Find the set of all elements related to 1 in each case.

Solution

(i) R = {(a,b):|a-b| is a multiple of 4}

For any element a ∈A, we have (a, a) ∈ R |a-a|=0 is a multiple of 4.

∴R is reflexive.

Now, let (a, b) ∈ R ⇒|a-a| is a multiple of 4.

⇒ (b, a) ∈ R

∴R is symmetric.

Now, let (a, b), (b, c) ∈ R.

ð  |a-b| is a multiple of 4 and |b-c| is a multiple of 4

ð  (a-b) is a multiple of 4 and (b-c) is a multiple of 4

ð  (a-c) = (a-b)+(b-c) is multiple of 4

ð  |a-c| is a multiple of 4

⇒ (a, c) ∈R

∴ R is transitive.

Hence, R is an equivalence relation.

(ii) R = {(a, b): a = b}

For any element a ∈A, we have (a, a) ∈ R, since a = a.

∴R is reflexive.

Now, let (a, b) ∈ R.

⇒ a = b

⇒ b = a

⇒ (b, a) ∈ R

∴R is symmetric.

Now, let (a, b) ∈ R and (b, c) ∈ R.

⇒ a = b and b = c

⇒ a = c

⇒ (a, c) ∈ R

∴ R is transitive.

Hence, R is an equivalence relation.

The elements in R that are related to 1 will be those elements from set A which are equal to 1.

Hence, the set of elements related to 1 is {1}.

Question 10:

Given an example of a relation. Which is

(i) Symmetric but neither reflexive nor transitive.

(ii) Transitive but neither reflexive nor symmetric.

(iii) Reflexive and symmetric but not transitive.

(iv)Reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive.

Solution

(i) Let A = {5, 6, 7}.

Define a relation R on A as R = {(5, 6), (6, 5)}.

Relation R is not reflexive as (5, 5), (6, 6), (7, 7) ∉ R.

Now, as (5, 6) ∈ R and also (6, 5) ∈ R, R is symmetric.

(5, 6), (6, 5) ∈ R, but (5, 5) ∉ R

∴R is not transitive.

Hence, relation R is symmetric but not reflexive or transitive.

(ii)Consider a relation R in R defined as:

R = {(a, b): a < b}

For any a ∈ R, we have (a, a) ∉ R since a cannot be strictly less than a itself. In fact, a =a.

∴ R is not reflexive.

Now,(1, 2) ∈ R (as 1 < 2)

But, 2 is not less than 1.

∴ (2, 1) ∉ R

∴ R is not symmetric.

Now, let (a, b), (b, c) ∈ R.

⇒ a < b and b < c

⇒ a < c

⇒ (a, c) ∈ R

∴R is transitive.

Hence, relation R is transitive but not reflexive and symmetric.

(iii)Let A = {4, 6, 8}.

Define a relation R on A as:

A = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}

Relation R is reflexive since for every a ∈ A, (a, a) ∈R i.e., (4, 4), (6, 6), (8, 8)} ∈ R.

Relation R is symmetric since (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ R.

Relation R is not transitive since (4, 6), (6, 8) ∈ R, but (4, 8) ∉ R.

Hence, relation R is reflexive and symmetric but not transitive.

(iv)Define a relation R in R as: R = {a, b): a³ ≥ b³}

Clearly (a, a) ∈ R as a³ = a³.

∴R is reflexive.

Now,

(2, 1) ∈ R (as 2³ ≥ 1³)

But,

(1, 2) ∉ R (as 1³ < 2³)

∴ R is not symmetric.

Now,

Let (a, b), (b, c) ∈ R.

⇒ a³ ≥ b³ and b³ ≥ c³

⇒ a³ ≥ c³

⇒ (a, c) ∈ R

∴R is transitive.

Hence, relation R is reflexive and transitive but not symmetric.

(v) Let A = {−5, −6}.

Define a relation R on A as:

R = {(−5, −6), (−6, −5), (−5, −5)}

Relation R is not reflexive as (−6, −6) ∉ R.

Relation R is symmetric as (−5, −6) ∈ R and (−6, −5} ∈ R.

It is seen that (−5, −6), (−6, −5) ∈ R. Also,(−5, −5) ∈ R.

