NCERT Solutions for Class 12 Maths Chapter 1 – Relations and Functions

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 $R = \{ (x,\;y):3x – y = 0\}$

SOLUTION

(i) A $= \{1, 2, 3, 4, 5, 6, \cdots, 13, 14\}$ is the given set

R $= \{(x, y) : 3x – y = 0\} \Rightarrow$ R $= \{(1, 3), (2, 6), (3, 9), (4, 12)\}$

Reflexive

Let $x \in A$ be any element.

Since, $(x,\;x) \notin R \;\;\;\therefore \;\; R \;\;\; is \;\; not \;\; reflexive$

Symmetric

$x,\;y \in A,\;\;(x,\;y) \in R\;\;but\;\;(y,\;x) \notin R$

$\therefore$ R is not symmetric.

Transitive

$x,\;y,\;z \in A\;\; \Rightarrow (x,\;y) \in R\;\;and\;\;(y,\;z) \in R\; \Rightarrow (x,\;z) \in R$

For example: $(1, 3) \in R$ and $(3, 9) \in R$ but $(1, 9) \notin R$, therefore, R is not transitive.

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

(ii) Relation R in the set N of natural numbers defined as $R = \{ (x,\;y):y = x + 5\;\;and\;\;x < 4\}$

SOLUTION

(ii) N is the set of natural numbers

R $= \{(x, y) : y = x + 5$ and $x < 4\}$ is the set of natural numbers.

$R = \{ (1,\;6),(2,\;7),\;(3,\;8)\}$

Reflexive

Let $x \in N$ be any element.

$(x,\;x) \notin R \;\;\;\therefore \;\; R$ is not reflexive.

Symmetric

$x,\;y \in N,\;\;(x,\;y) \in R\;\;\;but\;\;(y,\;x) \notin R$

$\therefore$ R is not symmetric.

Transitive

$(1, 6) \in R$ and $(6, 7) \notin R$ and $(1, 7) \notin R$

$\therefore$ R is not transitive.

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

(iii) Relation R in the set A $= \{1, 2, 3, 4, 5, 6\}$ defined as $R = \{ (x,\;y): y$ is divisible by $x\}$

SOLUTION

(iii) A $=$ {1, 2, 3, 4, 5, 6} is the given set

R $= \{(x, y) :$ y is divisible by x in A $\}$

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

Reflexive

Let $x \in A$ be any element.

Now, $(x,\;x) \in R\;\;\;i.e.\;\;(1,\;1) \in R,\;(2,\;2) \in R,\;(3,\;3) \in R,\;(4,\;4) \in R,\;(5,\;5) \in R,\;(6,\;6) \in R$

R is reflexive.

Symmetric

$x,\;y \in A,\;\;(x,\;y) \in R$ but $(y,\;x) \notin R$

i.e., $(1,\;2) \in R\;\;but\;\;(2,\;1) \notin R$

$\therefore$ R is not symmetric.

Transitive

$x,\;y,\;z \in A,\;\;(x,\;y) \in R,\;(y,\;z) \in R\;\; \Rightarrow (x,\;z) \in R$

i.e., $(1,\;2) \in R\;\;and\;\;(2,\;4) \in R\;\; \Rightarrow \;(1,\;4) \in R$

Thus, R is transitive.

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

(iv) Relation R in the set Z of all integers defined as R $= \{(x, y) : x – y$ is an integer$\}$

SOLUTION

(iv) Z is the set of all integers

R $= \{(x, y) : x – y$ is an integer$\}$

Reflexive

Let $x \in Z$, be any element, $(x, x)$ i.e., $(1, 1) = 1 – 1 = 0 \in Z$.

$\therefore$ R is reflexive.

Symmetric

$x,\;y \in Z,\;\;(x,\;y) \in R\;\;\;\; \Rightarrow (y,\;x) \in R$

i.e., $x – y$ is an integer $\Rightarrow$ $(y,\;x) \in R$

i.e., $x – y$ is an integer $\Rightarrow$ $y – x$ is also an integer.

$\therefore$ R is symmetric.

Transitive

$(x,\;y) \in R\;\;and\;\;(y,\;z) \in R$

i.e., $(x – y)$ is integer and $(y – z)$ is integer

$\Rightarrow$ $(x – z) = (x – y + y – z) \in$ integer $\Rightarrow$ $(x,\;z) \in R$

Hence, R is reflexive, symmetric and transitive.

(v) Relation R in the set A of human beings in 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$\}$

SOLUTION

Relation R in the set A of human beings in a town at a particular time.

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

Reflexive

$(x,\;x) \in R$ because x and x work at the same place. Thus, R is reflexive.

Symmetric

Let $(x,\;y) \in R \Rightarrow$ x and y work at the same place

$\Rightarrow$ y and x work at the same place $\Rightarrow$ $(y,\;x) \in R$

Thus, R is symmetric.

Transitive

$(x, y) \in R$ and $(y, z) \in R$

$\Rightarrow$ x and y work at the same place and y and z work at the same place

$\Rightarrow$ x and z work at the same place $\Rightarrow$ $(x, z) \in R$

Thus, R is transitive.

Hence, R is reflexive, symmetric and transitive.

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

Reflexive

$(x, x) \in R$ because x and x live in the same locality.

$\therefore$ R is reflexive.

Symmetric

Let $(x, y) \in R \Rightarrow$ x and y live in the same locality

$\Rightarrow$ y and x also live in the same locality $\Rightarrow$ $(y, x) \in R$

Thus, R is symmetric.

Transitive

Let $(x, y) \in R$ and $(y, z) \in R$

$\Rightarrow$ x and y live in the same locality and y and z live in the same locality

$\Rightarrow$ x and z live in the same locality $\Rightarrow$ $(x, z) \in R$

Thus, R is transitive.

Hence, R is reflexive, symmetric and transitive.

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

Reflexive

x is not exactly 7 cm taller than x, so $(x, x) \notin R$, thus R is not reflexive.

Symmetric

If x is exactly 7 cm taller than y, then y is not exactly 7 cm taller than x.

So, if $(x, y) \in R$ then $(y, x) \notin R \Rightarrow$ R is not symmetric.

Transitive

If x is exactly 7 cm taller than y and if y is exactly 7 cm taller than z, then it does not imply that x is exactly 7 cm taller than z. Thus, R is not transitive.

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

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

Reflexive

x is not wife of x, therefore, $(x, x) \notin R$ and thus R is not reflexive.

Symmetric

If x is wife of y, then y is not wife of x.

If $(x, y) \in R$, then $(y, x) \notin R$.

So, R is not symmetric.

Transitive

If x is the wife of y, then y is not wife of z, and R is transitive as transitivity is not contradicted in this case.

$(x,\;y) \in R$ and $(y,\;z) \notin R,$ then $(x,\;z) \notin R,$ for any z [if x is wife of y, then y is a male and male cannot be wife].

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

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

Reflexive

x is not father of x, so $(x, x) \notin R$, so R is not reflexive.

Symmetric

If x is father of y, then y is not father of x.

If $(x, y) \in R$, then $(y, x) \notin R$, so R is not symmetric.

Transitive

If $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \notin R$.

i.e., x is father of y, y is father of z, then x is not father of z.

So, R is not transitive.

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

Show that the relation R in the set R of real numbers, defined as $A = \{ (a,\;b):a \le {b^2}\},$ is neither reflexive nor symmetric nor transitive.

SOLUTION

We have $R = \{ (a,\;b);\;a \le {b^2}\},$ where $a,\;b \in R$

(i) Reflexivity

We observe that $\cfrac{1}{3} \le {\left( {\cfrac{1}{3}} \right)^2}$ is not true.

$\therefore$ $\left( {\cfrac{1}{3},\;\cfrac{1}{3}} \right) \notin R.$ So, R is not reflexive.

(ii) Symmetry

We observe that $1 \le {(2)^2}$ but $2 \not\le {1^2}$

i.e., $(1,\;2) \in R$ but $(2,\;1) \notin R$

So, R is not symmetric.

(iii) Transitivity

We observe that $10 \le {4^2}$ and $4 \le {3^2}$ but $10 \not\le {(3)^2}$

i.e., $(10,\;4) \in R\;\;and\;\;(4,\;3) \in R\;\;but\;\;(10,\;3) \notin R$

So, R is not 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.

SOLUTION

Given R $= \{(a, b) : b = a + 1\}, a, b \in \{1, 2, 3, 4, 5, 6\}$

$\Rightarrow R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$

(i) Reflexive

Consider $a \in \{ 1,\;2,\;3,\;4,\;5,\;6\} \Rightarrow a = a + 1$ which is false.

$\therefore$ $(a,\;a) \notin R.$ Thus, R is not reflexive.

(ii) Symmetric

Let $a,\;b \in \{ 1,\;2,\;3,\;4,\;5,\;6\}$

Consider, $(a, b) \in R \Rightarrow b = a + 1$

and $(b, a) \in R \Rightarrow a = b + 1$ which is false.

$\therefore$ R is not symmetric.

(iii) Transitive

Let $a,\;b,\;c \in \{ 1,\;2,\;3,\;4,\;5,\;6\}$

Consider $(a,\;b) \in R \Rightarrow b = a + 1,\;(b,\;c) \in R$

$\Rightarrow c = b + 1$

$\Rightarrow c = a + 2 \Rightarrow (a,\;c) \notin R$

$\therefore$ R is not transitive.

Hence, R is neither reflexive nor symmetric nor transitive.

Show that the relation R in R defined as $R = \{ (a,\;b):a \le b\},$ is reflexive and transitive but not symmetric.

