# NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.4

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## Chapter 1 Ex.1.4 Question 1

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 $${{\bf{Z}}^ + }$$, define $${^*}$$ by $$a{^*}b = a - b$$

(ii) On $${{\bf{Z}}^ + }$$,define $${^*}$$ by $$a{^*}b = ab$$

(iii) On$${\bf{R}}$$, define $${^*}$$by $$a{^*}b = a{b^2}$$

(iv) On $${{\bf{Z}}^ + }$$, define $${^*}$$ by $$a{^*}b = \left| {a - b} \right|$$

(v) On $${{\bf{Z}}^ + }$$, define $${^*}$$ by$$a{^*}b = a$$

### Solution

(i) On $${{\bf{Z}}^ + }$$, define $${^*}$$ by $$a{^*}b = a - b$$

It is not a binary operation as the image of $$\left( {1,2} \right)$$under $${^*}$$ is

\begin{align} & \text{1}{^*}\text{2}=\text{1}-\text{2} \\ & \Rightarrow -\text{1}\notin ~{{\mathbf{Z}}^{+}}. \\ \end{align}

Therefore, $${^*}$$ is not a binary operation.

(ii) On $${{\bf{Z}}^ + }$$,define $${^*}$$ by $$a{^*}b = ab$$

It is seen that for each$$a,~b~\in ~{{\mathbf{Z}}^{+}}$$, there is a unique element $$ab$$ in $${{\bf{Z}}^ + }$$.

This means that $${^*}$$ carries each pair $$\left( {a,b} \right)$$to a unique element $$a{^*}b = ab$$in $${{\bf{Z}}^ + }$$.

Therefore, $${^*}$$ is a binary operation.

(iii) On$${\bf{R}}$$, define $${^*}$$$$a{^*}b = a{b^2}$$

It is seen that for each $$a,~b~\in ~\mathbf{R}$$, there is a unique element $$a{b^{\rm{2}}}$$ in $${\bf{R}}$$. This means that $${^*}$$carries each pair $$\left( {a,b} \right)$$to a unique element $$a{^*}b = a{b^2}$$in R.

Therefore, $${^*}$$is a binary operation.

(iv) On $${{\bf{Z}}^ + }$$, define $${^*}$$ by $$a{^*}b = \left| {a - b} \right|$$
It is seen that for each $$a,~b~\in ~{{\mathbf{Z}}^{+}}$$,there is a unique element $$\left| {a - b} \right|$$ in $${{\bf{Z}}^ + }$$. This means that $${^*}$$ carries each pair $$\left( {a,b} \right)$$ to a unique element $$a{^*}b = \left| {a - b} \right|$$in $${{\bf{Z}}^ + }$$. Therefore,$${^*}$$is a binary operation.

(v) On $${{\bf{Z}}^ + }$$, define $${^*}$$ by$$a{^*}b = a$$

$${^*}$$carries each pair (a, b) to a unique element in $$a{^*}b = a$$ in $${{\bf{Z}}^ + }$$.

Therefore, $${^*}$$ is a binary operation.

## Chapter 1 Ex.1.4 Question 2

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

(i) On $${{\bf{Z}}^ + }$$, define $$a{^*}b = a - b$$

(ii) On $${\bf{Q}}$$, define $$a{^*}b = ab + 1$$

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

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

(v) On $${{\bf{Z}}^ + }$$, define $$a{^*}b = {a^b}$$

(vi) On $${\bf{R}} - \left\{ { - 1} \right\}$$, define $$a{^*}b = \frac{a}{{b + 1}}$$

### Solution

(i) On $${{\bf{Z}}^ + }$$, define $$a{^*}b = a - b$$

It can be observed that $${\rm{1}}{^*}{\rm{2}} = {\rm{1}} - {\rm{2}} = - {\rm{1}}$$and $${\rm{2}}{^*}{\rm{1}} = {\rm{2}} - {\rm{1}} = {\rm{1}}$$.

$$\therefore {\rm{1}}{^*}{\rm{2}} \ne {\rm{2}}{^*}{\rm{1}}$$; where $$\text{1},\text{ 2}\in ~\mathbf{Z}$$

Hence, the operation $${^*}$$ is not commutative.