∴The relation R is transitive.

Hence, relation R is symmetric and transitive but not reflexive.

Question 11:

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. Further, show that the set of all point related to a point P ≠ (0,

0) is the circle passing through P with origin as centre.

Solution

R = {(P, Q): distance of point P from the origin is the same as the distance of point Q from the origin}

Clearly, (P, P) ∈ R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.

∴R is reflexive.

Now,

Let(P, Q) ∈ R.

⇒ The distance of point P from the origin is the same as the distance of point Q from the origin.

⇒ The distance of point Q from the origin is the same as the distance of point P from the origin.

⇒ (Q, P) ∈ R

∴R is symmetric.

Now,

Let (P, Q), (Q, S) ∈ R.

⇒ The distance of points P and Q from the origin is the same and also, the distance of

points Q and S from the origin is the same.

⇒ The distance of points P and S from the origin is the same.

⇒ (P, S) ∈ R

∴R is transitive.

Therefore, R is an equivalence relation.

The set of all points related to P ≠ (0, 0) will be those points whose distance from the origin is the same as the distance of point P from the origin.

In other words, if O (0, 0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin.

Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P.

Question 12:

Show that the relation R defined in the set A of all triangles as R = {(T1, T2): T1 is similar to T2}, is equivalence relation. Consider three right angle triangles T1 with sides 3, 4, 5, T2 with sides 5, 12, 13 and T3 with sides 6, 8, 10. Which triangles among T1, T2 and T3 are related?

Solution

R = {(T1, T2): T1 is similar to T2}

R is reflexive since every triangle is similar to itself.

Further, if (T1, T2) ∈ R, then T1 is similar to T2.

⇒ T2 is similar to T1.

⇒ (T2, T1) ∈R

∴R is symmetric.

Now,

Let (T1, T2), (T2, T3) ∈ R.

⇒ T1 is similar to T2 and T2 is similar to T3.

⇒ T1 is similar to T3.

⇒ (T1, T3) ∈ R

∴ R is transitive.

Thus, R is an equivalence relation.

Now, we can observe that:

∴The corresponding sides of triangles T1 and T3 are in the same ratio.

Then, triangle T1 is similar to triangle T3.

Hence, T1 is related to T3.

Question 13:

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. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Solution

R = {(P1, P2): P1 and P2 have same the number of sides}

R is reflexive since (P1, P1) ∈ R as the same polygon has the same number of sides with

itself.

Let (P1, P2) ∈ R.

⇒ P1 and P2 have the same number of sides.

⇒ P2 and P1 have the same number of sides.

⇒ (P2, P1) ∈ R

∴R is symmetric.

Now,

Let (P1, P2), (P2, P3) ∈ R.

⇒ P1 and P2 have the same number of sides. Also, P2 and P3 have the same number of

sides.

⇒ P1 and P3 have the same number of sides.

⇒ (P1, P3) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

The elements in A related to the right-angled triangle (T) with sides 3, 4, and 5 are

those polygons which have 3 sides (since T is a polygon with 3 sides).

Hence, the set of all elements in A related to triangle T is the set of all triangles.

Question 14:

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}. Show that R is an equivalence relation. Find the set of all lines

related to the line y = 2x + 4.

Solution

R = {(L1, L2): L1 is parallel to L2}

R is reflexive as any line L1 is parallel to itself i.e., (L1, L1) ∈ R.

Now,

Let (L1, L2) ∈ R.

⇒ L1 is parallel to L2.

⇒ L2 is parallel to L1.

⇒ (L2, L1) ∈ R

∴ R is symmetric.

Now,

Let (L1, L2), (L2, L3) ∈R.

⇒ L1 is parallel to L2. Also, L2 is parallel to L3.

⇒ L1 is parallel to L3.

∴R is transitive.

Hence, R is an equivalence relation.

The set of all lines related to the line y = 2x + 4 is the set of all lines that are parallel to

the line y = 2x + 4.

Slope of line y = 2x + 4 is m = 2

It is known that parallel lines have the same slopes.

The line parallel to the given line is of the form y = 2x + c, where c ∈R.

Hence, the set of all lines related to the given line is given by y = 2x + c, where c ∈ R.