SOLUTION

(i) Reflexive

Let $a \in R,\;a \le a$ which is true. $\therefore$ $(a,\;a) \in R$

Thus, R is reflexive.

(ii) Symmetric

Let $a,\;b \in R\;\;\&\;\;(a,\;b) \in R$

Consider $a \le b$ does not imply $b \le a$

$\Rightarrow$ $(a,\;b) \in R\;\;but\;\;(b,\;a) \notin R$

$\therefore$ R is not symmetric.

(iii) Transitive

Let $a,\;b,\;c \in R$

If $(a,\;b) \in R \Rightarrow a \le b\;\;and\;\;(b,\;c) \in R \Rightarrow b \le c \Rightarrow a \le c$

$\Rightarrow$ $(a,\;c) \in R$

Thus, R is transitive.

Hence, R is reflexive and transitive but not symmetric.

Check whether the relation R in R defined by $R = \{ (a,\;b):a \le {b^3}\}$ is reflexive, symmetric or transitive.

SOLUTION

We have $R = \{ (a,\;b);\;\;a \le {b^3}\}$ where $a,\;b \in R.$

(i) Reflexive

We observe that $\cfrac{1}{2} \le {\left( {\cfrac{1}{2}} \right)^3}$ is not true.

$\therefore$ $\left( {\cfrac{1}{2},\;\cfrac{1}{2}} \right) \notin R.$ So R is not reflexive.

(ii) Symmetric

We observe that $1 \le {(3)^3}$ but $3 \not\le {1^3}$, i.e., $(1,\;3) \in R$ but $(3,\;1) \notin R$

So, R is not symmetric.

(iii) Transitive

We observe that $10 \le {3^3}$ and $3 \le {2^3}$ but $10 \not\le {2^3}$

i.e., $(10,\;3) \in R\;\;and\;\;(3,\;2) \in R\;\;but\;\;(10,\;2) \notin R$

So, R is not transitive.

R is neither reflexive nor symmetric nor 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.

SOLUTION

Given the set $\{1, 2, 3\}$ where R $= \{(1, 2), (2, 1)\}$

(i) Reflexive

$1,\;2,\;3 \in \{ 1,\;2,\;3\},\;(1,\;1) \notin R,\;(2,\;2) \notin R,\;(3,\;3) \notin R$

$\therefore$ R is not reflexive.

(ii) Symmetric

$1,\;2 \in \{ 1,\;2,\;3\},\;(1,\;2) \in R \Rightarrow (2,\;1) \in R$

$\therefore$ R is symmetric.

(iii) Transitive

$1,\;2,\;3 \in \{ 1,\;2,\;3\},$

Consider, $(1,\;2) \in R,\;(2,\;3) \notin R,\;(1,\;3) \notin R$

R is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

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

R $= \{(x, y) :$ x and y have same number of pages$\}$

(i) Reflexive

Books x and x have same number of pages.

$\therefore$ $(x,\;x) \in R$

$\therefore$ R is reflexive.

(ii) Symmetric

If $(x,\;y) \in R,$ i.e. Books x and y have same number of pages.

$\Rightarrow$ Books y and x have same number of pages.

$\Rightarrow$ $(y,\;x) \in R$

$\therefore$ R is symmetric.

(iii) Transitive

If $(x,\;y) \in R\;\;and\;\;(y,\;z) \in R$

$\Rightarrow$ Books x and y have same number of pages and books y and z have same number of pages.

$\Rightarrow$ Books x and z have same number of pages.

$\Rightarrow$ $(x,\;z) \in R\;\;\;\therefore \;\;R$ is transitive.

Hence, R is an equivalence relation.

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

We have A $= \{1, 2, 3, 4, 5\}$

$R = \{ (a,\;b):|a – b|\;\;is\;\;even\},\;a,\;b \in A$

(i) Reflexive

For any $a \in A$, we have $|a – a| = 0,$ which is even

$\Rightarrow$ $(a,\;a) \in R \;\forall\; a \in A$

So, R is reflexive.

(ii) Symmetry

Let $a,\;b \in A.$

Let $(a,\;b) \in R,\;\,then\;\;|a – b|\;\,is\;\;even \Rightarrow \;|b – a|\;\;is\;even$

$\Rightarrow$ $(b,\;a) \in R$

Thus, $(a,\;b) \in R \Rightarrow (b,\;a) \in R$

So, R is symmetric.

(iii) Transitive

Let $a,\;b,\;c \in A.$ Let $(a,\;b) \in R\;\;and\;\;(b,\;c) \in R$

$\Rightarrow$ $|a – b|$ is even and $|b – c|$ is even

$\Rightarrow$ (a and b both are even or both are odd) and (b and c both are even or both are odd)

Case I: When b is even

Let $(a,\;b) \in R$ and $(b,\;c) \in R$

$\Rightarrow$ $|a – b|$ is even and $|b – c|$ is even

$\Rightarrow$ a is even and c is even [$\therefore$ b is even]

$\Rightarrow$ $|a – c|$ is even $\Rightarrow$ $(a,\;c) \in R$

Case II: When b is odd

Let $(a,\;b) \in R$ and $(b,\;c) \in R \Rightarrow |a – b|$ is even and $|b – c|$ is even

$\Rightarrow$ a is odd and c is odd [$\because$ b is odd]

$\Rightarrow$ $|a – c|\;\;is\;even \Rightarrow (a,\;c) \in R$

Thus, $(a, b) \in R$ and $(b,\;c) \in R \Rightarrow (a,\;c) \in R$

So, R is transitive.

Hence, R is an equivalence relation.

We know that the difference of any two odd (even) natural numbers is always an even natural number.

$\therefore$ All the elements of set $\{1, 3, 5\}$ are related to each other and all the elements of $\{2, 4\}$ are related to each other.

We know that the difference of an even natural number and an odd natural number is an odd number.

$\therefore$ No element of $\{1, 3, 5\}$ is related to any element of $\{2, 4\}$.

Show that each of the relation R in the set $A = \{ x \in Z:0 \le x \le 12\},$ 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) $A = \{ x \in Z:0 \le x \le 12\}$

$\therefore$ $A = \{ 0,\;1,\;2,\;3,\;…,\;12\}$

We have $R = \{ (a,\;b):|a – b|\;\;is\;multiple\;of\;4\}$

(a) Reflexive

For any $a \in A,\;|a – a| = 0$ is a multiple of 4

Thus, $(a,\;a) \in R \;\;\;\therefore$ R is reflexive.

(b) Symmetry

For any $a, b \in A$, let $(a, b) \in R$

$\Rightarrow$ $|a – b|$ is multiple of 4 $\Rightarrow$ $|b – a|$ is multiple of 4

$\Rightarrow$ $(b,\;a) \in R$

i.e., $(a,\;b) \in R \Rightarrow (b,\;a) \in R$

$\therefore$ R is symmetric.

(c) Transitive

For any $a, b, c \in A$, let $(a,\;b) \in R$ and $(b, c) \in R$

$\Rightarrow$ $|a – b|$ is multiple of 4 and $|b – c|$ is multiple of 4

$\Rightarrow$ $|a – c| = |a – b + b – c|$

$\Rightarrow$ $|a – c| = |4{k_1} + 4{k_2}|$ where $a – b = 4{k_1}$ and $b – c = 4{k_2}$

$\Rightarrow$ $|a – c| = 4|{k_1} + {k_2}|$

$\Rightarrow$ $|a – c|$ is multiple of 4

$\Rightarrow$ $(a, c) \in R$

$\therefore$ R is transitive.

Hence, R is an equivalence relation.

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

$\Rightarrow$ $R = \{ (0,\;0),\;(1,\;1),\;…….,\;(12,\;12)\}$ and $A = \{0,\;1,\;2,\;…………,\;12\}$

(a) Reflexive

$a \in A \Rightarrow a = a \Rightarrow (a,\;a) \in R \Rightarrow R$ is reflexive.

(b) Symmetry

$a, b \in A$

Let $(a,\;b) \in R \Rightarrow a = b \Rightarrow b = a \Rightarrow (b,\;a) \in R$

$\Rightarrow$ R is symmetric.

(c) Transitive

$a,\;b,\;c \in A,\;\;\;Let\;(a,\;b) \in R \Rightarrow a = b$

$(b,\;c) \in R \Rightarrow b = c \Rightarrow a = c \Rightarrow (a,\;c) \in R$

$\Rightarrow$ R is transitive.

Hence, R is an equivalence relation.

Now set of all elements related to 1 in each case:

(i) Required set $= \{(5, 1), (1, 5), (9, 1), (1, 9)\}$

(ii) Required set $= \{(1, 1)\}$

Give 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) Relation R “is perpendicular to”

i.e., $R = \{ (x,\;y):x\;\;is\;\;perpendicular\;\;to\;y\}$

$l_1$ is not perpendicular to $l_1 \Rightarrow R$ is not reflexive.

If $l_1 \bot l_2,$ then $l_2 \bot l_1 \Rightarrow$ R is symmetric.

If $l_1 \bot l_2\;\;and\;\;l_2 \bot l_3,$ then $l_1$ is not perpendicular to $l_3.$

$\Rightarrow$ R is not transitive.

Clearly, R “is perpendicular to” is symmetric but neither reflexive nor transitive.

(ii) Relation R $= \{(x, y) : x > y\}$

We know that $x > x$ is false. So, R is not reflexive.

If $x > y,$ then it does not imply that $y > x$. So, R is not symmetric.

If $x > y, y > z$ imply $x > z$. So, R is transitive.

Thus, R is transitive but neither reflexive nor symmetric.

