NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.4


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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