Question 1: Determine whether each of the following operations define binary operation on the given set or not:  

(i) '\ast'  \text{ on } N \text{ defined by } a  \ast b = a^b \text{ for all } a, b \in N

(ii) ' \bigcirc'  \text{ on } Z \text{ defined by } a  \bigcirc b = a^b \text{ for all } a, b \in Z

(iii) '\ast'  \text{ on } N \text{ defined by } a  \ast b = a+b-2 \text{ for all } a, b \in N

\text{(iv)  } '\times_6 \ '  \text{ on } S = \{ 1, 2, 3, 4, 5 \} \text{ defined by } a  \times_6 b = \text{ Remainder when } ab \\ \\ \text{ is divided by 6. }

\text{(v)  } '+_6 \ '  \text{ on } S = \{ 0, 1, 2, 3, 4, 5 \} \text{ defined by } a  +_6 b = \Bigg\{ \begin{array}{lcl} a+b & ,  & \text{ if } \ a+b < 6 \\ a+b-6 & , & \text{ if } \ a+b \geq 6      \end{array}

(vi) ' \odot '  \text{ on } N \text{ defined by } a  \odot b = a^b + b^a \text{ for all } a, b \in N

\displaystyle \text{(vii)  }'\ast'  \text{ on } Q \text{ defined by } a  \ast b = \frac{a-1}{b+1} \text{ for all } a, b \in Q

Answer:

(i) '\ast'  \text{ on } N \text{ defined by } a  \ast b = a^b \text{ for all } a, b \in N

Given that ' \ast '  is an operation that is valid in the Natural Numbers 'N'  and it is defined as given:

\Rightarrow  a \ast b = ab, \text{ where } a,b \in N

\text{Since } a \in N \text{ and } b \in N,

According to the problem it is given that on applying the operation ' \ast '  for two given natural numbers it gives a natural number as a result of the operation,

\Rightarrow  a \ast b \in N \ldots \ \ldots (1)

\text{We also know that } pq>0 \text{ if } p>0 \text{ and } q>0.

So, we can state that,

\Rightarrow  ab>0

\Rightarrow  ab \in N \ldots \ \ldots (2)

From (1) and (2) we can see that both L.H.S and R.H.S gave only Natural numbers as a result.

Thus we can clearly state that ' \ast '  is a Binary Operation on ' N ' .

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(ii) ' \bigcirc'  \text{ on } Z \text{ defined by } a  \bigcirc b = a^b \text{ for all } a, b \in Z

Given that ' \bigcirc' is an operation that is valid in the Integers 'Z' and it is defined as given:

\Rightarrow  a \bigcirc b = ab, \text{ where } a,b \in Z

\text{Since } a \in Z \text{ and } b \in Z,

According to the problem it is given that on applying the operation ' \ast ' for two given integers it gives Integers as a result of the operation,

\Rightarrow  a \bigcirc b \in Z \ldots \ \ldots (1)

Let us values of a = 2 \text{ and } b = - 2 on substituting in the R.H.S side we get,

\Rightarrow  ab = 2^{- 2} 

\displaystyle \Rightarrow  ab = \frac{1}{4}

\Rightarrow  ab \notin Z \ldots \ \ldots (2)

From (2), we can see that ab doesn’t give only Integers as a result. So, this cannot be stated as a binary function.

Therefore the operation ' \bigcirc'  does not define a binary function on Z.

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(iii) '\ast'  \text{ on } N \text{ defined by } a  \ast b = a+b-2 \text{ for all } a, b \in N

Given that ' \ast ' is an operation that is valid in the Natural Numbers 'N'  and it is defined as given:

\Rightarrow  a \ast b = a + b - 2, \text{ where } a,b \in N

\text{Since } a \in N \text{ and } b \in N,

According to the problem it is given that on applying the operation ' \ast '  for two given natural numbers it gives a natural number as a result of the operation,

\Rightarrow  a \ast b \in N \ldots \ \ldots (1)

Let us take the values of a = 1 \text{ and } b = 1, substituting in the R.H.S side we get,

\Rightarrow  a + b - 2 = 1 + 1 - 2

\Rightarrow  a + b - 2 = 0 \notin N \ldots \ \ldots (2)

From (2), we can see that a + b - 2 doesn’t give only Natural numbers as a result.

So, this cannot be stated as a binary function.

Therefore the operation ' \ast ' does not define a binary operation on 'N' .