Also,

\begin{align}\left( {{\rm{1}}{^*}{\rm{2}}} \right){^*}{\rm{3}} = \left( {{\rm{1}} - {\rm{2}}} \right){^*}{\rm{3}} = - {\rm{1}}{^*}{\rm{3}} = - {\rm{1}} - {\rm{3}} = - {\rm{4}}\\{\rm{1}}{^*}\left( {{\rm{2}}{^*}{\rm{3}}} \right) = {\rm{1}}{^*}\left( {{\rm{2}} - {\rm{3}}} \right) = {\rm{1}}{^*} - {\rm{1}} = {\rm{1}} - \left( { - {\rm{1}}} \right) = {\rm{2}}\end{align}

$$\therefore \left( {{\rm{1}}{^*}{\rm{2}}} \right){^*}{\rm{3}} \ne {\rm{1}}{^*}\left( {{\rm{2}}{^*}{\rm{3}}} \right)$$ where $$\text{1},\text{2},\text{3}\in ~\mathbf{Z}$$

Hence, the operation $${^*}{\rm{ }}$$is not associative.

(ii) On $${\bf{Q}}$$, define $$a{^*}b = ab + 1$$

\begin{align}&ab = ba{\rm{ for all }}a,b \in Q\\& \Rightarrow ab + 1 = ba + 1 \qquad {\text{ for all }}a,b \in Q\\& \Rightarrow a{^*}b = b{^*}a \qquad {\text{for all }}a,b \in Q\end{align}

Hence, the operation $${^*}$$ is commutative.

\begin{align}&\left( {{\rm{1}}{^*}{\rm{2}}} \right){^*}{\rm{3}} = \left( {{\rm{1}} \times {\rm{2 + 1}}} \right){^*}{\rm{3}} = 3{^*}{\rm{3}} = 3 \times 3 + 1 = 10\\&{\rm{1}}{^*}\left( {{\rm{2}}{^*}{\rm{3}}} \right) = {\rm{1}}{^*}\left( {{\rm{2}} \times {\rm{3 + 1}}} \right) = {\rm{1}}{^*}7 = {\rm{1}} \times {\rm{7 + 1}} = 8\end{align}

$$\therefore \left( {{\rm{1}}{^*}{\rm{2}}} \right){^*}{\rm{3}} \ne {\rm{1}}{^*}\left( {{\rm{2}}{^*}{\rm{3}}} \right)$$ where $$\text{1},\text{2},\text{3}\in ~\mathbf{Q}$$

Hence, the operation $${^*}{\rm{ }}$$is not associative.

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

\begin{align}ab = ba{\rm{ for all }}a,b \in Q\\ \Rightarrow \frac{{ab}}{2} = \frac{{ab}}{2}\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ for all }}a,b \in Q\\ \Rightarrow a{^*}b = b{^*}a{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{for all }}a,b \in Q\end{align}

Hence, the operation $${^*}$$ is commutative.

$$\left( {a{^*}b} \right){^*}c = \left( {\frac{{ab}}{2}} \right){^*}c = \frac{{\left( {\frac{{ab}}{2}} \right)c}}{2} = \frac{{abc}}{4}$$

And

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

$$\therefore \left( {a{^*}b} \right){^*}c = a{^*}\left( {b{^*}c} \right)$$ where $$a,b,c\in ~\mathbf{Q}$$

Hence, the operation $${^*}{\rm{ }}$$is associative.

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

\begin{align}ab = ba{\rm{ for all }}a,b \in Z\\ \Rightarrow {2^{ab}} = {2^{ba}}\,\,\,{\rm{ for all }}a,b \in Z\\ \Rightarrow a{^*}b = b{^*}a{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{for all }}a,b \in Z\end{align}

Hence, the operation $${^*}$$ is commutative.

\begin{align}\left( {{\rm{1}}{^*}{\rm{2}}} \right){^*}{\rm{3}} = {2^{1 \times 2}}{^*}{\rm{3}} = 4{^*}{\rm{3}} = {2^{4 \times 3}} = {2^{12}}\\{\rm{1}}{^*}\left( {{\rm{2}}{^*}{\rm{3}}} \right) = {\rm{1}}{^*}{2^{2 \times 3}} = {\rm{1}}{^*}{2^6} = {\rm{1{^*}64}} = {2^{64}}\end{align}

$$\therefore \left( {{\rm{1}}{^*}{\rm{2}}} \right){^*}{\rm{3}} \ne {\rm{1}}{^*}\left( {{\rm{2}}{^*}{\rm{3}}} \right)$$ where $$\text{1},\text{2},\text{3}\in ~{{\mathbf{Z}}^{+}}$$

Hence, the operation $${^*}{\rm{ }}$$is not associative.

