exercise 1.4 class 12 maths ncert solutions || class 12 maths ncert solutions chapter 1 ex 1.4 || ex 1.4 class 12 maths ncert solutions|| relations and functions class 12 ncert solutions (English Medium)
NCERT Solutions for Class 12 Maths – Chapter 1 Exercise 1.4 (Relations and Functions)
Exercise 1.4 of Class 12 Maths NCERT Solutions focuses on the concept of binary operations and functions, which is a key part of the chapter Relations and Functions. This exercise helps students understand how to perform operations on sets and verify important properties like commutativity, associativity, identity, and invertibility. The Class 12 Maths Exercise 1.4 NCERT Solutions are explained with clarity and step-by-step reasoning to aid students in mastering complex problems with ease. Aligned with the latest CBSE syllabus and designed for English Medium students, these solutions are ideal for building a strong foundation in abstract algebraic concepts. By studying the Class 12 Maths NCERT Solutions Chapter 1 Exercise 1.4, students can enhance their analytical and logical thinking skills. The Exercise 1.4 Class 12 Maths NCERT Solutions are an excellent resource for board exam preparation and for strengthening overall understanding in mathematics.
class 12 maths ncert solutions chapter 1 ex 1.4 || exercise 1.4 class 12 maths ncert solutions || ex 1.4 class 12 maths ncert solutions || relations and functions class 12 ncert solutions
Exercise 1.4
1 A. 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.
On \( \mathrm{Z}^{+} \), define \( * \) by \( \mathrm{a} * \mathrm{b}=\mathrm{a}-\mathrm{b} \)
On \( \mathrm{Z}^{+} \), define \( * \) by \( \mathrm{a} * \mathrm{b}=\mathrm{a}-\mathrm{b} \)
Answer
It is given that On \( \mathrm{Z}^{+} \), define \( * \) by \( \mathrm{a} * \mathrm{b}=\mathrm{a}-\mathrm{b} \)
It is not a binary operation as the image of \( (1,2) \) under \(*\) is \( 1 * 2=1-2=-1 \notin Z^{+} \).
1 B. 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.
On \( \mathrm{Z}^{+} \), define \( * \) by \( \mathrm{a} * \mathrm{b}=\mathrm{ab} \)
On \( \mathrm{Z}^{+} \), define \( * \) by \( \mathrm{a} * \mathrm{b}=\mathrm{ab} \)
Answer
It is given that on \( \mathrm{Z}^{+} \), define \( * \) by \( \mathrm{a} * \mathrm{b}=\mathrm{ab} \)
We can see that for each \( \mathrm{a}, \mathrm{b} \in \mathrm{Z}^{+} \), there is a unique element ab in \( \mathrm{Z}^{+} \).
\( \Rightarrow * \) carries each pair \( (\mathrm{a}, \mathrm{b}) \) to a unique element \( \mathrm{a} * \mathrm{b}=\mathrm{ab} \) in \( \mathrm{Z}^{+} \).
Therefore, \(*\) is a binary operation.
1 C. 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.
On \(R ,\) define \( * \) by \( \mathrm{a} * \mathrm{b}=\mathrm{ab}^{2} \)
On \(R ,\) define \( * \) by \( \mathrm{a} * \mathrm{b}=\mathrm{ab}^{2} \)
Answer
It is given that On \(R \), define \( * \) by \( \mathrm{a} * \mathrm{b}=\mathrm{ab}^{2} \)
We can see that for each \( a, b \in R \), there is a unique element \( a b^{2} \) in \( R \).
\( \Rightarrow * \) carries each pair \( \left(\mathrm{a}, \mathrm{b}\right)\) to a unique element \( \mathrm{a} * \mathrm{b}=\mathrm{ab}^{2} \) in \(R \).
Therefore, \(*\) is a binary operation.
1 D. 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.
On \( \mathrm{Z}^{+} \), define \( * \) by \( \mathrm{a} * \mathrm{b}=|\mathrm{a}-\mathrm{b}| \)
On \( \mathrm{Z}^{+} \), define \( * \) by \( \mathrm{a} * \mathrm{b}=|\mathrm{a}-\mathrm{b}| \)
Answer
It is given On \( \mathrm{Z}^{+}\), define \( * \) by \( a * b=|a-b| \)
We can see that for each \( a, b \in Z^{+} \), there is a unique element \( |a-b| \) in \( Z^{+} \).