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\text{(iv)  } '\times_6 \ '  \text{ on } S = \{ 1, 2, 3, 4, 5 \} \text{ defined by } a  \times_6 b = \text{ Remainder when } ab \\ \\ \text{ is divided by 6. }

Given that '\times_6 \ ' is an operation that is valid for the numbers in the Set S = \{1,2,3,4,5\} and it is defined as given:

\Rightarrow  a\times_6 \ b = \text{ Remainder when } ab \text{ is divided by } 6, \text{ where } a,b \in S

\text{ Since } a \in S \text{ and } b \in S,

According to the problem it is given that on applying the operation ' \ast ' for two given numbers in the set 'S' it gives one of the numbers in the set 'S'  as a result of the operation,

\Rightarrow  a\times_6 \ b \in S \ldots \ \ldots (1)

Let us take the values of a = 3, b = 4,

\Rightarrow  ab = 3 \times 4

\Rightarrow  ab = 12

We know that 12 is a multiple of 6. So, on dividing 12 with 6 we get 0 as remainder which is not in the given set 'S'.

The operation '\times_6 \ ' does not define a binary operation on set S.

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\text{(v)  } '+_6 \ '  \text{ on } S = \{ 0, 1, 2, 3, 4, 5 \} \text{ defined by } a  +_6 b = \Bigg\{ \begin{array}{lcl} a+b & ,  & \text{ if } \ a+b < 6 \\ a+b-6 & , & \text{ if } \ a+b \geq 6      \end{array}

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(vi) ' \odot '  \text{ on } N \text{ defined by } a  \odot b = a^b + b^a \text{ for all } a, b \in N

Given that ' \odot ' is an operation that is valid in the Natural Numbers 'N' and it is defined as given:

\Rightarrow  a \odot b = ab + ba, \text{ where } a,b \in N

\text{Since } a \in N \text{ and } b \in N,

According to the problem it is given that on applying the operation ' \odot ' for two given natural numbers it gives a natural number as a result of the operation,

\Rightarrow  a \odot b \in N \ldots \ \ldots (1)

\text{We know that } pq>0 if p>0 \text{ and } q>0.

\Rightarrow  ab>0 \text{ and }  ba>0.

We also know that the sum of two natural numbers is a natural number.

\Rightarrow  ab + ba \in  N

Therefore the operation ' \odot ' defines the binary operation on N.

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\displaystyle \text{(vii)  }'\ast'  \text{ on } Q \text{ defined by } a  \ast b = \frac{a-1}{b+1} \text{ for all } a, b \in Q

Given that ' \ast ' is an operation that is valid in the Rational Numbers ' Q ' and it is defined as given:

\displaystyle \Rightarrow a \ast b = \frac{a-1}{b+1}, \text{ where } a, b \in Q

Since a \in  Q \text{ and } b \in  Q,

According to the problem it is given that on applying the operation ' \ast ' for two given rational numbers it gives a rational number as a result of the operation,

\Rightarrow  a \ast b \in  Q \ldots \ \ldots (1)

\text{Let the value of } b = - 1 \text{ and } a = 2

\displaystyle \Rightarrow \frac{a-1}{b+1} = \frac{2-1}{-1+1}

\displaystyle \Rightarrow \frac{a-1}{b+1} = \frac{1}{0} = \text{ undefined }

\displaystyle \Rightarrow  \frac{a-1}{b+1} \notin Q

Therefore the operation ' \ast ' does not define a binary operation on Q.

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Question 2: Determine whether or not each of the definition of \ast given below gives a binary operation. In the event that \ast is not a binary operation give justification of this.

\text{(i) On } Z^+ , \text{defined } \ast \text{by } a \ast b = a - b

\text{(ii) On } Z^+, \text{defined } \ast \text{by } a \ast b =ab

\text{(iii) On } R, \text{define by } a \ast b =ab^2

\text{(iv) On } Z^+ \text{define } \ast \text{by } a \ast b = |a-b |

\text{(v) On } Z^+ , \text{define } \ast \text{by } a\ast b =a

\text{(vi) On } R, \text{define } \ast \text{by } a \ast b = a + 4b^2

Here, Z^+ denotes the set of all non-negative integers.

