Question 1: Let $\ast$ be a binary operation on the set $\text{ I }$ of integers, defined by $a \ast b = 2a +b - 3$. Find the value of $3\ast 4.$ [CBSE 2011]

Given that $\ast$ is a binary operation on the set I of integers.

The operation is defined by $a \ast b = 2a + b - 3.$

We need to find the value of $3 \ast 4.$

Since 3 and 4 belongs to the set of integers we can use the binary operation.

$\Rightarrow 3 \ast 4 = (2 \times 3) + 4 - 3$

$\Rightarrow 3 \ast 4 = 6 + 1$

$\Rightarrow 3 \ast 4 = 7$

Therefore the value of $3 \ast 4$ is 7.

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Question 2: The binary operation $\ast : R \times R \rightarrow R$ is defined as $a \ast b = 2a + b. \text{ Find } ( 2 \ast 3) \ast 4.$ [CBSE 2012]

Given that $\ast$ is an operation that is valid for the following Domain and Range $R \times R \rightarrow R$ and is defined by $a \ast b = 2ab.$

We need to find the value of $(2 \ast 3) \ast 4$

According to the problem the binary operation involving $\ast$  is true for all real values of a and b.

$\Rightarrow (2 \ast 3) \ast 4 = ((2 \times 2) + 3) \ast 4$

$\Rightarrow (2 \ast 3) \ast 4 = (4 + 3) \ast 4$

$\Rightarrow (2 \ast 3) \ast 4 = 7 \ast 4$

$\Rightarrow (2 \ast 3) \ast 4 = (2 \times 7) + 4$

$\Rightarrow (2 \ast 3) \ast 4 = 14 + 4$

$\Rightarrow (2 \ast 3) \ast 4 = 18$

Therefore the value of $(2 \ast 3) \ast 4 \text{ is } 18.$

$\\$

Question 3: Let $\ast$ be a binary operation on $N \text{ given by } a \ast b =LCM (a, b) \text{ for all } a, b \in N. \text{ Find } 5 \ast 7.$   [CBSE 2012]

Given that $\ast$ is an operation that is valid for the natural numbers $'N'$ and is defined by $a \ast b = LCM(a, b).$

We need to find the value of $5 \ast 7.$

According to the Problem, Binary operation is assumed to be true for the values of $a \text{ and } b$ to be natural.

$\Rightarrow 5 \ast 7 = LCM(5,7)$

We know that LCM of two prime numbers is the product of that given two prime numbers.

$\Rightarrow 5 \ast 7 = 5 \times 7$

$\Rightarrow 5 \ast 7 = 35$

Therefore the value of $5 \ast 7 \text{ is } 35.$

Question 4: Let $S$ be the set of all rational numbers except $1$ and $\ast$ be defined on $S$ by  $a \ast b =a+b-ab,$ for all, $a,b \in S.$

Prove that : (i) $latex \ast$ is a binary operation on $S$

(ii) $\ast$ is commutative as well as associative     [CBSE 2014]

(i) Sum, difference and product of rational numbers is a unique rational number.

Therefore for each $(a, b) \in S \times S,$ there exists a unique image $( a + b - ab) \text{ in } S.$

$\Rightarrow *$ is a function

$\Rightarrow *$ is a binary operation on $S$

(ii) $a \ast b = a + b - ab = b + a - ba = b \ast a$

Therefore $\ast$ is commutative

$\Rightarrow a \ast ( b \ast c ) = a \ast ( b + c - bc)$

$\Rightarrow a \ast ( b \ast c ) = a + ( b + c - bc) - a ( b + c - bc)$

$\Rightarrow a \ast ( b \ast c ) = a + b + c - bc - ab - ac + abc \hspace{1.0cm}\ldots (i)$

$\Rightarrow( a \ast b ) * c = ( a + b - ab ) \ast c$

$\Rightarrow( a \ast b ) * c = ( a + b - ab) + c - ( a + b - ab) c$

$\Rightarrow( a \ast b ) * c = a + b - ab + c - ac -bc+abc \hspace{1.0cm}\ldots (ii)$

From (i) and (ii) we get $a \ast ( b \ast c ) = ( a \ast b ) * c$

Therefore $\ast$ is associative.