(iii) Relation “is friend of” R $= \{(x, y) : x$ is a friend of y$\}$

x is a friend of x. $\therefore$ R is reflexive.

If x is a friend of y, then y is a friend of x. $\therefore$ R is symmetric.

If x is a friend of y and y is a friend of z, then x cannot be friend of z.

$\therefore$ R is reflexive and symmetric but not transitive.

(iv) R is relation “is greater or equal to” i.e.,

$R = \{ (x,\;y):x \ge y\}$

$x \ge x$ is true. $\therefore$ R is reflexive.

If $x \ge y$ then it does not imply $y \ge x$. $\therefore$ R is not symmetric.

If $x \ge y,\;y \ge z \Rightarrow x \ge z$. $\therefore$ R is transitive.

Hence, R is reflexive and transitive but not symmetric.

(v) R is relation “is brother of” i.e.,

R $= \{(x, y) : x$ is a brother of y$\}$

x is not a brother of x. So, R is not reflexive.

If x is a brother of y, then y is a brother of x. So, R is symmetric.

If x R y and y R z, i.e., x is brother of y and y is brother of z $\Rightarrow$ x is brother of z $\Rightarrow$ x R z. So R is transitive.

Hence, R is symmetric, 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. Further, show that the set of all points related to a point $P \ne (0,\;0)$ is the circle passing through P with origin as centre.

SOLUTION

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

Let $P(x_1,\;y_1),\;Q(x_2,\;y_2)\;\;and\;\;O(0,\;0).$

$\therefore$ $OP = OQ = \sqrt{x_1^2 + y_1^2} = \sqrt{x_2^2 + y_2^2} \Rightarrow x_1^2 + y_1^2 = x_2^2 + y_2^2$

(i) Reflexive

$P \in A \Rightarrow (P,\;P) \in R$ ($\because$ $OP = OP$)

Distance of the point P from origin is same as the distance of the point P from origin.

$\therefore$ R is reflexive.

(ii) Symmetric

$P,\;Q \in A;\;\;(P,\;Q) \in R$

$\Rightarrow$ Distance of the point P from origin is same as the distance of point Q from origin, i.e., OP $=$ OQ

$\Rightarrow$ OQ $=$ OP $\Rightarrow$ $(Q,\;P) \in R$. $\therefore$ R is symmetric.

(iii) Transitive

$P,\;Q,\;S \in R,\;\;(P,\;Q) \in R\;\;\;and\;\;(Q,\;S) \in R,$

$\Rightarrow$ $OP = OQ$ and $OQ = OS \Rightarrow OP = OS \Rightarrow (P,\;S) \in R$

$\Rightarrow$ R is transitive. Hence, R is an equivalence relation.

We have to find the set of points related to $P \ne (0,\;0)$

As $x_1^2 + y_1^2 = x_2^2 + y_2^2 = r^2 \Rightarrow x^2 + y^2 = r^2$

Which represents a circle with centre $(0, 0)$ and radius $= r$.

Show that the relation R defined in the set A of all triangles as $R = \{ (T_1,\;T_2): T_1$ is similar to $T_2\},$ is an equivalence relation. Consider three right angle triangles $T_1$ with sides 3, 4, 5; $T_2$ with sides 5, 12, 13 and $T_3$ with sides 6, 8, 10. Which triangles among $T_1,\;T_2$ and $T_3$ are related?

SOLUTION

$R = \{ (T_1,\;T_2): T_1$ is similar to $T_2\}$ and $T_1$, $T_2$ are triangles.

(i) Reflexivity

We know that each triangle is similar to itself and thus $(T_1,\;T_1) \in R.$ $\therefore$ R is reflexive.

(ii) Symmetry

Also, two triangles are similar. Then $T_1 \sim T_2 \Rightarrow T_2 \sim T_1.$ $\therefore$ R is symmetric.

(iii) Transitivity

Again, if $T_1 \sim T_2$ and $T_2 \sim T_3 \Rightarrow T_1 \sim T_3$. $\therefore$ R is transitive.

Hence, R is an equivalence relation.

We are given three right angled triangles $T_1,\;\;T_2\;\;and\;\;T_3.$

$T_1$ with sides 3, 4, 5; $T_2$ with sides 5, 12, 13 and $T_3$ with sides 6, 8, 10.

We know that two triangles are similar if corresponding sides are proportional. We observe that $T_1$ and $T_3$ are similar because $\cfrac{3}{6} = \cfrac{4}{8} = \cfrac{5}{10}\left( = \cfrac{1}{2} \right)$

Hence, triangles $T_1$ and $T_3$ are related.

Show that the relation R defined in the set A of all polygons as $R = \{ (P_1,\;P_2): P_1$ and $P_2$ have same number of sides$\}$, is an equivalence relation. What is the set of the elements in A related to the right angle triangle T with sides 3, 4 and 5?

SOLUTION

R $= \{(P_1, P_2) : P_1$ and $P_2$ have same number of sides$\}.$

(i) Reflexive

Let $P_1 \in A.$ Consider the element $(P_1,\;P_1).$ It shows that $P_1$ and $P_1$ have same number of sides.

$\Rightarrow$ $(P_1,\;P_1) \in R.$ Hence, R is reflexive.

(ii) Symmetric

Let $P_1,\;P_2 \in A.$ If $(P_1,\;P_2) \in R,$

$\Rightarrow$ $P_1$ and $P_2$ have same number of sides.

$\Rightarrow$ $P_2$ and $P_1$ have same number of sides

$\Rightarrow$ $(P_2,\;P_1) \in R \Rightarrow R$ is symmetric.

(iii) Transitive

Let $P_1,\;P_2,\;P_3 \in A.$ If $(P_1,\;P_2) \in R$ and $(P_2,\;P_3) \in R,$

$\Rightarrow$ $P_1$ and $P_2$ have same number of sides and $P_2$ and $P_3$ have same number of sides

$\Rightarrow$ $P_1$ and $P_3$ have same number of sides $\Rightarrow$ $(P_1,\;P_3) \in R$

Thus, R is transitive.

Hence, R is an equivalence relation.

We know that if 3, 4, 5 are the sides of a triangle, then the triangle is right-angled. Now, the set of elements in A related to T is the set of right angled triangles.

Let L be the set of all lines in XY-plane and R be the relation in L defined as $R = \{ (L_1,\;L_2): L_1$ is parallel to $L_2\}.$ Show that R is an equivalence relation. Find the set of all lines related to the line y $=$ 2x + 4.

SOLUTION

$R = \{ (L_1,\;L_2): L_1$ is parallel to $L_2\}$

(i) Reflexive

Let $L_1 \in L.$ $L_1 \parallel L_1$, i.e., $(L_1,\;L_1) \in R.$ Thus R is reflexive.

(ii) Symmetric

$L_1,\;L_2 \in L$

Let $(L_1,\;L_2) \in R \Rightarrow L_1 \parallel L_2 \Rightarrow L_2 \parallel L_1 \Rightarrow (L_2,\;L_1) \in R$

Thus, R is symmetric.

(iii) Transitive

$L_1,\;L_2,\;L_3 \in L.$ Let $(L_1,\;L_2) \in R$ and $(L_2,\;L_3) \in R$

$\Rightarrow$ $L_1 \parallel L_2$ and $L_2 \parallel L_3 \Rightarrow L_1 \parallel L_3$

Thus, R is transitive. Hence, R is an equivalence relation.

All lines related to the line y $=$ 2x + 4 are y $=$ 2x + c, where c is a real number.

$L = \{ (y = 2x + 4,\;y = 2x + c): x,\;y \in R\}.$

Let R be the relation in the set $\{1, 2, 3, 4\}$ given by R $= \{(1, 2), (2, 2),(1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\}$. Choose the correct answer.

(A) R is reflexive and symmetric but not transitive.

(B) R is reflexive and transitive but not symmetric.

(C) R is symmetric and transitive but not reflexive.

(D) R is an equivalence relation.

SOLUTION

(B) R is reflexive for all $1, 2, 3, 4 \in \{1, 2, 3, 4\}$

R is not symmetric for all $1,\;2 \in \;\{ 1,\;2,\;3,\;4\}$ and $(3,\;2) \in R$

$\Rightarrow$ $(1,\;2) \in R$ for all $1, 2, 3 \in \{1, 2, 3, 4\}$

Let R be the relation in the set N given by $R = \{ (a,\;b): a = b – 2,\;b > 6\}.$ Choose the correct answer.

(A) $(2, 4) \in R$

(B) $(3, 8) \in R$

(C) $(6, 8) \in R$

(D) $(8, 7) \in R$

SOLUTION

(C) $(a,\;b) \in R$ only if $a = b – 2$ and $b > 6$

$(6,\;8) \in R$ as $6 = 8 – 2$ and $8 > 6.$

Show that the function $f:R \to R,$ defined by $f(x) = \cfrac{1}{x}$ 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

$f:R_* \to R_*,$ defined by $f(x) = \cfrac{1}{x}$

Injectivity

$f(x_1) = \cfrac{1}{x_1}\;\;and\;\;f(x_2) = \cfrac{1}{x_2}$

If $f(x_1) = f(x_2) \Rightarrow \cfrac{1}{x_1} = \cfrac{1}{x_2} \Rightarrow x_1 = x_2$

Thus, f is one-one.

Surjectivity

Since $f: R_* \to R_*$

Given any element $y \in R_*$ (co-domain of $R_*$), then there exist any element $x \in R_*$ (domain of $R_*$) such that

$f(x) = y\;\;and\;\;we\;\;have,\;\;f(x) = \cfrac{1}{x} \Rightarrow \cfrac{1}{x} = y \Rightarrow x = \cfrac{1}{y}$

$\Rightarrow f\left(\cfrac{1}{y}\right) = y$

Thus, f is onto. Hence, f is a one-one and onto function.