(v) On $${{\bf{Z}}^ + }$$, define $$a{^*}b = {a^b}$$

\begin{align}1{^*}2 = {1^2} = 1\\2{^*}1 = {2^1} = 2\end{align}

$$\therefore 1{^*}2 \ne 2{^*}1$$ where $$\text{1},\text{2},\in ~{{\mathbf{Z}}^{+}}$$

Hence, the operation $${^*}$$ is not commutative.

\begin{align}\left( {{\rm{2{^*}3}}} \right){^*}4 = {2^3}{^*}4 = 8{^*}4 = {8^4} = {2^{12}}\\2{^*}\left( {3{^*}4} \right) = 2{^*}{3^4} = 2{^*}81 = {2^{81}}\end{align}

$$\therefore \left( {{\rm{2{^*}3}}} \right){^*}4 \ne 2{^*}\left( {3{^*}4} \right)$$ where $$\text{2},\text{3,4}\in ~{{\mathbf{Z}}^{+}}$$

Hence, the operation $${^*}{\rm{ }}$$is not associative.

(vi) On $${\bf{R}} - \left\{ { - 1} \right\}$$, define $$a{^*}b = \frac{a}{{b + 1}}$$

\begin{align}1{^*}2 = \frac{1}{{2 + 1}} = \frac{1}{3}\\2{^*}1 = \frac{2}{{1 + 1}} = \frac{2}{2} = 1\end{align}

$$\therefore 1{^*}2 \ne 2{^*}1$$ where $$\text{1},\text{2},\in ~\mathbf{R}-\left\{ -1 \right\}$$

Hence, the operation $${^*}$$ is not commutative.

\begin{align}&\left( {{\rm{1}}{^*}{\rm{2}}} \right){^*}{\rm{3}} = \frac{1}{3}{^*}3 = \frac{{\frac{1}{3}}}{{3 + 1}} = \frac{1}{{12}}\\&{\rm{1}}{^*}\left( {{\rm{2}}{^*}{\rm{3}}} \right) = {\rm{1}}{^*}\frac{2}{{3 + 1}} = {\rm{1}}{^*}\frac{2}{4} = {\rm{1{^*}}}\frac{1}{2} = \frac{1}{{\frac{1}{2} + 1}} = \frac{1}{{\frac{3}{2}}} = \frac{2}{3}\end{align}

$$\therefore \left( {{\rm{1}}{^*}{\rm{2}}} \right){^*}{\rm{3}} \ne {\rm{1}}{^*}\left( {{\rm{2}}{^*}{\rm{3}}} \right)$$ where $$\text{1},\text{2},\text{3}\in ~\mathbf{R}-\left\{ -1 \right\}$$

Hence, the operation $${^*}$$ is not associative.

## Chapter 1 Ex.1.4 Question 3

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

### Solution

The binary operation $$\wedge$$on the set $$\left\{ {1,2,3,4,5} \right\}$$is defined by $$a \wedge b = \min \left\{ {a,b} \right\}$$ for all $$a,b \in \left\{ {1,2,3,4,5} \right\}$$.

The operation table for the given operation $$\wedge$$ can be given as:

 $$\wedge$$ $$1$$ $${\rm{2}}$$ $$3$$ $$4$$ $$5$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $${\rm{2}}$$ $$1$$ $${\rm{2}}$$ $${\rm{2}}$$ $${\rm{2}}$$ $${\rm{2}}$$ $$3$$ $$1$$ $${\rm{2}}$$ $$3$$ $$3$$ $$3$$ $$4$$ $$1$$ $${\rm{2}}$$ $$3$$ $$4$$ $$4$$ $$5$$ $$1$$ $${\rm{2}}$$ $$3$$ $$4$$ $$5$$

## Chapter 1 Ex.1.4 Question 4

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

(i) Compute $$\left( {{\rm{2}}*{\rm{3}}} \right)*{\rm{4 }}$$and $${\rm{2}}*\left( {{\rm{3}}*{\rm{4}}} \right)$$

(ii) Is $$^*$$commutative?