Question 5: If the binary operation $\ast$ on the set $Z$ is defined by $a \ast b =a \ast b -5,$ then element with respect to $\ast .$   [CBSE 2012]

Given that binary operation $\ast$ is valid for the set $Z$ defined by $a \ast b = a + b - 5 \text{ for all } a,b \in Z.$

Let us assume $a \in Z$ and the identity element that we need to compute is $e \in Z.$

We know that he Identity property is defined as follows:

$\Rightarrow a \ast e = e \ast a = a$

$\Rightarrow a + e - 5 = a$

$\Rightarrow e - 5 = a - a$

$\Rightarrow e = 5$

Therefore the required Identity element w.r.t  $\ast$  is 5.

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Question 6: On the set $Z$ of integers, if the binary operation $\ast$ is defined by $a \ast b = a +b + 2,$ then find the identity element.     [CBSE 2012]

Given that binary operation $\ast$ is valid for the set $Z$ of integers defined by $a \ast b = a + b \text{ for all } a,b \in Z.$

Let us assume $a \in Z$ and the identity element that we need to compute be $e \in Z.$

We know that he Identity property is defined as follows:

$\Rightarrow a \ast e = e \ast a = a$

$\Rightarrow a + e + 2 = a$

$\Rightarrow e + 2 = a - a$

$\Rightarrow e = - 2$

Therefore the required Identity element w.r.t  $\ast \text{ is } - 2.$

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Question 7: Let $A = R \times R \text{ and } \ast$ be a binary operation on $A$ defined by $(a, b) \ast (c, d) = (a + c, b + d).$Show that $\ast$ is commutative and associative. Find the binary element for $\ast \text{ on } A,$ if any.     [CBSE 2017]

$A = R \times R$

$\text{For the } ' \ast ' \text{ to be commutative, } p \ast q = p \ast q \text{ must be true for all } p, q \text{ belong to } A.$

Let’s check. Note: Here $p, q$ represent the ordered pairs $(a, b) \text{ and } (c, d)$ respectively.

$p \ast q = (a, b) \ast (c, d) = (a + c, b + d) \\ \\ q \ast p = (c, d) \ast (a, b) = (c + a, d + b) = (a + c, b + d) \Rightarrow p \ast q = q \ast p \\ \\ \Rightarrow \text{ (binary operation } \ast \text{is commutative) }$

$\text{For the } ' \ast ' \text{ to be associative, } a \ast (b \ast c) = (a \ast b) \ast c \text{ must hold for every } a, b, c \in A. \text{ Here } r = (e, f)$

$p \ast (q \ast r) = (a, b) \ast ((c, d) \ast (e, f)) = (a, b) \ast (c + e, d + f) = (a + c + e, b + d + f) \\ \\ (p \ast q) \ast r = ((a, b) \ast (c, d)) \ast (e, f) = (a + c, b + d) \ast (e, f) = (a + c + e, b + d + f) \\ \\ \Rightarrow p \ast (q \ast r) = (p \ast q) \ast r \text{ (Associative) }$

Binary elements:

$\text{Identity Element: Given a binary operation } \ast : A \times A \rightarrow A, \text{ an element } e \in A, \text{ if it exists, is called an identity of the operation } \ast , \text{ if } p \ast e = p = e \ast p \forall p \in A.$

$\text{Here } p = (a, b) \text{ and } e = (x, y).$

$\text{For the element } e \text{ to exist, } (a, b) \ast (x, y) = (a, b)\Rightarrow (a + x, b + y) = (a, b)\Rightarrow a + x = a, b + y = b$

$\text{Since, ordered pairs are only equal when both the first and second terms are equal } \Rightarrow x = 0, y = 0$

$\text{Hence the identity element } e = (x, y) = (0, 0)$

$\text{Given a binary operation } \ast : A \times A \rightarrow A \text{ with the identity element } e \text{ in } A, \text{ an element } a \text{ in } \\ \\ A\text{ is said to be invertible with respect to the operation, if there exists an element } b \text{ in } \\ \\ A \text{ such that } a \ast b = e = b \ast a \text{ and } b \text{ is called the inverse of } a \text{ and is denoted by } a^{ -1}.$

$\text{Let } i = (r, s) \text{ be the inverse of } p = (a, b) \text{ in } A.$

$\text{Therefore, } (r, s) \ast (a, b) = e = (0, 0)\Rightarrow (r + a, s + b) = (0, 0) \\ \\ \therefore r + a = 0, r + b = 0 \Rightarrow r = - a, s = - b$