The result is not true, since if domain R* is replaced by N and co-domain being same $R_*$, N does not have inverse.

Check the injectivity and surjectivity of the following functions:

(i) $f:N \to N\;\;given\;\;by\;\;f(x) = x^2$

(ii) $f:Z \to Z\;\;given\;\;by\;\;f(x) = x^2$

(iii) $f:R \to R\;\;given\;\;by\;\;f(x) = x^2$

(iv) $f:N \to N\;\;given\;\;by\;\;f(x) = x^3$

(v) $f:Z \to Z\;\;given\;\;by\;\;f(x) = x^3$

SOLUTION

(i) $f:N \to N$ given by $f(x) = x^2$

Injectivity

$f(x_1) = f(x_2) \Rightarrow x_1^2 = x_2^2 \Rightarrow x_1 = x_2$

$\therefore$ f is one-one i.e., f is injective.

Surjectivity

There are many such numbers of co-domain which have no image in domain N. e.g., $3 \in$ co-domain N, but there is no pre-image in domain of f.

Thus, f is not onto i.e., f is not surjective.

Hence, f is injective but not surjective.

(ii) $f:Z \to Z$ given by $f(x) = x^2 \Rightarrow Z = \{0,\; \pm 1,\; \pm 2,\; \pm 3,\;…..\}$

Injectivity

Let $-1,\;1 \in Z,\;\;\;f(-1) = f(1) \Rightarrow 1 = 1$

But, $-1 \ne 1.$ $\therefore$ f is not one-one i.e., f is not injective.

Surjectivity

There are many such elements belonging to codomain which have no preimage in its domain Z.

$2 \in Z$ (co-domain). But $2^{1/2} \notin Z$ (domain).

$\therefore$ Element 2 has no pre-image in its co-domain Z.

$\therefore$ f is not onto i.e., f is not surjective.

Hence, f is neither injective nor surjective.

(iii) $f:R \to R$ given by $f(x) = x^2$

Injectivity

Let $-1,\;1 \in R,\;\;\;f(-1) = f(1) \Rightarrow 1 = 1$

But, $-1 \ne 1 \Rightarrow$ f is not injective.

Surjectivity

$-2$ belongs to co-domain R of f but $\sqrt{-2}$ does not belong to domain R of f. $\therefore$ f is not surjective.

Hence, f is neither injective nor surjective.

(iv) $f:N \to N$ given by $f(x) = x^3$

Injectivity

Let $x_1,\;x_2 \in N,\;\;f(x_1) = f(x_2) \Rightarrow x_1^3 = x_2^3 \Rightarrow x_1 = x_2$

i.e., for every $x \in N,$ f has a unique image in its co-domain.

$\therefore$ f is one-one and thus f is injective.

Surjectivity

There are many such members of co-domain of f which do not have pre-image in its domain e.g., 2, 3 etc.

$\therefore$ f is not onto and thus f is not surjective.

Hence, f is injective but not surjective.

(v) $f:Z \to Z,$ given by $f(x) = x^3.$ Here Z is the set of integers.

Injectivity

$f(x_1) = f(x_2) \Rightarrow x_1^3 = x_2^3 \Rightarrow x_1 = x_2$

$\therefore$ f is one-one and thus it is injective.

Surjectivity

Many members of co-domain of f have no pre-image in its domain. e.g., 2 belonging to its co-domain has no pre-image in its domain of Z. Therefore, f is not surjective.

Hence, f is injective but not surjective.

Prove that the Greatest Integer Function $f:R \to R,$ given by $f(x) = [x],$ is neither one-one nor onto, where $[x]$ denotes the greatest integer less than or equal to x.

SOLUTION

$f:R \to R,\;\;f(x) = [x]$

Injectivity

We have, for $1 \le x < 2,$

$f(x) = 1$

Thus, f is not one-one.

Surjectivity

$f:R \to R$ does not attain non-integral values.

$\therefore$ Non-integer points in R do not have their pre-images in the domain.

$\therefore$ f is not onto.

Hence, f is neither one-one nor onto.

Show that Modulus Function $f:R \to R,$ 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

$f:R \to R,\;\;given\;\;by\;\;f(x) = |x|$

Where $|x| = \left\{ \begin{array}{l} x,\;\;if\;\;x \ge 0\\ -x,\;\;if\;\;x < 0 \end{array} \right.$

Injectivity

Clearly, f contains $(-1, 1), (1, 1), (-2, 2), (2, 2) \cdots$

We have, $f(-1) = f(1) \Rightarrow |-1| = |1| \Rightarrow 1 = 1$

But, $(-1) \ne (1)$

Thus, negative integers are not images of any elements.

$\therefore$ f is not one-one.

Surjectivity

We have, for $f:R \to R,\;f(x) = |x|$ assumes only non-negative values. So, negative real numbers in R (co-domain) do not have their pre-images in R (domain).

$\therefore$ f is not onto. Hence, f is neither one-one nor onto.

Show that the Signum Function $f:R \to R,$ given by

$$f(x) = \left\{ \begin{array}{l} 1,\;\;if\;\;x > 0\\ 0,\;\;if\;\;x = 0\\ -1,\;\;if\;\;x < 0 \end{array} \right.$$

is neither one-one nor onto.

SOLUTION

$$f(x) = \left\{ \begin{array}{l} 1,\;\;\;\;if\;\;\;\;x > 0\\ 0,\;\;\;\;if\;\;\;\;x = 0\\ -1,\;\;if\;\;\;\;x < 0 \end{array} \right.$$

Injectivity

We have, $f(1) = f(2) = 1$ but $1 \ne 2$

i.e., $f(x_1) = f(x_2) = 1\;\;\;for\;\;x > 0\;\;\;But,\;\;x_1 \ne x_2$

Also, $f(-2) = f(-3) = -1.\;\;\;But,\;\;-2 \ne -3$

i.e., $f(x_1) = f(x_2) = -1\;\;for\;\;x < 0.\;\;But,\;\;x_1 \ne x_2$

$\Rightarrow$ f is not one-one.

Surjectivity

Except the numbers $-1,\;0,\;1,$ no other members of co-domain of f have any pre-image in its domain.

$\therefore$ f is not onto.

Hence, f is neither one-one nor onto.

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

A $= \{1, 2, 3\},$ B $= \{4, 5, 6, 7\}$ and f $= \{(1, 4), (2, 5), (3, 6)\}$

We have, f(1) $=$ 4, f(2) $=$ 5 and f(3) $=$ 6. Distinct elements of A have distinct images in B. Hence, f is a one-one function.

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

(i) $f:R \to R,$ defined by $f(x) = 3 – 4x$

(ii) $f:R \to R,$ defined by $f(x) = 1 + x^2$

SOLUTION

(i) $f:R \to R$ defined by $f(x) = 3 – 4x.$

Injectivity

$f(x_1) = f(x_2) \Rightarrow 3 – 4x_1 = 3 – 4x_2$

$\Rightarrow -4x_1 = -4x_2 \Rightarrow x_1 = x_2$

$\therefore$ f is one-one.

Surjectivity

Now, $f:R \to R$ given for every $y \in R$ (co-domain of f), there exists an element $x \in R$ (domain of f) such that $f(x) = y$

$\Rightarrow y = 3 – 4x \Rightarrow x = \cfrac{3 – y}{4}$

$\therefore f\left(\cfrac{3 – y}{4}\right) = 3 – 4\left(\cfrac{3 – y}{4}\right) = 3 – 3 + y = y$

Hence, f is onto.

Thus, f is one-one and onto i.e., a bijective function.

(ii) $f:R \to R$ defined by $f(x) = 1 + x^2$

Injectivity

Let $x_1,\;\;x_2 \in R,\;\;then\;\;f(x_1) = 1 + x_1^2\;\;and\;\;f(x_2) = 1 + x_2^2$

$f(x_1) = f(x_2) \Rightarrow x_1^2 = x_2^2 \Rightarrow x_1 = \pm x_2$

Thus, f is not one-one.

Surjectivity

Now, $f:R \to R,$ given for every $y \in R$ (co-domain of f), there exists an element $x \in R$ (domain of f) such that $f(x) = y$.

Thus, elements less than 1 have no pre-image.

$\therefore$ f is not onto.

Hence, f is neither one-one nor onto and hence not bijective.

Let A and B be sets. Show that $f:A \times B \to B \times A$ such that $f(a,\;b) = (b,\;a)$ is a bijective function.

SOLUTION

Injectivity

Let $(a_1,\;b_1)\;\;and\;\;(a_2,\;b_2) \in A \times B$ such that,

$f(a_1,\;b_1) = f(a_2,\;b_2)$

$\Rightarrow (b_1,\;a_1) = (b_2,\;a_2) \Rightarrow b_1 = b_2\;\;and\;\;a_1 = a_2$

$\Rightarrow (a_1,\;b_1) = (a_2,\;b_2)$

Thus, $f(a_1,\;b_1) = f(a_2,\;b_2) \Rightarrow (a_1,\;b_1) = (a_2,\;b_2)$ [For all $(a_1,\;b_1),\;(a_2,\;b_2) \in A \times B$]

So, f is an injective function.

Surjectivity

Let $(b, a)$ be an arbitrary element of $B \times A$, where $b \in B$ and $a \in A \Rightarrow (a, b) \in A \times B$

Thus, for all $(b, a) \in B \times A$, there exists $(a, b) \in A \times B$ such that $f(a,\;b) = (b,\;a)$

So, $f:A \times B \to B \times A$ is an onto function.