(iii) Compute $$\left( {{\rm{2}}*{\rm{3}}} \right)*\left( {4*5} \right)$$.

(Hint: Use the following table)

 $$^*$$ $$1$$ $${\rm{2}}$$ $$3$$ $$4$$ $$5$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $${\rm{2}}$$ $$1$$ $${\rm{2}}$$ $$1$$ $${\rm{2}}$$ $$1$$ $$3$$ $$1$$ $$1$$ $$3$$ $$1$$ $$1$$ $$4$$ $$1$$ $${\rm{2}}$$ $$1$$ $$4$$ $$1$$ $$5$$ $$1$$ $$1$$ $$1$$ $$1$$ $$5$$

### Solution

(i) \begin{align}\left( {{\rm{2}}*{\rm{3}}} \right)*{\rm{4 = 1*4 = 1}}\\{\rm{2*}}\left( {3*4} \right) = 2*1 = 1\end{align}

(ii) For every $$a,b \in \left\{ {1,2,3,4,5} \right\}$$, we have $$a*b = b*a$$. Therefore, $$^*$$ is commutative.

(iii) $$\left( {{\rm{2}}*{\rm{3}}} \right)*\left( {4*5} \right)$$

\begin{align}&\left( {2*3} \right) = 1\,\,and\,\,\left( {4*5} \right) = 1\\&\therefore \left( {2*3} \right)*\left( {4*5} \right) = 1*1 = 1\end{align}

## Chapter 1 Ex.1.4 Question 5

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

### Solution

The binary operation on the set $$\left\{ {1,2,3,4,5} \right\}$$is defined by $$a^{*\prime} b$$= H.C.F. of $$a$$and$$b$$.

The operation table for the operation $$^{*\prime}$$ can be given as:

 $$^{*\prime}$$ $$1$$ $${\rm{2}}$$ $$3$$ $$4$$ $$5$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $${\rm{2}}$$ $$1$$ $${\rm{2}}$$ $$1$$ $${\rm{2}}$$ $$1$$ $$3$$ $$1$$ $$1$$ $$3$$ $$1$$ $$1$$ $$4$$ $$1$$ $${\rm{2}}$$ $$1$$ $$4$$ $$1$$ $$5$$ $$1$$ $$1$$ $$1$$ $$1$$ $$5$$

The operation table for the operations $$^{*\prime}$$ and $${^*}$$ are same.

operation $$^{*\prime}$$ is same as operation $${^*}$$.

## Chapter 1 Ex.1.4 Question 6

Let $${^*}$$ be the binary operation on N defined 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

The binary operation on N is defined by $$a{^*}b$$= L.C.M. of $$a$$and$$b$$.

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

$$20{^*}16$$=LCM of $$20$$and $$16$$=$$80$$

(ii) L.C.M. of $$a$$and$$b$$=LCM of $$b$$and $$a$$for all $$a,b \in N$$

$$\therefore a{^*}b = b{^*}a$$

Operation $${^*}$$is commutative.

(iii) For $$a,b,c \in N$$

$$\left( {a{^*}b} \right){^*}c$$= ( L.C.M. of $$a$$and$$b$$)$${^*}$$$$c$$= L.C.M. of $$a,b,c$$

$$a{^*}\left( {b{^*}c} \right)$$=$$a$$$${^*}$$( L.C.M. of $$b$$and $$c$$)= L.C.M. of $$a,b,c$$

$$\therefore \left( {a{^*}b} \right){^*}c = a{^*}\left( {b{^*}c} \right)$$

Operation $${^*}$$is associative.

(iv) L.C.M. of $$a$$and $$1$$=$$a$$= L.C.M. of $$1$$ and $$a$$for all $$a \in N$$

$$a{^*}1 = a = 1{^*}a$$ for all $$a \in N$$

Therefore,$$1$$is the identity of $${^*}$$in N.