Exercise 1.2

Question 1:

Show that the function f: R* → R* defined by f(x) =  is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R*?

Solution

It is given that f: R* → R* is defined by f(x) = One-one: f(x) = f(y) =>  =

ð  x = y

∴f is one-one.

Onto:

It is clear that for y∈ R*, there exists x= ∈ R such that f(x) =  = y

∴f is onto.

Thus, the given function (f) is one-one and onto.

Now, consider function g: N → R*defined by g(x) =  

We have, g(x₁) = g(x₂) =>  =

ð  x₁ = x₂

∴g is one-one.

Further, it is clear that g is not onto as for 1.2 ∈R* there does not exit any x in N such

That g(x) = .

Hence, function g is one-one but not onto.

Question 2:

Check the injectivity and surjectivity of the following functions:

(i) f: N → N given by f(x) = x²

(ii) f: Z → Z given by f(x) = x²

(iii) f: R → R given by f(x) = x²

(iv) f: N → N given by f(x) = x³

(v) f: Z → Z given by f(x) = x³

Solution

(i) f: N → N is given by,

f(x) = x²

It is seen that for x, y ∈N, f(x) = f(y) ⇒ x² = y² ⇒ x = y.

∴f is injective.

Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x² = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.

(ii) f: Z → Z is given by,

f(x) = x²

It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.

∴ f is not injective.

Now,−2 ∈ Z. But, there does not exist any element x ∈Z such that f(x) = x² = −2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

(iii) f: R → R is given by,

f(x) = x²

It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.

∴ f is not injective.

Now,−2 ∈ R. But, there does not exist any element x ∈ R such that f(x) = x² = −2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

(iv) f: N → N given by, f(x) = x³

It is seen that for x, y ∈N, f(x) = f(y) ⇒ x³ = y³ ⇒ x = y.

∴f is injective.

Now, 2 ∈ N. But, there does not exist any element x in domain N such that f(x) = x³ =2.

∴ f is not surjective

Hence, function f is injective but not surjective.

(v) f: Z → Z is given by, f(x) = x³

It is seen that for x, y ∈ Z, f(x) = f(y) ⇒x³ = y³

⇒ x = y.

∴ f is injective.

Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x³ = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.

Question 3:

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.

Solution

f: R → R is given by,

f(x) = [x]

It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1.

∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9.

∴ f is not one-one.

Now, consider 0.7 ∈ R.

It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x

∈ R such that f(x) = 0.7.

∴ f is not onto.

Hence, the greatest integer function is neither one-one nor onto.

Question 4:

Show that the Modulus Function f: R → R given by f(x)=|x|, is neither one-one nor

onto, where |x| is x, if x is positive or 0 and |x| is − x, if x is negative.

Solution

f: R → R is given by,

It is seen that.

∴f(−1) = f(1), but −1 ≠ 1.

∴ f is not one-one.

Now, consider −1 ∈ R.

It is known that f(x) = |x| is always non-negative. Thus, there does not exist any

element x in domain R such that f(x) =|x| = −1.

∴ f is not onto.

Hence, the modulus function is neither one-one nor onto.

Question 5:

Show that the Signum Function f: R → R, given by f(x) =  is neither one-one nor onto.

Solution

f: R → R is given by f(x) = ,It is seen that f(1) = f(2) = 1, but 1 ≠ 2.

∴f is not one-one.

Now, as f(x) takes only 3 values (1, 0, or −1) for the element −2 in co-domain R, there does not exist any x in domain R such that f(x) = −2.

∴ f is not onto.

Hence, the signum function is neither one-one nor onto.

Question 6:

Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

Solution

It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.

f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.

∴ f (1) = 4, f (2) = 5, f (3) = 6

It is seen that the images of distinct elements of A under f are distinct.

Hence, function f is one-one.

Question 7:

In each of the following cases, state whether the function is one-one, onto or bijective.

Justify your Solution.

(i) f: R → R defined by f(x) = 3 − 4x

(ii) f: R → R defined by f(x) = 1 + x²

Solution

(i) f: R → R is defined as f(x) = 3 − 4x.

Let x₁, x₂ ∈ R such that f(x₁) = f(x₂)

ð  3-4x₁ = 3-4x₂

ð  -4x₁ = -4x₂

ð  x₁ =x₂

∴ f is one-one.