Hence, f is a bijective function.

Let $f:N \to N$ be defined by $f(n) = \left\{ \begin{array}{l} \cfrac{n+1}{2},\;\;if\;\;n\;\;is\;\;odd\\ \cfrac{n}{2},\;\;\;\;\;\;\;if\;\;n\;\;is\;\;even \end{array} \right.$ for all $n \in N.$ State whether the function f is bijective. Justify your answer.

SOLUTION

Injectivity

Here, $f(1) = \cfrac{1+1}{2} = 1,\;\;\;f(2) = \cfrac{2}{2} = 1,\;\;\;f(3) = \cfrac{3+1}{2} = 2,\;\;\;f(4) = \cfrac{4}{2} = 2$

Thus $f(2k-1) = \cfrac{(2k-1)+1}{2} = k\;\;\;and\;\;\;f(2k) = \cfrac{2k}{2} = k$

$\Rightarrow f(2k-1) = f(2k),\;\;where\;\;k \in N$

But, $2k-1 \ne 2k,$ where $k \in N \Rightarrow$ f is not one-one.

Surjectivity

But, f is onto because range of f $=$ N

$\Rightarrow$ f is onto.

Hence, f is not bijective.

Let $A = R – \{3\}\;\;and\;\;B = R – \{1\}.$ Consider the function $f:A \to B$ defined by $f(x) = \left(\cfrac{x-2}{x-3}\right).$ Is f one-one and onto? Justify your answer.

SOLUTION

$A = R – \{3\},\;\;B = R – \{1\}\;\;and\;\;f(x) = \cfrac{x-2}{x-3}$

Let $x_1,\;x_2 \in A \Rightarrow f(x_1) = \cfrac{x_1-2}{x_1-3}\;\;and\;\;f(x_2) = \cfrac{x_2-2}{x_2-3}$

Injectivity

$f(x_1) = f(x_2) \Rightarrow \cfrac{x_1-2}{x_1-3} = \cfrac{x_2-2}{x_2-3}$

$\Rightarrow (x_1-2)(x_2-3) = (x_2-2)(x_1-3)$

$\Rightarrow x_1 x_2 – 3x_1 – 2x_2 + 6 = x_1 x_2 – 2x_1 – 3x_2 + 6$

$\Rightarrow -3x_1 – 2x_2 = -2x_1 – 3x_2 \Rightarrow x_1 = x_2$

$\therefore$ f is a one-one function.

Surjectivity

$f:A \to B,$ let $y \in B$ (co-domain of f) be any element, then there exist $x \in A$ (domain of f) such that $f(x) = y$

$\Rightarrow y = \cfrac{x-2}{x-3} \Rightarrow y(x-3) = x-2 \Rightarrow xy – 3y = x – 2$

$\Rightarrow x(y-1) = 3y – 2 \Rightarrow x = \cfrac{3y-2}{y-1}$

$\therefore f\left(\cfrac{3y-2}{y-1}\right) = \cfrac{\cfrac{3y-2}{y-1} – 2}{\cfrac{3y-2}{y-1} – 3} = \cfrac{3y-2-2y+2}{3y-2-3y+3} = y$

$\therefore$ f is an onto function.

Hence, f is one-one and onto.

Let $f:R \to R$ be defined as $f(x) = x^4.$ Choose the correct answer.

(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

SOLUTION

(D) $f:R \to R$ defined as $f(x) = x^4.$

Injectivity

Let $x_1,\;x_2 \in R,\;\;\;f(x_1) = f(x_2)$

$\Rightarrow x_1^4 = x_2^4 \Rightarrow x_1^2 = x_2^2 \Rightarrow x_1 = \pm x_2$

$\therefore$ f is not one-one.

Surjectivity

$f:R \to R.$ Let $y \in R$ (co-domain of f) be any element, then there exist $x \in R$ (domain of f) such that

$f(x) = y \Rightarrow y = x^4 \Rightarrow x = \pm\; y^{1/4}$

Now, $f(y^{1/4}) = (y^{1/4})^4 = y$ and $f(-y^{1/4}) = (-y^{1/4})^4 = y$

$\therefore$ f is not onto.

Thus, f is neither one-one nor onto.

Let $f:R \to R$ be defined as $f(x) = 3x.$ Choose the correct answer.

(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

SOLUTION

(A)

Injectivity

Let $x_1,\;x_2 \in R$ such that $f(x_1) = f(x_2),$

$\Rightarrow 3x_1 = 3x_2 \Rightarrow x_1 = x_2 \Rightarrow$ f is one-one.

Surjectivity

For any $y \in R$ (co-domain of f), there exist $x \in R$ (domain of f) such that

$f(x) = y \Rightarrow 3x = y \Rightarrow x = \cfrac{y}{3}$

$\Rightarrow f(x) = f\left(\cfrac{y}{3}\right) = 3 \cdot \cfrac{y}{3} = y$

$\Rightarrow$ f is onto.

Thus, f is one-one and onto.

Let $f:\{1,\;3,\;4\} \to \{1,\;2,\;5\}$ and $g:\{1,\;2,\;5\} \to \{1,\;3\}$ be given by $f = \{(1,\;2),\;(3,\;5),\;(4,\;1)\}$ and $g = \{(1,\;3),\;(2,\;3),\;(5,\;1)\}.$ Write down gof.

SOLUTION

$f = \{(1,\;2),\;(3,\;5),\;(4,\;1)\}$ and $g = \{(1,\;3),\;(2,\;3),\;(5,\;1)\}$

Now, $f(1) = 2,\;f(3) = 5,\;f(4) = 1\;\;and\;\;g(1) = 3,\;g(2) = 3,\;\;g(5) = 1$

$(gof)(x) = g[f(x)] \Rightarrow g[f(1)] = g(2) = 3$

$g[f(3)] = g(5) = 1,\;\;g[f(4)] = g(1) = 3$

Hence, $gof = \{(1,\;3),\;(3,\;1),\;(4,\;3)\}.$

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

SOLUTION

First, we show that $(f + g)oh = foh + goh$

L.H.S. $= (f + g)oh = (f + g)[h(x)] = f[h(x)] + g[h(x)] = foh + goh =$ R.H.S.

Now, we show that $(f \cdot g)oh = (foh) \cdot (goh)$

L.H.S. $= (f \cdot g)oh = (f \cdot g)[h(x)] = f[h(x)] \cdot g[h(x)] = (foh) \cdot (goh) =$ R.H.S.

Find gof and fog, if

(i) $f(x) = |x|\;\;and\;\;g(x) = |5x – 2|$

(ii) $f(x) = 8x^3\;\;and\;\;g(x) = x^{1/3}$

SOLUTION

(i) $f(x) = |x|\;\;and\;\;g(x) = |5x – 2|$

$gof = g(f(x)) = g(|x|) = |5|x| – 2|$

$fog = f(g(x)) = f(|5x – 2|) = ||5x – 2|| = |5x – 2|$

(ii) $f(x) = 8x^3\;\;and\;\;g(x) = x^{1/3}$

$gof = g[f(x)] = g(8x^3) = (8x^3)^{1/3} = 2x$

and $fog = f(g(x)) = f(x^{1/3}) = 8(x^{1/3})^3 = 8x.$

If $f(x) = \cfrac{4x + 3}{6x – 4},\;x \ne \cfrac{2}{3},$ show that $fof(x) = x,$ for all $x \ne \cfrac{2}{3}.$ What is the inverse of f?

SOLUTION

$f(x) = \cfrac{4x + 3}{6x – 4},\;x \ne \cfrac{2}{3}$

L.H.S. $= fof(x) = f(f(x))$

$= f\left[\cfrac{4x + 3}{6x – 4}\right] = \cfrac{4\left(\cfrac{4x + 3}{6x – 4}\right) + 3}{6\left(\cfrac{4x + 3}{6x – 4}\right) – 4}$

$= \cfrac{16x + 12 + 18x – 12}{24x + 18 – 24x + 16} = \cfrac{34x}{34} = x =$ R.H.S.

Now, $y = \cfrac{4x + 3}{6x – 4} \Rightarrow 6xy – 4y = 4x + 3$

$\Rightarrow 6xy – 4x = 4y + 3 \Rightarrow x(6y – 4) = 4y + 3$

$\Rightarrow x = \cfrac{4y + 3}{6y – 4} \Rightarrow y = \cfrac{4x + 3}{6x – 4}$

Hence, inverse of f is f itself.

State with reason whether following functions have inverse:

(i) $f:\{1,\;2,\;3,\;4\} \to \{10\}\;\;with\;\;f = \{(1,\;10),\;(2,\;10),\;(3,\;10),\;(4,\;10)\}$

(ii) $g:\{5,\;6,\;7,\;8\} \to \{1,\;2,\;3,\;4\}\;\;with\;\;g = \{(5,\;4),\;(6,\;3),\;(7,\;4),\;(8,\;2)\}$

(iii) $h:\{2,\;3,\;4,\;5\} \to \{7,\;9,\;11,\;13\}\;\;with\;\;h = \{(2,\;7),\;(3,\;9),\;(4,\;11),\;(5,\;13)\}$

SOLUTION

(i) A function is invertible if it is one-one and onto.

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

It is a many-one function. Hence, f has no inverse.

(ii) $g = \{(5,\;4),\;(6,\;3),\;(7,\;4),\;(8,\;2)\}$

g is a many-one function. Hence, g has no inverse.