(v) An element a in N is invertible with respect to the operation $${^*}$$ if there exists an element b in N, such that $$a{^*}b = e = b{^*}a$$

$$e = 1$$

L.C.M. of $$a$$and$$b$$=$$1$$=LCM of $$b$$and $$a$$ possible only when $$a$$ and $$b$$ are equal to $$1$$.

$$1$$ is the only invertible element of N with respect to the operation $${^*}$$.

## Chapter 1 Ex.1.4 Question 7

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

### Solution

The operation $${^*}$$ on the set $$\left\{ {1,2,3,4,5} \right\}$$is defined by $$a{^*}b$$= LCM of $$a$$and$$b$$.

The operation table for the operation $${^*}\prime$$ can be given as:

 $${^*}$$ $$1$$ $${\rm{2}}$$ $$3$$ $$4$$ $$5$$ $$1$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $${\rm{2}}$$ $$2$$ $${\rm{2}}$$ $$6$$ $$4$$ $$10$$ $$3$$ $$3$$ $$6$$ $$3$$ $$12$$ $$15$$ $$4$$ $$4$$ $$4$$ $$12$$ $$4$$ $$20$$ $$5$$ $$5$$ $$10$$ $$15$$ $$20$$ $$5$$

\begin{align}&3{^*}2 = 2{^*}3 = 6 \notin A,\\&5{^*}2 = 2{^*}5 = 10 \notin A,\\&3{^*}4 = 4{^*}3 = 12 \notin A,\\&3{^*}5 = 5{^*}3 = 15 \notin A,\\&4{^*}5 = 5{^*}4 = 20 \notin A\end{align}

The given operation $${^*}$$is not a binary operation.

## Chapter 1 Ex.1.4 Question 8

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

The binary operation $${^*}$$ on N defined by $$a{^*}b$$= H.C.F. of$$a$$ and$$b$$.

$$\therefore a{^*}b = b{^*}a$$

Operation $${^*}$$ is commutative.

For all $$a,b,c \in N$$,

$$\left( {a{^*}b} \right){^*}c$$= ( HCF of $$a$$and$$b$$)$${^*}$$$$c$$= HCF of $$a,b,c$$

$$a{^*}\left( {b{^*}c} \right)$$=$$a$$$${^*}$$( HCF. of $$b$$and $$c$$)= HCF of $$a,b,c$$

$$\therefore \left( {a{^*}b} \right){^*}c = a{^*}\left( {b{^*}c} \right)$$

Operation $${^*}$$is associative.

$$e \in N$$will be the identity for the operation$${^*}$$if $$a{^*}e = a = e{^*}a$$for all $$a \in N$$. But this relation is not true for any $$a \in N$$.

Operation $${^*}$$ does not have any identity in N.

## Chapter 1 Ex.1.4 Question 9

Let $${^*}$$be the binary operation on $$Q$$ of rational numbers as follows:

(i) $$a{^*}b = a - b$$

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

(iii) $$a{^*}b = a + ab$$

(iv) $$a{^*}b = {\left( {a - b} \right)^2}$$

(v) $$a + b = \frac{{ab}}{4}$$

(vi) $$a{^*}b = a{b^2}$$

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

### Solution

(i) On Q, the operation $${^*}$$ is defined as $$a{^*}b = a - b$$

$$\frac{1}{2}{^*}\frac{1}{3} = \frac{1}{2} - \frac{1}{3} = \frac{{3 - 2}}{3} = \frac{1}{6}$$

And

$$\frac{1}{3}{^*}\frac{1}{2} = \frac{1}{3} - \frac{1}{2} = \frac{{2 - 3}}{6} = \frac{{ - 1}}{6}$$

$$\therefore \left( {\frac{1}{2}{^*}\frac{1}{3}} \right) \ne \left( {\frac{1}{3}{^*}\frac{1}{2}} \right)$$ where$$\frac{1}{2},\frac{1}{3} \in Q$$

Operation $${^*}$$ is not commutative.