For any real number (y) in R, there exists in R such that f() = 3 – 4()

ð  y

∴f is onto.

Hence, f is bijective.

(ii) f: R → R is defined as f(x) = 1+x²

Let x₁, x₂ ∈ R such that f(x₁) = f(x₂)

ð  1+x₁² = 1+x₂²

ð  x₁² = x₂²

ð  x₁= ± x₂

∴ f(x₁) = f(x₂)does not imply that x₁ =x₂

For instance, f(1) = f(-1) = 2

∴ f is not one-one.

Consider an element −2 in co-domain R.

It is seen that f(x) = 1+x²  is positive for all x ∈ R.

Thus, there does not exist any x in domain R such that f(x) = −2.

∴ f is not onto.

Hence, f is neither one-one nor onto.

Question 8:

Let A and B be sets. Show that f: A × B → B × A such that (a, b) = (b, a) is bijective function.

Solution

f: A × B → B × A is defined as f(a, b) = (b, a).

let (a₁,b₁), (a₂,b₂) ∈ A x B such that f(a₁,b₁) = f(a₂,b₂)

ð  (b₁,a₁) = (b₂,a₂)

ð  b₁=b₂ and a₁­=a₂

ð  (a₁,b₁) = (a₂,b₂)

∴ f is one-one.

Now, let (b, a) ∈ B × A be any element.

Then, there exists (a, b) ∈A × B such that f(a, b) = (b, a). [By definition of f]

∴ f is onto.

Hence, f is bijective.

Question 9:

Let f: N →N be defined by f(n) =  for all n  N

State whether the function f is bijective. Justify your Solution.

Solution

It can be observed that: f(1) =  = 1 and f(2) =  = 1

f(1) = f(2), where 1 2

∴ f is not one-one.

Consider a natural number (n) in co-domain N.

Case I: n is odd

∴n = 2r + 1 for some r ∈ N. Then, there exists 4r + 1∈N such that f(4r + 1) = = 2r +1

Case II: n is even

∴n = 2r for some r ∈ N. Then, there exists 4r ∈N such that .

∴ f is onto.

Hence, f is not a bijective function.

Question 10:

Let A = R − {3} and B = R − {1}. Consider the function f: A → B defined by. Is f one-one and onto? Justify your Solution.

Solution:

Let x,y  A such that f(x) = f(y)

ð 

ð  (x-2)(y-3) = (y-2)(x-3)

ð  xy-3x-2y+6 = xy-3y-2x+6

ð  -3x-2y = -3y-2x

ð  3x-2y = 3y-2y

ð  x = y

∴ f is one-one.

Let y ∈ B = R − {1}. Then, y ≠ 1.

The function f is onto if there exists x ∈A such that f(x) = y.

Now, f(x) = y

ð 

ð  x-2 = xy-3y

ð  x(1-y) = -3y+2

ð  x =  

Thus, for any y ∈ B, there exists such that 

=  

f is onto.

Hence, function f is one-one and onto.

Exercise 1.3

Question 1:

Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5),

(4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.

Solution

The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as

f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}.

gof(1)=g(f(1))=g(2)=3

gof(3)=g(f(3))=g(5)=1

gof(4)=g(f(4))=g(1)=3

gof ={(1,3),(3,1),(4,3)}

Question 2:

Let f, g and h be functions from R to R. Show that (f + g)oh=foh + goh

(f . g)oh =(foh).(goh)

Solution:

i)              ((f + g )oh)(x)

ð  (f + g)(h(x))

ð  f(h(x)) + g(h(x))

ð  (foh)(x) +( goh)(x)

ð  ((f + g )oh)(x) = (foh)(x) +( goh)(x)

Hence proved

ii)           ((f.g)oh)(x)

ð  (f.g)(h(x)

ð  f(h(x).g(h(x))

ð  (foh)(x).( goh)(x)

ð  ((f.g)oh)(x) = (foh)(x).( goh)(x)

Hence proved

Question 3:

Find gof and fog, if

(i)                  f(x) =|x| and g(x) = |5x-2|

(ii)                 f(x) = 8x³ and g(x) =

Solution

(i)                  (gof)(x) = g(f(x)) =g(|x|) = |5|x|-2|

  (fog)(x) = f(g(x)) = f(|5x-2|) = ||5x-2|| = |5x-2|

(ii)                 (gof)(x) = g(f(x)) =g(8x³) = = 2x

(fog)(x) = f(g(x)) = f() = 8() = 8x

Question 4:

If  , show that f o f(x) = x, for all . What is the inverse of f?