(iii) $h = \{(2,\;7),\;(3,\;9),\;(4,\;11),\;(5,\;13)\}$

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

Show that $f:[-1,\;1] \to R,$ given by $f(x) = \cfrac{x}{x+2}$ is one-one. Find the inverse of the function $f:[-1,\;1] \to \text{Range}\;f.$

(Hint: For $y \in \text{Range}\;f,\;y = f(x) = \cfrac{x}{x+2},$ for some $x$ in $[-1,\;1],$ i.e., $x = \cfrac{2y}{1-y}$.)

SOLUTION

$f:[-1,\;1] \to R$ is given by $f(x) = \cfrac{x}{x+2},\;x \ne -2$

Let $x_1,\;x_2 \in [-1,\;1] \Rightarrow f(x_1) = \cfrac{x_1}{x_1+2}\;\;and\;\;f(x_2) = \cfrac{x_2}{x_2+2}$

$f(x_1) = f(x_2)$

$\Rightarrow \cfrac{x_1}{x_1+2} = \cfrac{x_2}{x_2+2} \Rightarrow x_1 x_2 + 2x_1 = x_1 x_2 + 2x_2$

$\Rightarrow 2x_1 = 2x_2 \Rightarrow x_1 = x_2.$ Thus, f is one-one.

Now, $f:[-1,\;1] \to R,$ given for every $y \in R$ (co-domain of f), there exist $x \in [-1,\;1]$ (domain of f) such that $f(x) = y$

$\Rightarrow y = f(x) = \cfrac{x}{x+2}\;\;for\;\;some\;\;x\;\;in\;\;[-1,\;1]$

As, $y = \cfrac{x}{x+2} \Rightarrow yx + 2y = x \Rightarrow 2y = x(1-y) \Rightarrow x = \cfrac{2y}{1-y}$

$\Rightarrow f\left[\cfrac{2y}{1-y}\right] = \cfrac{\cfrac{2y}{1-y}}{\cfrac{2y}{1-y}+2} = \cfrac{2y}{2y+2-2y} = y$

$\Rightarrow f(x)$ is onto $\Rightarrow$ f is bijective and hence invertible.

Now, $x = \cfrac{2y}{1-y} \Rightarrow f^{-1}(y) = \cfrac{2y}{1-y} \Rightarrow f^{-1}(x) = \cfrac{2x}{1-x}$

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

SOLUTION

We know that $f:R \to R,\;f(x) = 4x + 3$

Injectivity

Now, let $x_1,\;x_2 \in R.$ $\therefore f(x_1) = 4x_1 + 3\;\;and\;\;f(x_2) = 4x_2 + 3$

For one-one, $f(x_1) = f(x_2) \Rightarrow 4x_1 + 3 = 4x_2 + 3 \Rightarrow x_1 = x_2$

$\therefore f(x)$ is one-one.

Surjectivity

$f:R \to R,$ given for every $y \in R$ (co-domain of f) there exist $x \in R$ (domain of f) such that $f(x) = y$

$\therefore y = 4x + 3 \Rightarrow x = \cfrac{y-3}{4}$

Now, $f\left(\cfrac{y-3}{4}\right) = 4\left(\cfrac{y-3}{4}\right) + 3 = y$ $\therefore f(x) = y$

Thus, f(x) is onto. Thus, f is bijective and hence invertible.

Now, we find the inverse of f. We have $f(x) = y$

$\Rightarrow x = f^{-1}(y) = \cfrac{y-3}{4}\;\;or\;\;f^{-1}(x) = \cfrac{x-3}{4}$

Consider $f:R_+ \to [4, \infty)$ given by $f(x) = x^2 + 4.$ Show that f is invertible with the inverse $f^{-1}$ of f given by $f^{-1}(y) = \sqrt{y-4},$ where R+ is the set of all non-negative real numbers.

SOLUTION

$f:R_+ \to [4, \infty)$ and $f(x) = x^2 + 4$

Injectivity

Consider $x_1, x_2 \in R_+$

Now, $f(x_1) = x_1^2 + 4$ and $f(x_2) = x_2^2 + 4$

If $f(x_1) = f(x_2) \Rightarrow x_1^2 + 4 = x_2^2 + 4 \Rightarrow x_1 = x_2$ (since both $x_1, x_2 > 0$)

$\Rightarrow f(x)$ is one-one.

Surjectivity

$f:R_+ \to [4, \infty)$ is given, let $y \in [4, \infty)$ (co-domain of f), then there exists an element $x \in R_+$ (domain of f) such that $f(x) = y$

Now, $f(x) = y \Rightarrow y = x^2 + 4 \Rightarrow x = \sqrt{y-4}$

For $f(x):$ $f(\sqrt{y-4}) = (\sqrt{y-4})^2 + 4 = y – 4 + 4 = y$

$\Rightarrow f^{-1}(y) = \sqrt{y-4}$ or $f^{-1}(x) = \sqrt{x-4}$

Consider $f:R_+ \to [-5, \infty)$ given by $f(x) = 9x^2 + 6x – 5.$ Show that f is invertible with $f^{-1}(y) = \left(\cfrac{\sqrt{y+6} – 1}{3}\right)$

SOLUTION

$f:R_+ \to [-5, \infty)$ and $f(x) = 9x^2 + 6x – 5$

Injectivity

Let $x_1, x_2 \in R_+$

$f(x_1) = 9x_1^2 + 6x_1 – 5$ and $f(x_2) = 9x_2^2 + 6x_2 – 5$

If $f(x_1) = f(x_2)$

$\Rightarrow 9x_1^2 + 6x_1 – 5 = 9x_2^2 + 6x_2 – 5 \Rightarrow 9x_1^2 + 6x_1 = 9x_2^2 + 6x_2$

$\Rightarrow 9(x_1^2 – x_2^2) + 6(x_1 – x_2) = 0 \Rightarrow (x_1 – x_2)[9(x_1 + x_2) + 6] = 0$

$\Rightarrow x_1 – x_2 = 0 \Rightarrow x_1 = x_2$ $\therefore f(x)$ is one-one.

Surjectivity

$f:R_+ \to [-5, \infty)$ is given, let $y \in [-5, \infty)$ (co-domain of f), then there exists an element $x \in R_+$ (domain of f) such that $f(x) = y$

$\Rightarrow y = 9x^2 + 6x – 5 \Rightarrow 9x^2 + 6x – (5 + y) = 0$

$\Rightarrow x = \cfrac{-(6) \pm \sqrt{(6)^2 + 4 \times 9(5+y)}}{18}$

$\Rightarrow x = \cfrac{-6 \pm 6\sqrt{1+(5+y)}}{18}$

$\Rightarrow x = \cfrac{-6 + 6\sqrt{y+6}}{18} \Rightarrow x = \cfrac{\sqrt{y+6} – 1}{3}$

$\Rightarrow f(x) = f\left(\cfrac{\sqrt{y+6}-1}{3}\right)$

$= 9\left[\cfrac{\sqrt{y+6}-1}{3}\right]^2 + 6\left[\cfrac{\sqrt{y+6}-1}{3}\right] – 5$

$= 9\left[\cfrac{y+6+1-2\sqrt{y+6}}{9}\right] + 2(\sqrt{y+6}-1) – 5$

$= y + 7 – 2\sqrt{y+6} + 2\sqrt{y+6} – 2 – 5$

$\Rightarrow f(x) = y.$ $\Rightarrow f(x)$ is onto.

Thus, f(x) is bijective and hence invertible.

We have $f(x) = y \Rightarrow f^{-1}(y) = x$

$\Rightarrow f^{-1}(y) = \cfrac{\sqrt{y+6}-1}{3}$ or $f^{-1}(x) = \cfrac{\sqrt{x+6}-1}{3}$

Let $f:X \to Y$ be an invertible function. Show that f has a unique inverse.

(Hint: Suppose $g_1$ and $g_2$ are two inverses of f. Then for all $y \in Y,\;fog_1(y) = I_Y(y) = fog_2(y).$ Use one-one-ness of f.)

SOLUTION

Given $f:X \to Y$ be invertible.

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

Let $g_1$ and $g_2$ be the two inverses of f. Now for all $y \in Y,$

$fog_1(y) = I_Y(y) = fog_2(y)$

$\Rightarrow fog_1(y) = fog_2(y) \Rightarrow f[g_1(y)] = f[g_2(y)] \Rightarrow g_1(y) = g_2(y)$

Hence, f has a unique inverse.

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

SOLUTION

$f = \{(1,\;a),\;(2,\;b),\;(3,\;c)\}.$ Clearly f is one-one and onto.

Also, $f^{-1} = \{(a,\;1),\;(b,\;2),\;(c,\;3)\}$

$\Rightarrow f^{-1}(a) = 1,\;f^{-1}(b) = 2,\;f^{-1}(c) = 3$

Also, $(f^{-1})^{-1} = \{(1,\;a),\;(2,\;b),\;(3,\;c)\} = f$

Hence, the result $(f^{-1})^{-1} = f.$

Let $f:X \to Y$ be an invertible function. Show that the inverse of $f^{-1}$ is f, i.e., $(f^{-1})^{-1} = f.$

SOLUTION

We know that $f:X \to Y$

As f is invertible $\Rightarrow$ f is one-one and onto $\Rightarrow$ $f^{-1}$ exists.

Also, $f^{-1}$ is one-one and onto $\Rightarrow$ $f^{-1}$ is invertible.