\begin{align}\left( {\frac{1}{2}{^*}\frac{1}{3}} \right){^*}\frac{1}{4} = \left( {\frac{1}{2} - \frac{1}{3}} \right){^*}\frac{1}{4} = \frac{1}{6}{^*}\frac{1}{4} = \frac{1}{6} - \frac{1}{4} = \frac{{2 - 3}}{{12}} = \frac{{ - 1}}{{12}}\\\frac{1}{2}{^*}\left( {\frac{1}{3}{^*}\frac{1}{4}} \right) = \frac{1}{2}{^*}\left( {\frac{1}{3} - \frac{1}{4}} \right) = \frac{1}{2}{^*}\frac{1}{{12}} = \frac{1}{2} - \frac{1}{{12}} = \frac{{6 - 1}}{{12}} = \frac{5}{{12}}\end{align}

$$\therefore \left( {\frac{1}{2}{^*}\frac{1}{3}} \right){^*}\frac{1}{4} \ne \frac{1}{2}{^*}\left( {\frac{1}{3}{^*}\frac{1}{4}} \right)$$ where$$\frac{1}{2},\frac{1}{3},\frac{1}{4} \in Q$$

Operation $${^*}$$ is not associative.

(ii) On $$Q$$, the operation $${^*}$$ is defined as $$a{^*}b = {a^2} + {b^2}$$

For $$a,b \in Q$$

\begin{align}a{^*}b = {a^2} + {b^2} = {b^2} + {a^2} = b{^*}a\\\therefore a{^*}b = b{^*}a\end{align}

Operation $${^*}$$ is commutative.

\begin{align}\left( {1{^*}2} \right){^*}3 = \left( {{1^2} + {2^2}} \right){^*}3 = \left( {1 + 4} \right){^*}3 = 5{^*}3 = {5^2} + {3^2} = 25 + 9 = 34\\1{^*}\left( {2{^*}3} \right) = 1{^*}\left( {{2^2} + {3^2}} \right) = 1{^*}\left( {4 + 9} \right) = 1{^*}13 = {1^2} + {13^2} = 1 + 169 = 170\end{align}

$$\therefore \left( {1{^*}2} \right){^*}3 \ne 1{^*}\left( {2{^*}3} \right)$$ where$$1,2,3 \in Q$$

Operation $${^*}$$ is not associative.

(iii) On $$Q$$, the operation $${^*}$$ is defined as $$a{^*}b = a + ab$$

\begin{align}1{^*}2 = 1 + 1 \times 2 = 1 + 2 = 3\\2{^*}1 = 2 + 2 \times 1 = 2 + 2 = 4\end{align}

$$\therefore 1{^*}2 \ne 2{^*}1$$ where$$1,2 \in Q$$

Operation $${^*}$$ is not commutative.

\begin{align}\left( {1{^*}2} \right){^*}3 = \left( {1 + 1 \times 2} \right){^*}3 = 3{^*}3 = 3 + 3 \times 3 = 3 + 9 = 12\\1{^*}\left( {2{^*}3} \right) = 1{^*}\left( {2 + 2 \times 3} \right) = 1{^*}8 = 1 + 1 \times 8 = 1 + 8 = 9\end{align}

$$\therefore \left( {1{^*}2} \right){^*}3 \ne 1{^*}\left( {2{^*}3} \right)$$ where$$1,2,3 \in Q$$

Operation $${^*}$$ is not associative.

(iv) On $$Q$$, the operation $${^*}$$ is defined as $$a{^*}b = {\left( {a - b} \right)^2}$$

For $$a,b \in Q$$

\begin{align}a{^*}b = {\left( {a - b} \right)^2}\\b{^*}a = {\left( {b - a} \right)^2} = {\left[ { - \left( {a - b} \right)} \right]^2} = {\left( {a - b} \right)^2}\\\therefore a{^*}b = b{^*}a\end{align}

Operation $${^*}$$ is commutative.

\begin{align}\left( {1{^*}2} \right){^*}3 = {\left( {1 - 2} \right)^2}{^*}3 = {\left( { - 1} \right)^2}{^*}3 = 1{^*}3 = {\left( {1 - 3} \right)^2} = {\left( { - 2} \right)^2} = 4\\1{^*}\left( {2{^*}3} \right) = 1{^*}{\left( {2 - 3} \right)^2} = 1{^*}{\left( { - 1} \right)^2} = 1{^*}1 = {\left( {1 - 1} \right)^2} = 0\end{align}

$$\therefore \left( {1{^*}2} \right){^*}3 \ne 1{^*}\left( {2{^*}3} \right)$$ where$$1,2,3 \in Q$$

Operation $${^*}$$ is not associative.