$\text{Therefore, } (r, s) = ( - a, - b) \text{ is the inverse pair. }$

$\\$

$\text{Question 8:Consider the binary operation } \ast : R \times R \rightarrow R \text{ and } o:R \times R \rightarrow R \\ \\ \text{defined as } a \ast b =|a-b| \text{ and } a \circ b = a \text{ for all } a,b \in R. \text{ Show that } \ast \text{ is commutative } \\ \\ \text{but not associative, } \circ \text{ is associative but not commutative. } \text{Further, show that } \\ \\ \ast \text{ is distributive over } \circ . \text{ Does } \circ \text{ distribute over } \ast ? \text{ Justify your answer. } \\ {\hspace{12.0cm} \textbf{[CBSE 2012] } }$

For any $a,b \in R,$ we have

$a \ast b =| a -b | \text{ and } b \ast a = |b - a | \\ \\ \because | a- b | = | b-a| \text{ for all } a. b \in R$

$\text{Therefore } a \ast b = b \ast a \text{ for all } a, b \in R$

$\text{So, } \ast \text{ is commutative on } R. \text{ We have, }$

$(( - 2 ) \ast 3) \ast 4 = |-2-3| \ast 4 = 5 \ast 4 = | 5-4| = 1$

$\text{and } (-2) \ast ( 3 \ast 4 ) = (-2) \ast | 3-4| = ( -2) \ast 1 = | -2-1| = 3$

$\text{Therefore } ( ( -2) \ast 3 ) \ast 4 \neq ( -2) \ast ( 3 \ast 4 )$

$\text{So, } \ast \text{ is not associative on } R.$

$\text{We have, } 2 \circ 3 =2 \text{ and } 3 \circ 2 = 3$

$\text{Therefore } 2 \circ 3 \neq 3 \circ 2$

$\text{So, } \circ \text{ is not commutative on } R.$

$\text{For any } a, b, c \in R, \text{ we have }$

$(a \circ b) \circ c = a \circ c = a \text{ and } a \circ (b \circ c) = a \circ b = a$

$(a \circ b) \circ c = a \circ (b \circ c) \text{ for all } a, b , c \in R$

$\text{So, } \circ \text{ is associative on } R.$

$\text{For any } a,b, c \in R, \text{ we have }$

$a \ast (b \circ c) =a \ast b =| a -b|, a \ast b =| a -b|, | a \ast c| = | a - c|$

$\text{and, } (a \ast b) \circ (a \ast c) = |a -b| \circ | a- c | = | a - b |$

$\text{Therefore } a \ast (b \circ c) =(a \ast b) \circ (a \ast c) \text{ for all } a, b, c \in S$

$\text{So, } \ast \text{ distributive over } ' \circ '.$

$\text{Further, for any } a,b, c \in R, \text{ we have }$

$a \circ (b \ast c) =a \circ |b - c|=a, a \circ b =a, a \circ c=a \\ \\ \text{ and } (a \circ b) \ast (a \circ c) =a \ast a=|a-a|=0$

$\text{Therefore } a \circ (b \ast c) \neq (a \circ b) \ast (a \circ c)$

$\text{So, } \circ \text{ is not distributive over } ' \ast '$

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$\displaystyle \text{Question 9: If } \ast \text{ is defined on the set of } R_0 \text{ of all non zero real numbers by } \\ \\ a \ast b = \frac{3ab}{7}, \text{ find the identity element in } R \text{ for the binary operation } \ast . \\ \\ { \hspace{12.0cm} \textbf{ [CBSE 2012] } }$

Let $e$ be the identify element in $R$ for the binary operation $\ast$ on $R.$ Then,

$a \ast e = a = e \ast a \text{ for all } a \in R_0$

$\Rightarrow a \ast e = a \text{ and } e \ast a \text{ for all } a \in R_0$

$\displaystyle \Rightarrow \frac{3ae}{7} = a \text{ and } \frac{3ea}{7} \text{ for all } a \in R_0$

$\displaystyle \Rightarrow e = \frac{7}{3}$

$\displaystyle \text{ Hence, } \frac{7}{3} \text{ is the identity element in } R_0.$

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$\text{Question 10: Let } ' \ast ' \text{ be a binary operation on set } Q-\{1 \} \text{ defined by } \\ \\ a \ast b = a +b - ab \text{ for all } a , b \in Q - \{1\}. \text{ Find the identity element with respect to } \\ \\ \ast \text{ on } Q. \text{ Also, prove that every element of } Q - \{ 1\} \text{ is invertible. } \textbf{ [CBSE 2017] }$