$\Rightarrow (f^{-1})^{-1}$ exists $\Rightarrow (f^{-1})^{-1} = f$

If $f:R \to R$ be given by $f(x) = (3 – x^3)^{1/3},$ then fof(x) is

(A) $x^{1/3}$

(B) $x^3$

(C) x

(D) $(3 – x^3)$

SOLUTION

(C) $f:R \to R$ and $f(x) = (3 – x^3)^{1/3}$

$\Rightarrow f[f(x)] = [3 – \{f(x)\}^3]^{1/3}$

$\Rightarrow f[f(x)] = [3 – \{(3 – x^3)^{1/3}\}^3]^{1/3}$

$= [3 – (3 – x^3)]^{1/3} = (3 – 3 + x^3)^{1/3} = x$

Let $f:R – \left\{-\cfrac{4}{3}\right\} \to R$ be a function defined as $f(x) = \cfrac{4x}{3x+4}.$ The inverse of f is the map $g:\text{Range}\;f \to R – \left\{-\cfrac{4}{3}\right\}$ given by

(A) $g(y) = \cfrac{3y}{3-4y}$

(B) $g(y) = \cfrac{4y}{4-3y}$

(C) $g(y) = \cfrac{4y}{3-4y}$

(D) $g(y) = \cfrac{3y}{4-3y}$

SOLUTION

(B) Given: $f:R – \left\{-\cfrac{4}{3}\right\} \to R$

We have $f(x) = \cfrac{4x}{3x+4}$

Now, range of f is $R – \left\{-\cfrac{4}{3}\right\}.$ Let $y = f(x) \Rightarrow y = \cfrac{4x}{3x+4}$

$\Rightarrow 3xy + 4y = 4x \Rightarrow x(4 – 3y) = 4y$

$\Rightarrow x = \cfrac{4y}{4-3y}$ $\therefore f^{-1}(y) = g(y) = \cfrac{4y}{4-3y}$

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 $a * b = ab^2$

(iv) On Z+, define $*$ by $a * b = |a – b|$

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

SOLUTION

(i) $Z^+ = \{1, 2, 3,…\}$, we have $a * b = a – b$

Let $a = 1, b = 3 \Rightarrow a * b = 1 – 3 = -2 \notin Z^+$

Hence, the operation $*$ is not a binary operation on Z+.

(ii) $Z^+ = \{1, 2, 3,…\}$, we have $a * b = ab$

Let $a = 2, b = 4 \Rightarrow a * b = 2 * 4 = 8 \in Z^+$

Hence, the operation $*$ is a binary operation on Z+.

(iii) R (set of real numbers), we have $a * b = ab^2$

Let $a = 5.2, b = 3 \Rightarrow a * b = 5.2 \times (3)^2 = 46.8 \in R$

Hence, the operation $*$ is a binary operation on R.

(iv) $Z^+ = \{1, 2, 3,…\}$, we have $a * b = |a – b|$

Let $a = 3, b = 7 \Rightarrow a * b = |3 – 7| = |-4| = 4 \in Z^+$

Hence, the operation $*$ is a binary operation on Z+.

(v) $Z^+ = \{1, 2, 3,…\}$, we have $a * b = a$

Let $a = 5, b = 7 \Rightarrow a * b = 5 \in Z^+$

Hence, the operation $*$ is a binary operation on $Z^+.$

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

(i) On Z, define $a * b = a – b$

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

(iii) On Q, define $a * b = \cfrac{ab}{2}$

(iv) On Z+, define $a * b = 2^{ab}$

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

(vi) On $R – \{-1\}$, define $a * b = \cfrac{a}{b+1}$

SOLUTION

(i) $a * b = a – b$ on Z

For commutativity:

$a * b = a – b$ and $b * a = b – a = -(a – b) \ne a * b$

$\Rightarrow a * b \ne b * a$

For associativity:

$a * (b * c) = a * (b – c) = a – (b – c) = a – b + c$

And $(a * b) * c = (a – b) * c = a – b – c$

$\therefore a * (b * c) \ne (a * b) * c$

Thus, the operation $*$ is neither commutative nor associative.

(ii) $a * b = ab + 1$ on Q

For commutativity:

$a * b = ab + 1$ and $b * a = ba + 1 = ab + 1$

$\therefore a * b = b * a$

For associativity:

$a * (b * c) = a * (bc + 1) = a(bc + 1) + 1 = abc + a + 1$

And $(a * b) * c = (ab + 1) * c = (ab + 1)c + 1 = abc + c + 1$

$\therefore a * (b * c) \ne (a * b) * c$

Thus, the operation $*$ is commutative but not associative.

(iii) $a * b = \cfrac{ab}{2}$ on Q

For commutativity:

$a * b = \cfrac{ab}{2}$ and $b * a = \cfrac{ba}{2} = \cfrac{ab}{2}$

$\therefore a * b = b * a$

For associativity:

$a * (b * c) = a * \left(\cfrac{bc}{2}\right) = \cfrac{a \cdot \cfrac{bc}{2}}{2} = \cfrac{abc}{4}$

And $(a * b) * c = \left(\cfrac{ab}{2}\right) * c = \cfrac{\cfrac{ab}{2} \cdot c}{2} = \cfrac{abc}{4}$

$\therefore a * (b * c) = (a * b) * c$

Thus, the operation $*$ is commutative and also associative.

(iv) $a * b = 2^{ab}$ on Z+

For commutativity:

$a * b = 2^{ab}$ and $b * a = 2^{ba} = 2^{ab}$. $\therefore a * b = b * a$

For associativity:

$a * (b * c) = a * (2^{bc}) = 2^{a \cdot 2^{bc}}$ and $(a * b) * c = (2^{ab}) * c = 2^{2^{ab} \cdot c}$

$\therefore a * (b * c) \ne (a * b) * c$

Hence, the operation $*$ is commutative but not associative.

(v) $a * b = a^b$ on Z+

For commutativity:

$a * b = a^b$ and $b * a = b^a$. $\therefore a * b \ne b * a$

For associativity:

$a * (b * c) = a * (b^c) = a^{b^c}$ and $(a * b) * c = (a^b) * c = (a^b)^c = a^{bc}$

Thus, $a * (b * c) \ne (a * b) * c$

Hence, the operation $*$ is neither commutative nor associative.

(vi) $a * b = \cfrac{a}{b+1}$ on $R – \{-1\}$

For commutativity:

$a * b = \cfrac{a}{b+1}$ and $b * a = \cfrac{b}{a+1}$. $\therefore a * b \ne b * a$

For associativity:

$a * (b * c) = a * \left(\cfrac{b}{c+1}\right) = \cfrac{a}{\cfrac{b}{c+1}+1} = \cfrac{a(c+1)}{b+c+1}$

And $(a * b) * c = \left(\cfrac{a}{b+1}\right) * c = \cfrac{\cfrac{a}{b+1}}{c+1} = \cfrac{a}{(b+1)(c+1)}$

$\therefore a * (b * c) \ne (a * b) * c$

Hence, the operation $*$ is neither commutative nor associative.

Consider the binary operation $\wedge$ on the set $\{1, 2, 3, 4, 5\}$ defined by $a \wedge b = \min\{a, b\}$. Write the operation table of the operation $\wedge$.

SOLUTION

Let $A = \{1, 2, 3, 4, 5\}$

$a \wedge b =$ minimum of $a$ and $b$

Consider a binary operation $*$ on the set $\{1, 2, 3, 4, 5\}$ given by the following multiplication table.

(i) Compute $(2 * 3) * 4$ and $2 * (3 * 4).$

(ii) Is $*$ commutative?

(iii) Compute $(2 * 3) * (4 * 5).$

SOLUTION

(i) $2 * 3 = 1$ and $3 * 4 = 1$

Now, $(2 * 3) * 4 = 1 * 4 = 1$ and $2 * (3 * 4) = 2 * 1 = 1$

(ii) $2 * 3 = 1$ and $3 * 2 = 1$ $\therefore 2 * 3 = 3 * 2$

Hence, the operation is commutative.

(iii) $(2 * 3) * (4 * 5) = 1 * 1 = 1.$

Let $*’$ be a binary operation on the set $\{1, 2, 3, 4, 5\}$ defined by $a *’ b =$ H.C.F. of $a$ and $b$. Is the operation $*’$ the same as the operation $*$ defined in Q. 4 above? Justify your answer.

SOLUTION

Let $A = \{1, 2, 3, 4, 5\}$

$a *’ b =$ HCF of $a$ and $b$

We observe that the operation $*’$ is the same as the operation $*$ in Q. 4.

Let $*$ be the binary operation on N given by $a * b =$ L.C.M. of $a$ and $b$. Find

(i) $5 * 7,\; 20 * 16$

(ii) Is $*$ commutative?

(iii) Is $*$ associative?

(iv) Find the identity of $*$ in N.

(v) Which elements of N are invertible for the operation $*$?

SOLUTION

$a * b =$ L.C.M. of $a$ and $b$.

(i) $5 * 7 =$ L.C.M. of 5 and 7 $= 35$

$20 * 16 =$ L.C.M. of 20 and 16 $= 80$

(ii) $a * b =$ L.C.M. of $a$ and $b$ $=$ L.C.M. of $b$ and $a$ $= b * a.$

Thus, operation $*$ is commutative.

(iii) $a * (b * c) = a *$ (L.C.M. of $b$ and $c$)

$=$ L.C.M. of ($a$ and (L.C.M. of $b$ and $c$)) $=$ L.C.M. of $a$, $b$ and $c$.

Similarly, $(a * b) * c =$ (L.C.M. of $a$ and $b$) $* c$

$=$ L.C.M. of ((L.C.M. of $a$ and $b$) and $c$) $=$ L.C.M. of $a$, $b$ and $c$.

Thus, $a * (b * c) = (a * b) * c$

Hence, the operation $*$ is associative.