(v) On $$Q$$, the operation $${^*}$$ is defined as $$a + b = \frac{{ab}}{4}$$

For $$a,b \in Q$$

\begin{align}a{^*}b = \frac{{ab}}{4} = \frac{{ba}}{4} = b{^*}a\\\therefore a{^*}b = b{^*}a\end{align}

Operation $${^*}$$ is commutative.

For $$a,b,c \in Q$$

\begin{align}\left( {a{^*}b} \right){^*}c = \frac{{ab}}{4}{^*}c = \frac{{\frac{{ab}}{4} \cdot c}}{4} = \frac{{abc}}{{16}}\\a{^*}\left( {b{^*}c} \right) = a{^*}\frac{{ab}}{4} = \frac{{a \cdot \frac{{ab}}{4}}}{4} = \frac{{abc}}{{16}}\end{align}

$$\therefore \left( {a{^*}b} \right){^*}c = a{^*}\left( {b{^*}c} \right)$$ where$$a,b,c \in Q$$

Operation $${^*}$$ is associative.

(vi) On $$Q$$, the operation $${^*}$$ is defined as $$a{^*}b = a{b^2}$$

\begin{align}\frac{1}{2}{^*}\frac{1}{3} = \frac{1}{2} \cdot {\left( {\frac{1}{3}} \right)^2} = \frac{1}{2} \cdot \frac{1}{9} = \frac{1}{{18}}\\\frac{1}{3}{^*}\frac{1}{2} = \frac{1}{3} \cdot {\left( {\frac{1}{2}} \right)^2} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{{12}}\end{align}

$$\therefore \left( {\frac{1}{2}{^*}\frac{1}{3}} \right) \ne \left( {\frac{1}{3}{^*}\frac{1}{2}} \right)$$ where$$\frac{1}{2},\frac{1}{3} \in Q$$

Operation $${^*}$$ is not commutative.

\begin{align}\left( {\frac{1}{2}{^*}\frac{1}{3}} \right){^*}\frac{1}{4} = \left( {\frac{1}{2} \cdot {{\left( {\frac{1}{3}} \right)}^2}} \right){^*}\frac{1}{4} = \frac{1}{{18}}{^*}\frac{1}{4} = \frac{1}{{18}} \cdot {\left( {\frac{1}{4}} \right)^2} = \frac{1}{{18 \times 16}}\\\frac{1}{2}{^*}\left( {\frac{1}{3}{^*}\frac{1}{4}} \right) = \frac{1}{2}{^*}\left( {\frac{1}{3} \cdot {{\left( {\frac{1}{4}} \right)}^2}} \right) = \frac{1}{2}{^*}\frac{1}{{48}} = \frac{1}{2} \cdot {\left( {\frac{1}{{48}}} \right)^2} = \frac{1}{{2 \times {{\left( {48} \right)}^2}}}\end{align}

$$\therefore \left( {\frac{1}{2}{^*}\frac{1}{3}} \right){^*}\frac{1}{4} \ne \frac{1}{2}{^*}\left( {\frac{1}{3}{^*}\frac{1}{4}} \right)$$ where$$\frac{1}{2},\frac{1}{3},\frac{1}{4} \in Q$$

Operation $${^*}$$ is not associative.

Operations defined in ($$ii$$), ($$iv$$), ($$v$$) are commutative and the operation defined in ($$v$$) is associative.

## Chapter 1 Ex.1.4 Question 10

Find which of the operations given above has identity.

### Solution

An element $$e \in Q$$ will be the identity element for the operation $${^*}$$ if

$$a{^*}e = a = e{^*}a,$$ for all $$a \in Q$$

\begin{align}&a{^*}b = \frac{{ab}}{4}\\& \Rightarrow\; a{^*}e = a\\ &\Rightarrow \;\frac{{ae}}{4} = a\\ &\Rightarrow \;e = 4\end{align}

Similarly, it can be checked for $$e {^*} a = a$$, we get $$e = 4$$ is the identity.