$\text{Let the identity element } e \text{ exists in } Q - \{1\} \text{ with respect to } \ast \text{ on } Q - \{1\}. \text{ Then, }$

$a \ast e = a = e \ast a \text{ for all } a \in Q - \{ 1\}$

$\Rightarrow a \ast e = a \text{ for all } a \in Q - \{ 1\} \hspace{1.0cm} [\because '\ast ' \text{ is commutative on } Q-\{1\} ]$

$\Rightarrow a + e - ae = a \text{ for all } a \in Q - \{ 1\}$

$\Rightarrow e(1-a) = 0 \text{ for all } a \in Q - \{ 1\}$

$\Rightarrow e= 0 \hspace{1.0cm} [\because a \in Q - \{ 1\} \therefore a \neq 1 \Rightarrow a - 1 \neq 0 ]$

$\text{Thus, } 0 \text{ is the identity element for } * \text{ on } Q - \{1\}$

$\text{Let } a \text{ be an arbitrary element of } Q - {1} \text{ and let } b \text{ (if exists) } \text{ be the inverse of } a. \text{ Then, }$

$a \ast b = 0 = b \ast a \hspace{1.0cm} [\because 0 \text{ is the identity element } ]$

$\Rightarrow a \ast b = 0 \hspace{1.0cm} [\because '\ast ' \text{ is commutative } ]$

$\Rightarrow a + b - ab = 0$

$\Rightarrow b( 1- a ) = - a$

$\displaystyle \Rightarrow b = \frac{a}{a-1} \hspace{1.0cm} [\because a \in Q - \{ 1 \} \therefore a - 1 \neq 0 ]$

$\displaystyle \text{Since, } a \in Q - \{ 1 \}. \text{ Therefore, } b = \frac{a}{a-1} \in Q - \{1\}$

$\displaystyle \text{Thus, every element of } Q - \{1\} \text{ is invertible and the inverse of an element } a \text{ is } \frac{a}{a-1}.$

$\\$

Question 11:  On the set $R- \{-1 \}$ a binary operation $\ast$ is defined by $a* b= a + b + ab$ for all $a, b \in R - \{ - 1 \}.$ Prove that \ast commutative as well as associative on $R - \{ -1 \}.$ Find the identity element and prove that every element of $R - \{ -1\}$ is invertible.            [CBSE 2015, 2016]

We observe the following properties of $\ast$ on $R -\{-1\}.$

Commutativity: For any $a, b \in R - \{ -1\},$ we have

$a \ast b = a + b + ab \text{ and } b \ast a = b + a + ba$

$\because a + b + ab = b + a + ba$

[ By commutativity of addition and multiplication on   $R - \{ -1 \}$ ]

$\Rightarrow a \ast b = b \ast a$

Hence, $\ast$ is commutative on $R - \{ -1 \}$

Associativity: For any $a, b, c \in R - \{-1 \} ,$ we have

$(a \ast b ) \ast c = ( a + b + ab ) \ast c \\ = ( a + b + ab) + c + ( a + b + ab) c = a + b + c + ab + bc+ ac + abc \hspace{1.0cm} \ldots (i)$

$a \ast ( b + c ) = a \ast ( b + c + bc) \\ = a + ( b + c + bc) + a ( b + c + bc) = a + b + c + ab + bc+ ac + abc \hspace{1.0cm} \ldots (ii)$

From (i) and (ii), we have

$( a \ast b ) \ast c = a \ast ( b \ast c) \text{ for all } a, b, c \in R - \{ -1\}$

Hence, $\ast$ is associative on $R - \{ -1\}$

Existence of identity: Let $e$ be the identity element. Then,

$a \ast e=a=e \ast a \text{ for all } a \ast R - \{ -1\}$

$\Rightarrow a + e + ae=a \text{ and } e+a+ea=a \text{ for all } a \in R - \{ -1\}$

$\Rightarrow e(1+a)=0 \text{ for all } a \in R - \{ -1\}$

$\Rightarrow e = 0$

$\text{Also, } 0 \in R - \{ -1\}$

So, $0$ is the identity element for $\ast$ defined on $R - \{ -1\}$

Existence of inverse: Let $a \in R - \{- 1 \}$ and let $b$ be the inverse of $a.$ Then,