\( \Rightarrow * \) carries each pair \( (\mathrm{a}, \mathrm{b}) \) to a unique element \( \mathrm{a}* \mathrm{b}=|\mathrm{a}-\mathrm{b}| \) in \( \mathrm{Z}^{+} \).
Therefore, \(*\) is a binary operation.
1 E. 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.
On \( \mathrm{Z}^{+} \) define \( * \) by \( \mathrm{a} * \mathrm{b}=\mathrm{a} \)
On \( \mathrm{Z}^{+} \) define \( * \) by \( \mathrm{a} * \mathrm{b}=\mathrm{a} \)
Answer
It is given that On \( Z^{+} \) define \( * \) by \( \mathrm{a} * \mathrm{b}=\mathrm{a} \)
We can see that for each \( \mathrm{a}, \mathrm{b} \in \mathrm{Z}^{+} \), there is a unique element a in \( \mathrm{Z}^{+} \).
\( \Rightarrow * \) carries each pair \( (\mathrm{a}, \mathrm{b}) \) to a unique element \( \mathrm{a} * \mathrm{b}=\mathrm{a} \) in \( \mathrm{Z}^{+} \).
Therefore, \(*\) is a binary operation.
2 A. For each operation \( * \) defined below, determine whether \( * \) is binary, commutative or associative.
On \( Z \), define \( a * b=a-b \)
On \( Z \), define \( a * b=a-b \)
Answer
It is given that On \( Z \), define \( a * b=a-b \)
\( \mathrm{a}-\mathrm{b} \in \mathrm{Z} \). so the operation \(*\) is binary.
We can see that \( 1 * 2=1-2=-1 \) and \( 2 * 1=2-1=1 \).
\( \Rightarrow 1 * 2 \neq 2 * 1 \), where \( 1,2 \in \mathrm{Z} \).
\( \Rightarrow \) the operation \(*\) is not commutative.
2 B. For each operation \( * \) defined below, determine whether \( * \) is binary, commutative or associative.
On Q, define \( \mathrm{a} * \mathrm{b}=\mathrm{ab}+1 \)
On Q, define \( \mathrm{a} * \mathrm{b}=\mathrm{ab}+1 \)
Answer
It is given that On Q, define \( \mathrm{a} * \mathrm{b}=\mathrm{ab}+1 \)
\( \mathrm{ab}+1 \in \mathrm{Q} \), so operation \(*\) is binary
We know that \( \mathrm{ab}=\mathrm{ba} \) for \( \mathrm{a}, \mathrm{b} \in \mathrm{Q} \)
\(
\Rightarrow \mathrm{ab}+1=\mathrm{ba}+1 \text { for } \mathrm{a}, \mathrm{b} \in \mathrm{Q}\)
\(\Rightarrow \mathrm{a} * \mathrm{b}=\mathrm{a} * \mathrm{b} \text { for } \mathrm{a}, \mathrm{b} \in \mathrm{Q}\)
\(\Rightarrow 1* 2 \neq 2 * 1, \text { where } 1,2 \in Z
\)
\( \Rightarrow \) The operation \(*\) is commutative.
Also, we get,
\(
(1 * 2) * 3=(1 \times 2) * 3=3 * 3=3 \times 3+1=10\)
\(1 *(2 * 3)=1 *(2 \times 3)=1 * 7=1 \times 7+1=8\)
\(\Rightarrow(1 * 2) * 3 \neq 1 *(2 * 3)
\)
\( \Rightarrow \) the operation \(*\) is not associative.
2 C. For each operation \( * \) defined below, determine whether \( * \) is binary, commutative or associative.
On \( \mathrm{Z}^{+} \), define \( \mathrm{a} * \mathrm{b}=2 \mathrm{ab} \)
On \( \mathrm{Z}^{+} \), define \( \mathrm{a} * \mathrm{b}=2 \mathrm{ab} \)
Answer
It is given that on \( \mathrm{Z}^{+} \), define \( \mathrm{a} * \mathrm{b}=2 \mathrm{ab} \)
\(2ab \in \mathrm{Z}^{+} \), so operation \(*\) is binary
We know that \( \mathrm{ab}= \) ba for \( \mathrm{a}, \mathrm{b} \in \mathrm{Z}^{+} \)
\( \Rightarrow 2 \mathrm{ab}=2 \mathrm{ba} \) for \( \mathrm{a}, \mathrm{b} \in \mathrm{Z}^{+} \)
\( \Rightarrow \mathrm{a} * \mathrm{b}=\mathrm{a} * \mathrm{b} \) for \( \mathrm{a}, \mathrm{b} \in \mathrm{Z}^{+} \)
\( \Rightarrow \) The operation \(*\) is commutative.