Solution

It is given that fof(x) = f(f(x)) = .

=  =  

Therefore, fof(x) = x   

ð  fof = I

Hence, the given function f is invertible and the inverse of f is f itself.

Question 5:

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)}

Solution

(i) f: {1, 2, 3, 4} → {10}defined as:

f = {(1, 10), (2, 10), (3, 10), (4, 10)}

From the given definition of f, we can see that f is a many one function as: f(1) = f(2) = f(3) = f(4) = 10

∴f is not one-one.

Hence, function f does not have an inverse.

(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} defined as: g = {(5, 4), (6, 3), (7, 4), (8, 2)}

From the given definition of g, it is seen that g is a many one function as: g(5) = g(7) =4.

∴g is not one-one,

Hence, function g does not have an inverse.

(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} defined as:

h = {(2, 7), (3, 9), (4, 11), (5, 13)}

It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h.

∴Function h is one-one.

Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an

element x in the set {2, 3, 4, 5}such that h(x) = y.

Thus, h is a one-one and onto function. Hence, h has an inverse.

Question 6:

Show that f: [−1, 1] → R, given by is one-one. Find the inverse of the function f: [−1, 1] → Range f.

Solution

f: [−1, 1] → R is given as

Let f(x) = f(y).

 

xy+2x = xy+ 2y

2x = 2y

x=y

∴ f is a one-one function.

Let g: Range f → [−1, 1] be the inverse of f.

Let y be an arbitrary element of range f.

we have: y = f(x) for same x  [-1, 1]

ð   

ð 

ð  X(1-y) = x

ð  X =

Now, let us define g: Range f → [−1, 1] as

 

gof(x) =g(f(x)) = g 

fog(y) = f(g(y)) = f 

∴gof = i_x and fog = i_y

f^-1 = g

Question 7:

Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Solution

f: R → R is given by,

f(x) = 4x + 3

One-one:

Let f(x) = f(y).

ð  4x+3+ = 4y+3

ð  4x = 4x

ð  x = y

∴ f is a one-one function.

Onto:

For y ∈ R, let y = 4x + 3.

X =   R

Therefore, for any y ∈ R, there exists X =   R such that f(x) = f

∴ f is onto.

Thus, f is one-one and onto and therefore, f^-1 exists.

Let us define g: R→ R by g(x) =   

Now, gof(x) = g(f(x)) =g(4x+3) =

∴fog(y) = f(g(y)) = f

gof = fog = I_R

Hence, f is invertible and the inverse of f is given by

Question 8:

Consider f: R+ → [4, ∞) given by f(x) = x² + 4. Show that f is invertible with the inverse f^-1 of given f by

where R+ is the set of all non-negative real numbers.

Solution

f: R+ → [4, ∞) is given as f(x) = x² + 4.

One-one:

Let f(x) = f(y).

ð  x²+4 = y²+4

ð  x² = y²

ð  x = y

∴ f is a one-one function.

Onto:

For y ∈ [4, ∞), let y = x² + 4

ð  x² = y-4

ð  x =

ð  Therefore, for any y ∈ R, there exists x =  such that f(x) = f() = ()² + 4 = y-4+4 = y

∴ f is onto.

Thus, f is one-one and onto and therefore, f^-1 exists.

Let us define g: [4, ∞) → R+ by g(y) =,

Now gof(x) = g(f(x)) = g(x²+4) =  = y-4+4 = y

Fog(y) = f(g(y)) = f() =()²+4 = (y-4)+4 =y

∴ gof = fog = I_R

Hence, f is invertible and the inverse of f is given by f^-1 =

Question 9:

Consider f: R+ → [−5, ∞) given by f(x) = 9x² + 6x − 5. Show that f is invertible with

Solution

f: R+ → [−5, ∞) is given as f(x) = 9x² + 6x − 5.