Answer:

\text{(i) On } Z^+ , \text{defined } \ast \text{by } a \ast b = a - b

Given that ' \ast ' is an operation that is valid in the Positive integers 'Z^+ \ ' and it is defined as given:

\Rightarrow  a \ast b = a - b, \text{ where } a,b \in Z^+ 

\text{Since } a \in Z^+ \text{ and } b \in Z^+,

According to the problem it is given that on applying the operation ' \ast ' for two given positive integers it gives a positive integer as a result of the operation,

\Rightarrow  a \ast b \in Z + \ \ldots \ \ldots (1) 

\text{Let us take the values } a = 1 \text{ and } b = 2

\Rightarrow  a - b = 1 - 2 

\Rightarrow  a - b = - 1 \notin Z^+ 

Therefore the operation ' \ast ' does not define a binary operation on Z^+

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\text{(ii) On } Z^+, \text{defined } \ast \text{by } a \ast b =ab

Given that ' \ast ' is an operation that is valid in the Positive integers 'Z^+ \ ' and it is defined as given:

\Rightarrow  a \ast b = ab, \text{ where } a,b \in Z^+ ,

\text{Since } a \in Z^+ \text{ and } b \in Z^+ ,

According to the problem it is given that on applying the operation ' \ast ' for two given positive integers it gives a Positive integer as a result of the operation,

\Rightarrow  a \ast b \in Z^{+} \ \ldots \ \ldots (1)

\text{ We know that } p q>0 \text{ if } p>0 \text{ and } q>0.

\Rightarrow  ab>0 \in N

\Rightarrow  ab \in Z^+

Therefore the operation ' \ast ' defines a binary operation on Z^+.

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\text{(iii) On } R, \text{define by } a \ast b =ab^2

Given that ' \ast ' is an operation that is valid in the Real Numbers ‘R’ and it is defined as given:

\Rightarrow  a \ast b = ab^2, \text{ where } a,b \in  R

\text{Since } a \in R \text{ and } b \in R,

According to the problem it is given that on applying the operation ' \ast ' for two given real numbers it gives a real number as a result of the operation,

\Rightarrow  a \ast b \in R \ \ldots \ \ldots (1)

\text{We know that } ab \in R \text{ if } a \in R \text{ and } b \in R

Therefore the operation ' \ast ' defines a binary operation on R.

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\text{(iv) On } Z^+ \text{define } \ast \text{by } a \ast b = |a-b |

Given that ' \ast ' is an operation that is valid in the Positive integers 'Z^+ \ ' and it is defined as given:

\Rightarrow  a \ast b = |a - b|, \text{ where } a,b \in Z^+ ,

\text{ Since } a \in Z + \text{ and } b \in Z + ,

According to the problem it is given that on applying the operation ' \ast ' for two given positive integers it gives a Positive integer as a result of the operation,

\Rightarrow  a \ast b \in Z ...... (1)

\text{ Let us take } a = 2 \text{ and } b = 2,

\Rightarrow  |a - b| = |2 - 2|

\Rightarrow  |a - b| = |0|

\Rightarrow  |a - b| = 0 \notin Z^+

Therefore the operation ' \ast ' does not define a binary function on Z^+ .

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\text{(v) On } Z^+ , \text{define } \ast \text{by } a\ast b =a

Given that ' \ast ' is an operation that is valid in the Positive integers 'Z^+ \ ' and it is defined as given:

\Rightarrow  a \ast b = a, \text{ where } a,b \in Z^+ ,

\text{ Since } a \in Z^+ \text{ and } b \in Z^+ ,

According to the problem it is given that on applying the operation ' \ast ' for two given positive integers it gives a Positive integer as a result of the operation,

\Rightarrow  a \ast b \in Z^+ \ \ldots \ \ldots (1)

It is told from the problem a \in Z^+ .

The operation ' \ast ' defines a binary operation on Z^+.

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\text{(vi) On } R, \text{define } \ast \text{by } a \ast b = a + 4b^2

Given that ' \ast ' is an operation that is valid in the Real Numbers 'R' and it is defined as given:

\Rightarrow  a \ast b = a + 4b^2, \text{ where } a,b \in R, 

\text{ Since } a \in R \text{ and } b \in R,

According to the problem it is given that on applying the operation ' \ast ' for two given real numbers it gives a Real number as a result of the operation,

\Rightarrow  a \ast b \in R \ \ldots \ \ldots (1)

\text{ Since } b \in R \text{ then } b^2 \in R,

We also know that the sum of two real numbers gives a real number. So,

\Rightarrow  a + 4b^2 \in R \ \ldots \ \ldots (2) 

From (1) and (2),

Therefore the operation ' \ast ' defines a binary operation on R.

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Question 3: 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 4: Is \ast defined on the set \{ 1,2,3,4,5 \}  \text{ by }  a \ast b = LCM of a and b a binary operation? Justify your answer.