Also, we get,
\((1 * 2) * 3=2(1 \times 2) * 3=4 * 3=2(4 \times 3)=212\)
\(1 *(2 * 3)=1 * 2(2 \times 3)=1 * 26=1 \times 64=264\)
\(\Rightarrow(1 * 2) * 3 \neq 1 *(2 * 3)
\)
\( \Rightarrow \) The operation \(*\) is not associative.
2 D. For each operation \( * \) defined below, determine whether \( * \) is binary, commutative or associative.
On \( \mathrm{Z}^{+} \), define \( \mathrm{a} * \mathrm{b}=\mathrm{a}^{\mathrm{b}} \)
On \( \mathrm{Z}^{+} \), define \( \mathrm{a} * \mathrm{b}=\mathrm{a}^{\mathrm{b}} \)
Answer
It is given that On \(Z^{+} \), define \( a * b=a^{b} \)
\( ab \in\mathrm{Z}^{+} \), so operation \(*\) is binary
We know that \( \mathrm{ab}= \mathrm{ba}\) for \( \mathrm{a}, \mathrm{b} \in \mathrm{Z}^{+} \)
\( \Rightarrow 1 * 2=1^{2} \) and \( 2 * 1=2^{1}=2 \)
\( \Rightarrow 1 * 2 \neq 2 * 1 \), where \( 1,2 \in \mathrm{Z}^{+} \)
\( \Rightarrow \) The operation \(*\) is not commutative.
Also, we get,
\( (1 * 2) * 3=2^{3} * 3=8 * 3=2^{4} \)
\( 1 *(2 * 3)=1 * 3^{2}=1 * 9=9 \)
\( \Rightarrow(1 * 2) * 3 \neq 1 *(2 * 3) \), where \( 1,2,3 \in \mathrm{Z}^{+} \)
\( \Rightarrow \) The operation \( * \) is not associative.
2 E. For each operation \( * \) defined below, determine whether \( * \) is binary, commutative or associative.
On \( \mathrm{R}-\{-1\} \), define \( \mathrm{a} * \mathrm{b}= \)
On \( \mathrm{R}-\{-1\} \), define \( \mathrm{a} * \mathrm{b}= \)
Answer
It is given that On \( \mathrm{R}, *-\{-1\} \), define \( \mathrm{a} * \mathrm{b}=\frac{a}{b+1} \)
\( \frac{a}{b+1} \in \mathrm{R} \) for \( \mathrm{b} \neq-1 \), so the operation \(*\) is binary.
We can see that \( 1 * 2=\frac{1}{2+1}=\frac{1}{3} \) and \( 2 * 1=\frac{2}{1+1}=\frac{2}{2}=1 \)
\( \Rightarrow 1 * 2 \neq 2 * 1 \); where \( 1,2 \in \mathrm{R}-\{-1\} \)
\( \Rightarrow \) the operation * is not commutative.
Now, we can have observed that
\( (1 * 2) * 3=\frac{1}{3} * 3=\frac{\frac{1}{3}}{3+1}=\frac{1}{12} \)
\( 1 *(2 * 3)=1 * \frac{2}{3+1}=1 * \frac{2}{4}=1 * \frac{1}{2}=\frac{1}{\frac{1}{2}+1}=\frac{1}{\frac{3}{2}}=\frac{2}{3} \)
\( \Rightarrow(1 * 2) * 3 \neq 1 *(2 * 3) \), where \( 1,2,3 \in R*-\{-1\} \)
\( \Rightarrow \) The operation \(*\) is not associative.
3. consider the binary operation \( \wedge \) on the set \( \{1,2,3,4,5\} \) defined by \( \mathrm{a}\wedge \mathrm{b} \) \( =\min \{\mathrm{a}, \mathrm{b}\} \). Write the operation table of the operation \( \wedge \).