(iv) Identity of $*$ in N $= 1$ because $a * 1 =$ L.C.M. of $a$ and $1 = a.$

(v) Only the element 1 in N is invertible for the operation $*$ because $1 * 1 = 1.$

Is $*$ defined on the set $\{1, 2, 3, 4, 5\}$ by $a * b =$ L.C.M. of $a$ and $b$ a binary operation? Justify your answer.

SOLUTION

Let $A = \{1, 2, 3, 4, 5\}$ and $a * b =$ L.C.M. of $a$ and $b$.

$2 * 3 =$ L.C.M. of 2 and 3 $= 6 \notin A$

Hence, the operation $*$ is not a binary operation.

Let $*$ be the binary operation on N defined by $a * b =$ H.C.F. of $a$ and $b$. Is $*$ commutative? Is $*$ associative? Does there exist identity for this binary operation on N?

SOLUTION

Commutativity

$a * b =$ H.C.F. of $a$ and $b$ $=$ H.C.F. of $b$ and $a$ $= b * a$

Thus, operation $*$ is commutative.

Associativity

$(a * b) * c =$ (H.C.F. of $a$ and $b$) $* c$

$=$ H.C.F. of [(H.C.F. of $a$ and $b$) and $c$] $=$ H.C.F. of $a$, $b$ and $c$

$a * (b * c) = a *$ [H.C.F. of $b$ and $c$]

$=$ H.C.F. of [$a$ and (H.C.F. of $b$ and $c$)] $=$ H.C.F. of [$a$, $b$ and $c$]

$\Rightarrow (a * b) * c = a * (b * c).$ Thus, operation $*$ is associative.

Identity

Now, $1 * a = a * 1 \ne a$

There does not exist any identity element.

Let $*$ be a binary operation on the set Q of rational numbers as follows:

(i) $a * b = a – b$

(ii) $a * b = a^2 + b^2$

(iii) $a * b = a + ab$

(iv) $a * b = (a – b)^2$

(v) $a * b = \cfrac{ab}{4}$

(vi) $a * b = ab^2$

Find which of the binary operations are commutative and which are associative.

SOLUTION

(i) For commutativity:

$a * b = a – b = -(b – a) = -(b * a)$

Thus, the operation $*$ is not commutative.

For associativity:

$(a * b) * c = (a – b) * c = (a – b) – c = a – b – c$

And $a * (b * c) = a * (b – c) = a – (b – c) = a – b + c$

$\therefore (a * b) * c \ne a * (b * c)$

Thus, the operation $*$ is not associative.

(ii) For commutativity:

$a * b = a^2 + b^2 = b^2 + a^2 = b * a$

Thus, the operation $*$ is commutative.

For associativity:

$(a * b) * c = (a^2 + b^2) * c = (a^2 + b^2)^2 + c^2$

And $a * (b * c) = a * (b^2 + c^2) = a^2 + (b^2 + c^2)^2$

Thus, $(a * b) * c \ne a * (b * c)$

Hence, the operation $*$ is not associative.

(iii) For commutativity:

$a * b = a + ab = a(1 + b)$ and $b * a = b + ba = b(1 + a)$

$\therefore a * b \ne b * a$

Thus, the operation $*$ is not commutative.

For associativity:

$(a * b) * c = (a + ab) * c = (a + ab) + (a + ab)c$

And $a * (b * c) = a * (b + bc) = a + a(b + bc)$

$\therefore (a * b) * c \ne a * (b * c)$

Hence, the operation $*$ is not associative.

(iv) For commutativity:

$a * b = (a – b)^2 = (b – a)^2 = b * a$

Thus, the operation $*$ is commutative.

For associativity:

$a * (b * c) = a * (b – c)^2 = [a – (b – c)^2]^2$

And $(a * b) * c = (a – b)^2 * c = [(a – b)^2 – c]^2$

$\therefore a * (b * c) \ne (a * b) * c$

Hence, the operation $*$ is not associative.

(v) For commutativity:

$a * b = \cfrac{ab}{4} = \cfrac{ba}{4} = b * a$

Thus, the operation $*$ is commutative.

For associativity:

$a * (b * c) = a * \cfrac{bc}{4} = \cfrac{a \cdot \cfrac{bc}{4}}{4} = \cfrac{abc}{16}$

And $(a * b) * c = \cfrac{ab}{4} * c = \cfrac{\cfrac{ab}{4} \cdot c}{4} = \cfrac{abc}{16}$

$\therefore a * (b * c) = (a * b) * c$

Thus, the operation $*$ is associative.

(vi) For commutativity:

$a * b = ab^2$ and $b * a = ba^2$

$\therefore a * b \ne b * a$

Thus, the operation $*$ is not commutative.

For associativity:

$a * (b * c) = a * (bc^2) = a(bc^2)^2 = ab^2c^4$

And $(a * b) * c = (ab^2) * c = (ab^2)c^2 = ab^2c^2$

$\therefore a * (b * c) \ne (a * b) * c$

Thus, the operation $*$ is not associative.

Find which of the operations given above has identity.

SOLUTION

(i) If $e$ is an identity element, then $a * e = a = e * a \;\forall a \in Q$

$\Rightarrow a – e = e – a = a \;\forall a \in Q$

$\Rightarrow a – e = a$ and $e – a = a$

$\Rightarrow e = 0$ and $e = 2a \;\forall a \in Q$

Which is not possible. Hence, identity element does not exist.

(ii) If $e$ is an identity element, then $a * e = a = e * a \;\forall a \in Q$

$\Rightarrow a^2 + e^2 = e^2 + a^2 = a \;\forall a \in Q$

$\Rightarrow e = \sqrt{a – a^2} \;\forall a \in Q$

Which is not possible. Hence, identity element does not exist.

(iii) If $e$ is an identity element, then $a * e = a = e * a \;\forall a \in Q$

$\Rightarrow a + ae = e + ae = a \;\forall a \in Q$

$\Rightarrow a + ae = a$ and $e + ae = a$

$\Rightarrow e = 0$ and $e = \cfrac{a}{1+a} \;\forall a \in Q$

Which is not possible. Hence, identity element does not exist.

(iv) If $e$ is an identity element, then $a * e = a = e * a \;\forall a \in Q$

$\Rightarrow (a – e)^2 = (e – a)^2 = a \;\forall a \in Q$

$\Rightarrow a – e = \pm\sqrt{a}$ and $e – a = \pm\sqrt{a}$

$\Rightarrow e = a \pm \sqrt{a}$ and $e = a \pm \sqrt{a} \;\forall a \in Q$

Which is not possible. Hence, there is no identity element.

(v) If $e$ is an identity element, then $a * e = a = e * a \;\forall a \in Q$

$\Rightarrow \cfrac{ae}{4} = \cfrac{ea}{4} = a \;\forall a \in Q$

So, $e = 4$ is the identity element.

(vi) If $e$ is an identity element, then $a * e = a = e * a \;\forall a \in Q$

$\Rightarrow ae^2 = ea^2 = a$

$\Rightarrow ae^2 = a$ and $ea^2 = a$

$\Rightarrow e = \pm 1$ and $e = \cfrac{1}{a}$

Which is not possible. Hence, identity element does not exist.

Let $A = N \times N$ and $*$ be the binary operation on A defined by $(a, b) * (c, d) = (a + c, b + d).$ Show that $*$ is commutative and associative. Find the identity element for $*$ on A, if any.

SOLUTION

$A = N \times N$ and $*$ is a binary operation defined on A.

For commutativity:

$(a, b) * (c, d) = (a + c, b + d) = (c + a, d + b) = (c, d) * (a, b)$

The operation $*$ is commutative.

For associativity:

$[(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f) = (a + c + e, b + d + f)$

Also, $(a, b) * [(c, d) * (e, f)] = (a, b) * (c + e, d + f) = (a + c + e, b + d + f)$

$\therefore [(a, b) * (c, d)] * (e, f) = (a, b) * [(c, d) * (e, f)]$

Hence, the operation $*$ is associative.

For identity:

Let identity element be $(e, f)$

$\therefore (a, b) * (e, f) = (a, b)$

$(a + e, b + f) = (a, b) \Rightarrow a + e = a,\; b + f = b$

$\Rightarrow e = 0,\; f = 0.$ But $0 \notin N$

Hence, identity element does not exist.

State whether the following statements are true or false. Justify.

(i) For an arbitrary binary operation $*$ on a set N, $a * a = a \;\forall a \in N$

(ii) If $*$ is a commutative binary operation on N, then $a * (b * c) = (c * b) * a$

SOLUTION

(i) False.

A binary operation on N is defined as $a * a = a \;\forall a \in N$. For example, $a * b = a + b \;\forall a, b \in N$, then

$a * a = a + a = 2a \ne a.$

Hence $*$ is not defined as such in general.

(ii) True.

$a * (b * c) = (b * c) * a = (c * b) * a$ ($b * c = c * b$ since $*$ is commutative)

Consider a binary operation $*$ on N defined as $a * b = a^3 + b^3.$ Choose the correct answer.

(A) Is $*$ both associative and commutative?

(B) Is $*$ commutative but not associative?

(C) Is $*$ associative but not commutative?

(D) Is $*$ neither commutative nor associative?

SOLUTION

(B)

For commutativity:

$a * b = a^3 + b^3 = b^3 + a^3 = b * a.$

$\therefore *$ is a commutative operation.

For associativity:

$a * (b * c) = a * (b^3 + c^3) = a^3 + (b^3 + c^3)^3$

And $(a * b) * c = (a^3 + b^3) * c = (a^3 + b^3)^3 + c^3$

$\Rightarrow a * (b * c) \ne (a * b) * c.$ $\therefore *$ is not associative.