Let y be an arbitrary element of [−5, ∞).

Let y = 9x² + 6x – 5

ð  Y =

ð  Y =

ð  Y+6 = (3x+1)²

ð  3x+1 =

ð  X =

∴f is onto, thereby range f = [−5, ∞).

Let us define g: [−5, ∞) → R+ as g(y) =

We now have: gof(x) = g(f(x)) = g(9x²+6x-5)

gof(x) = ((3x+1)²-6)

gof(x) =  

          =

        = x

fog(y) = f(g(y)) = f()

                       = [3()+1]² -6

                       = ()²-6

                       = y

∴ gof=I_R  and fog = I

Hence, f is invertible and the inverse of f is given by f^-1 = g(y) =

Question 10:

Let f: X → Y be an invertible function. Show that f has unique inverse.

(Hint: suppose g1 and g2 are two inverses of f. Then for all y ∈ Y,

Fog₁(y) = I_y(y) = fog₂(y). Use one-one ness of f).

Solution

Let f: X → Y be an invertible function.

Also, suppose f has two inverses (say g₁ and g₂).

Then, for all y ∈Y, we have: fog₁(y) = I_y(y) = fog₂(y)

ð  f(g₁(y)) = f(g₂(y))

ð  g₁(y) = g₂(y)

ð  g₁ = g₂

Hence, f has a unique inverse.

Question 11:

Consider f: {1, 2, 3} → {a, b, c} is given by f(1) = a, f(2) = b and f(3) = c. Find f^-1 and show that (f^-1)^-1 = f.

Solution

Function f: {1, 2, 3} → {a, b, c} is given by,

f(1) = a, f(2) = b, and f(3) = c

If we define g: {a, b, c} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have: fog(a) = f(g(a)) = f(1) = a

         fog(b) = f(g(b)) = f(2) = b

         fog(c) = f(g(c)) = f(3) = c

and gof(1) = g(f(1)) = g(a) = 1

      gof(2) = g(f(2)) = g(b) = 2

      gof(3) = g(f(3)) = g(c) = 3

∴ gof = I_x and fog = I_y , where X = {1, 2, 3} and Y= {a, b, c}.

Thus, the inverse of f exists and f^-1 = g.

∴f^-1: {a, b, c} → {1, 2, 3} is given by,

f^-1(a) = 1, f^-1(b) = 2, f^-1(c) = 3

Let us now find the inverse of f^-1 i.e., find the inverse of g.

If we define h: {1, 2, 3} → {a, b, c} as

h(1) = a, h(2) = b, h(3) = c, then we have:  goh(1) = g(h(1)) = g(a) = 1

          goh(2) = g(h(2)) = g(b) = 2

          goh(3) = g(h(3)) = g(c) = 3

and, hog(a) = h(g(a)) = h(1) = a

       hog(b) = h(g(b)) = h(2) = b

       hog(c) = h(g(c)) = h(3) = c

∴ goh = I_x and hog = I_y, where X = {1, 2, 3} and Y = {a, b, c}.

Thus, the inverse of g exists and g^-1 = h ⇒ (f^-1)^-1 = h.

It can be noted that h = f.

Hence, (f^-1)^-1 = f.

Question 12:

Let f: X → Y be an invertible function. Show that the inverse of f^-1 is f, i.e.,

(f^-1)^-1 = f.

Solution

Let f: X → Y be an invertible function.

Then, there exists a function g: Y → X such that gof = I_x and fog = I_y.

Here, f^-1 = g.

Now, gof = I_x and fog = I_y

⇒ f ^-1of = I_x and fof^-1= I_y

Hence, f^-1: Y → X is invertible and f is the inverse of f^-1

i.e., (f^-1)^-1 = f.

Exercise 1.4

Question 1:

Determine whether or not each of the definition of given below gives a binary operation.

In the event that * is not a binary operation, give justification for this.

(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

Solution

(i) On Z+, * is defined by a * b = a − b.

It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 − 2

= −1 ∉ Z+.

(ii) On Z+, * is defined by a * b = ab.

It is seen that for each a, b ∈ Z+, there is a unique element ab in Z+.

This means that * carries each pair (a, b) to a unique element a * b = ab in Z+.