## Chapter 1 Ex.1.4 Question 11

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

### Solution

$$A = N \times N$$and $$^*$$be the binary operation on A defined by

\begin{align}&\left( {a,b} \right){\rm{*}}\left( {c,d} \right){\rm{ = }}\left( {a + c,b + d} \right)\\&\left( {a,b} \right)*\left( {c,d} \right) \in A\\&a,b,c,d \in N\\&\left( {a,b} \right){\rm{*}}\left( {c,d} \right){\rm{ = }}\left( {a + c,b + d} \right)\\&\left( {c,d} \right)*\left( {a,b} \right) = \left( {c + a,d + b} \right) = \left( {a + c,b + d} \right)\end{align}

$$\therefore \left( {a,b} \right)*\left( {c,d} \right) = \left( {c,d} \right)*\left( {a,b} \right)$$
Operation $$^*$$ is commutative.

Now, let $$\left( {a,b} \right),\left( {c,d} \right),\left( {e,f} \right) \in A$$

$$a,b,c,d,e,f \in N$$

\begin{align}&\left[ {\left( {a,b} \right)*\left( {c,d} \right)} \right]*\left( {e,f} \right) = \left( {a + c,b + d} \right)*\left( {e,f} \right) = \left( {a + c + e,b + d + f} \right)\\&\left( {a,b} \right)*\left[ {\left( {c,d} \right)*\left( {e,f} \right)} \right] = \left( {a,b} \right)*\left( {c + e,d + f} \right) = \left( {a + c + e,b + d + f} \right)\\&\therefore \left[ {\left( {a,b} \right)*\left( {c,d} \right)} \right]*\left( {e,f} \right) = \left( {a,b} \right)*\left[ {\left( {c,d} \right)*\left( {e,f} \right)} \right]\end{align}

Operation $$^*$$ is associative.

An element $$e = \left( {{e_1},{e_2}} \right) \in A$$ will be an identity element for the operation $$^*$$ if $$a + e = a = e*a$$ for all $$a = \left( {{a_1},{a_2}} \right) \in A$$ i.e., $$\left( {{a_1} + {e_1},{a_2} + {e_2}} \right) = \left( {{a_1},{a_2}} \right) = \left( {{e_1} + {a_1},{e_2} + {a_2}} \right)$$, which is not true for any element in A.

Therefore, the operation $$^*$$ does not have any identity element.

## Chapter 1 Ex.1.4 Question 12

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

(i) For an arbitrary binary operation $$^*$$ on a set N,$$a*a = a$$ for all $$a \in N$$.

(ii) If $$^*$$ is a commutative binary operation on N, then $$a*\left( {b*c} \right) = \left( {c*b} \right)*a$$

### Solution

(i) Define operation $$^*$$ on a set N as $$a*a = a$$ for all $$a \in N$$.

In particular, for$$a = {\rm{3}}$$,

$${\rm{3*3 = 9}} \ne {\rm{3}}$$

Therefore, statement (i) is false.

(ii) R.H.S. = $$\left( {c*b} \right)*a$$

$$= \left( {b*c} \right)*a$$ [$$^*$$ is commutative]

$$= a*\left( {b*c} \right)$$ [Again, as $$^*$$ is commutative]

$$= \rm{L.H.S.}$$

$$\therefore a*\left( {b*c} \right) = \left( {c*b} \right)*a$$

Therefore, statement (ii) is true.

## Chapter 1 Ex.1.4 Question 13

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

On N, operation $$^*$$is defined as$$a*b = {a^3} + {b^3}$$.

For all $$a,b \in N$$

$$a*b = {a^3} + {b^3} = {b^3} + {a^3} = b*a$$

Operation $$^*$$ is commutative.

\begin{align}&\left( {{\rm{1}}*{\rm{2}}} \right)*3 = \left( {{1^3} + {2^3}} \right)*3 = \left( {1 + 8} \right)*3 = 9*3 = {9^3} + {3^3} = 729 + 27 = 756\\&1*\left( {2*3} \right) = 1*\left( {{2^3} + {3^3}} \right) = 1*\left( {8 + 27} \right) = 1*35 = {1^3} + {35^3} = 1 + 42875 = 42876\\&\therefore \left( {{\rm{1}}*{\rm{2}}} \right)*3 \ne 1*\left( {2*3} \right)\end{align}

Operation $$^*$$is not associative.

Therefore, Operation $$^*$$ is commutative, but not associative.

The correct answer is $$B$$.

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