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]

Answer:

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]

Answer:

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.

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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]

Answer:

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]

Answer:

(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]

Answer:

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]

Answer:

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]

Answer:

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. }

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\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] } }

Answer:

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] } }

Answer: 

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] }

Answer: 

\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}.

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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]

Answer:

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}

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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.

Answer:

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}

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\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]

Answer:

\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