Class 12: Binary Operations – Exercise 3.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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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(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]