Therefore, * is a binary operation.

(iii) On R, * is defined by a * b = ab².

It is seen that for each a, b ∈ R, there is a unique element ab² in R.

This means that * carries each pair (a, b) to a unique element a * b = ab² in R.

Therefore, * is a binary operation.

(iv) On Z+, * is defined by a * b = |a − b|.

It is seen that for each a, b ∈ Z+, there is a unique element |a − b| in Z+.

This means that * carries each pair (a, b) to a unique element a * b =

|a − b| in Z+.

Therefore, * is a binary operation.

(v) On Z+, * is defined by a * b = a.

* carries each pair (a, b) to a unique element a * b = a in Z+.

Therefore, * is a binary operation.

Question 2:

For each binary operation * defined below, determine whether * is commutative or

associative.

(i) On Z, define a * b = a − b

(ii) On Q, define a * b = ab + 1

(iii) On Q, define a * b =

(iv) On Z+, define a * b =

(v) On Z+, define a * b = a^b

(vi) On R − {−1}, define a*b =

Solution

(i) On Z, * is defined by a * b = a − b.

It can be observed that 1 * 2 = 1 − 2 = 1 and 2 * 1 = 2 − 1 = 1.

∴1 * 2 ≠ 2 * 1; where 1, 2 ∈ Z

Hence, the operation * is not commutative.

Also we have:

(1 * 2) * 3 = (1 − 2) * 3 = −1 * 3 = −1 − 3 = −4

1 * (2 * 3) = 1 * (2 − 3) = 1 * −1 = 1 − (−1) = 2

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Z

Hence, the operation * is not associative.

(ii) On Q, * is defined by a * b = ab + 1.

It is known that:

ab = ba  a, b ∈ Q

⇒ ab + 1 = ba + 1  a, b ∈ Q

⇒ a * b = a * b  a, b ∈ Q

Therefore, the operation * is commutative.

It can be observed that:

(1 * 2) * 3 = (1 × 2 + 1) * 3 = 3 * 3 = 3 × 3 + 1 = 10

1 * (2 * 3) = 1 * (2 × 3 + 1) = 1 * 7 = 1 × 7 + 1 = 8

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Q

Therefore, the operation * is not associative.

(iii) On Q, * is defined by a * b =

It is known that:

ab = ba  a, b ∈ Q

⇒ =  a, b ∈ Q

⇒ a * b = b * a  a, b ∈ Q

Therefore, the operation * is commutative.

For all a, b, c ∈ Q, we have:

(a * b) *c = (* c =  =  

a*(b*c) = a* () =  =

(a*b)*c =a*(b*c)

Therefore, the operation * is associative.

(iv) On Z+, * is defined by a * b = 2^ab.

It is known that:

ab = ba  a, b ∈ Z+

⇒ 2^ab = 2^ba  a, b ∈ Z+

⇒ a * b = b * a  a, b ∈ Z+

Therefore, the operation * is commutative.

It can be observed that:

(1*2)*3 = 2^1x2*3 = 4*3 = 2^4x3 =2¹²

1*(2*3) = 1*2^2x3 = 1*2⁶ = 1*64 = 2⁶⁴

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Z+

Therefore, the operation * is not associative.

(v) On Z+, * is defined by a * b = a^b.

It can be observed that:

1*2 = 1² = 1 and 2*1 = 2¹ = 2

∴ 1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ Z+

Therefore, the operation * is not commutative.

It can also be observed that:

(2*3)*4 = 2³*4 = 8*4 = 8⁴ = 2¹²

2*(3*4) = 2*3⁴ = 2*81 = 2⁸¹

∴(2 * 3) * 4 ≠ 2 * (3 * 4) ; where 2, 3, 4 ∈ Z+

Therefore, the operation * is not associative.

(vi) On R, * − {−1} is defined by a*b =

It can be observed that 1*2 = = and 2*1 =  =  = 1

∴1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ R − {−1}

Therefore, the operation * is not commutative.

It can also be observed that: (1*2)*3 = *3 =  =  

1*(2*3) = 1* = 1* = 1*  =  =

∴ (1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ R − {−1}

Therefore, the operation * is not associative.

Miscellaneous Questions:

Qrse of f is f itself.