$\Rightarrow a \ast b=e=b \ast a$

$\Rightarrow a \ast b=e$

$\Rightarrow a+b+ab=0$

$\displaystyle \Rightarrow b = \frac{-a}{a+1}$

$\displaystyle \text{Now, } a \in R - \{- 1 \} \Rightarrow a \neq -1 \Rightarrow a + 1 \neq 0 \Rightarrow b = \frac{-a}{a+1} \in R$

$\displaystyle \text{Also, } \frac{-a}{a+1} = -1 \Rightarrow -a = -a - 1 \Rightarrow -1 = 0 \text{ which is not possible. }$

$\displaystyle \text{Therefore, } \frac{-a}{a+1} \in R - \{- 1 \}$

$\displaystyle \text{Hence, every element of }R - \{- 1 \} \text{ is invertible and the inverse of an element } a \text{ is } \frac{-a}{a+1}$

$\\$

Question 12: Consider the infimum binary operation $\wedge$ on set $S = \{ 1, 2, 3,4 ,5 \}$ defined by $a \wedge b =$ minimum of $a$ and $b.$ Write the composition table of the operation $\wedge.$

We have,

$1 \wedge 1 = ( \text{ minimum of 1 and 1 } ) = 1, \\ \\ 1 \wedge 2 = ( \text{ minimum of 1 and 2 } ) = 1, \\ \\ 4 \wedge 3 = (\text{ minimum of 4 and 3 }) = 3 \text{ etc. }$

So, we have the following composition table for $\wedge \text{ on } S$

$\begin{array}{|c|c|c|c|c|c|} \hline \ \ \ \wedge \ \ \ & \ \ \ 1 \ \ \ & \ \ \ 2 \ \ \ & \ \ \ 3 \ \ \ & \ \ \ 4 \ \ \ & \ \ \ 5 \ \ \ \\ \hline 1 & 1 & 1 & 1 & 1 & 1 \\ \hline 2 & 1 & 2 & 2 & 2 & 2 \\ \hline 3 & 1 & 2 & 3 & 3 & 3 \\ \hline 4 & 1 & 2 & 3 & 4 & 4 \\ \hline 5 & 1 & 2 & 3 & 4 & 5 \\ \hline \end{array}$

$\\$

$\text{Question 13: Define a binary operation} \ast \text{ on the set } \{ 0, 1, 2,3,4,5 \} \text{ as }$

$a \ast b = \Bigg\{ \begin{array}{lll} a+b & , & \text{ if } a+b < 6 \\ a+b-6 & , & \text{ if } a+b \geq 6 \end{array}$

Show that $0$ is the identity for this operation and each element $a \neq 0$ of the set is invertible with $6 -a$ being the inverse of $a.$             [CBSE 2011]

$\text{Let } X = \{ 0, 1, 2,3,4,5 \}$

The operation $\ast \text{ on } X$ is defined as

$a \ast b = \Bigg\{ \begin{array}{lll} a+b & , & \text{ if } a+b < 6 \\ a+b-6 & , & \text{ if } a+b \geq 6 \end{array}$

$\text{ An element } e \in X \text{ is the identity element for the operation } \ast, \text{ if } a \ast e = a = e \ast a \text{ for all } a \in X$

$\text{ For } a \in X, \text{ we have }$

$a \ast 0 = a + 0 = a \hspace{1.0cm} [ a \in X \Rightarrow a + 0 < 6 ]$

$0 \ast a = 0 + a = a \hspace{1.0cm} [ a \in X \Rightarrow 0 + a < 6 ]$

$\text{ Therefore } a \ast 0=a=0 \ast a \text{ for all } a \in X$

Thus, $0$ is the identity element for the given operation $\ast .$

$\text{ An element } a \in X \text{ is invertible if there exists } b \in X \text{ such that } a \ast b = 0 = b \ast a.$

$\text{i.e. } \Bigg\{ \begin{array}{lll} a+b = 0 = b+a & , & \text{ if } a+b < 6 \\ a+b-6=0= b+a-6 & , & \text{ if } a+b \geq 6 \end{array}$

$\Rightarrow a = - b \text{ or } b = 6 - a$

$\text{ But, } X = \{ 0, 1, 2,3,4,5 \} \text{ and } a, b \in X. \text{ Then } a \neq -b$

$\text{ Therefore } b = 6 - a \text{ is the inverse of } a \text{ for all } a \in X$

$\text{ Hence, the inverse of an element } a \in X, a \neq 0 \text{ is } 6-a \text{ i.e., } a^{-1} = 6 - a$