Answer:

Given that \ast is an operation that is valid on the set S = \{1,2,3,4,5\} defined by a \ast b = \text{ LCM of } a \text{ and } b.

According to the problem it is given that on applying the operation \ast   for two given numbers in the set 'S' it gives a number in the set 'S' as a result of the operation.

Let us take the values of a = 2 \text{ and } b = 3. 

We know that L.C.M of two prime numbers is given by the product of that two prime numbers.

\Rightarrow \text{ L.C.M of } a \text{ and } b \text{ is } 2 \times 3 = 6.

\Rightarrow 6 \notin S 

Therefore the operation \ast does not define a binary operation on 'S' .

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Question 5: Let S = \{  a, b, c \}. Find the total number of binary operations on S.

Answer:

Given set S = \{a,b,c \}, we need to find the total number of binary operations possible for the set 'S'.

We know that the total number of binary operations on a set 'S' with 'n' elements is given by n^{n^2}.

Here n = 3,

\Rightarrow n^{n^2} = 3^{3^2}

\Rightarrow n^{n^2} = 3^{9}

Therefore the total number of binary operations possible on set 'S' is 3^9.

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Question 6: Find the total number of binary operations only \{ a, b\}.

Answer:

Given set S = \{a,b,\}, we need to find the total number of binary operations possible for the set 'S'.

We know that the total number of binary operations on a set 'S' with 'n' elements is given by n^{n^2}.

Here n = 2,

\Rightarrow n^{n^2} = 2^{2^2}

\Rightarrow n^{n^2} = 2^{4}

Therefore the total number of binary operations possible on set 'S' is 2^4=16.

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Question 7:Prove that the operation \ast on the set

M = \Bigg\{  \begin{bmatrix} a & 0 \\0 & b  \end{bmatrix} : a, b \in R - \{ 0\} \Bigg\}

defined by A \ast B = AB is a binary operation.

Answer:

Given that * is an operation that is valid on the set.

M = \Bigg\{  \begin{bmatrix} a & 0 \\0 & b  \end{bmatrix} : a, b \in R - \{ 0\} \Bigg\}

defined by A \ast B = AB is a binary operation.

According to the problem it is given that on applying the operation \ast for two given numbers in the set 'M'   it gives a number in the set 'M' as a result of the operation.

\Rightarrow A \ast B \in M \ \ldots \ \ldots (1)

\displaystyle \text{Let us take } A = \begin{bmatrix} a & 0 \\0 & b  \end{bmatrix} \text{ and } B = \begin{bmatrix} c & 0 \\0 & d  \end{bmatrix} \text{ here } a \in R, b \in R, c \in R \text{ and } d \in R \text{ then, }

\Rightarrow AB = \begin{bmatrix} a & 0 \\0 & b  \end{bmatrix} \times \begin{bmatrix} c & 0 \\0 & d  \end{bmatrix}

\Rightarrow AB = \begin{bmatrix} ac+0 & 0+0 \\0+0 & 0+bd  \end{bmatrix}

\Rightarrow AB = \begin{bmatrix} ac & 0 \\0 & bd  \end{bmatrix}

\text{Since } a \in R \text{ and } c \in R \text{ then } ac \in R \text{ And also } b \in R \text{ and } d \in R \text{ then } bd \in R.

\Rightarrow AB \in R

Therefore the operation \ast defines a binary operation on 'M'.

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Question 8: Let S be the set of all rational numbers of the form \frac{m}{n}, where m \in Z and n = 1, 2, 3. Prove that \ast on S defined by a \ast b = ab is not a binary operation.

Answer:

Given that \ast is an operation that is valid on the set S which consists of all rational numbers of the form \displaystyle \frac{m}{n}, \text{ here } m \in Z \text{ and } n = 1,2,3 \text{ and is defined by } a \ast b = ab. 

According to the problem it is given that on applying the operation \ast for two given numbers in the set 'S' it gives a number in the set 'S' as a result of the operation.

\Rightarrow a \ast  b \in S \ \ldots \ \ldots (1) 

\text{Since } a \in S \text{ and } b \in S, 

\displaystyle \text{Let us take the values of } a = \frac{5}{3} \text{ and } b = \frac{8}{3} \text{ then, }

\displaystyle \Rightarrow ab = \frac{5}{3} \times \frac{8}{3}

\displaystyle \Rightarrow ab = \frac{40}{9} \notin S \text{ as } 9 \notin n

Therefore the operation ' \ast ' does not define a binary operation on 'S' .

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Question 9: 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 10: 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.