Answer
The binary operation \( \wedge \) on the set \( \{1,2,3,4,5\} \) is defined as \( \mathrm{a}^{\wedge} \mathrm{b}=\min \) \( \{\mathrm{a}, \mathrm{b}\} \) for \( \mathrm{a}, \mathrm{b} \in\{1,2,3,4,5\} \)
Therefore, the operation table of the given operation \( { }^{\wedge} \) can be given as:
\(\begin{array}{|l|l|l|l|l|l|}
\hline { }^{\wedge} & 1 & 2 & 3 & 4 & 5 \\
\hline 1 & 1 & 1 & 1 & 1 & 1 \\
\hline 2 & 1 & 2 & 2 & 2 & 2 \\
\hline 3 & 1 & 2 & 3 & 3 & 3 \\
\hline 4 & 1 & 2 & 3 & 4 & 4 \\
\hline 5 & 1 & 2 & 3 & 4 & 5 \\
\hline
\end{array}\)
class 12 maths ncert solutions chapter 1 ex 1.4 || exercise 1.4 class 12 maths ncert solutions || ex 1.4 class 12 maths ncert solutions || relations and functions class 12 ncert solutions
4. Consider a binary operation \( * \) on the set \( \{1,2,3,4,5\} \) given by the following multiplication table (Table 1.2).
(i) Compute \( (2 * 3) * 4 \) and \( 2 *(3 * 4) \)
(ii) Is \( * \) commutative?
(iii) Compute \( (2 * 3) *(4 * 5) \).
Table 1.2
\(\begin{array}{|l|l|l|l|l|l|}
\hline * & 1 & 2 & 3 & 4 & 5 \\
\hline 1 & 1 & 1 & 1 & 1 & 1 \\
\hline 2 & 1 & 2 & 1 & 2 & 1 \\
\hline 3 & 1 & 1 & 3 & 1 & 1 \\
\hline 4 & 1 & 2 & 1 & 4 & 1 \\
\hline 5 & 1 & 1 & 1 & 1 & 5 \\
\hline
\end{array}\)
(i) Compute \( (2 * 3) * 4 \) and \( 2 *(3 * 4) \)
(ii) Is \( * \) commutative?
(iii) Compute \( (2 * 3) *(4 * 5) \).
Table 1.2
\(\begin{array}{|l|l|l|l|l|l|}
\hline * & 1 & 2 & 3 & 4 & 5 \\
\hline 1 & 1 & 1 & 1 & 1 & 1 \\
\hline 2 & 1 & 2 & 1 & 2 & 1 \\
\hline 3 & 1 & 1 & 3 & 1 & 1 \\
\hline 4 & 1 & 2 & 1 & 4 & 1 \\
\hline 5 & 1 & 1 & 1 & 1 & 5 \\
\hline
\end{array}\)
Answer
(i) \( (2 * 3) * 4=1 * 4=1 \)
\(
2 *(3 * 4)=2 * 1=1
\)
(ii) For every \( \mathrm{a}, \mathrm{b} \in\{1,2,3,4,5\} \),
We have, \( \mathrm{a} * \mathrm{b}=\mathrm{b} * \mathrm{a} \)
\( \Rightarrow \) the operation \( * \) is commutative.
(iii) \( (2 * 3)=1 \)
\(
\Rightarrow(2 * 3) *(4 * 5)=1 * 1=1
\)
5. Let \( *^{\prime} \) be the binary operation on the set \( \{1,2,3,4,5\} \) 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.
Answer
The binary operation \(*{ }^{\prime}\) on the set \( \{1,2,3,4,5\} \) is defined as \( a*{ }^{\prime} b= \) HCF of \( a \) and \( b \).
The operation table for the operation \( *{ }^{\prime} \) is given by:
\(\begin{array}{|l|l|l|l|l|l|}
\hline * & 1 & 2 & 3 & 4 & 5 \\
\hline 1 & 1 & 1 & 1 & 1 & 1 \\
\hline 2 & 1 & 2 & 1 & 2 & 1 \\
\hline 3 & 1 & 1 & 3 & 1 & 1 \\
\hline 4 & 1 & 2 & 1 & 4 & 1 \\
\hline 5 & 1 & 1 & 1 & 1 & 5 \\
\hline
\end{array}\)
From the above table we can see that the operation tables for the operation \(*\) and \(*{ }^{\prime}\) are the same.
Therefore, the operation \( *{ }^{\prime}\) is same as the operation \(*\)
6. Let \( * \) be the binary operation on N given by \( \mathrm{a} * \mathrm{b}= \) 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 \(*\) ?
(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 \(*\) ?
Answer
(i) It is given that the binary operation on N given by \( \mathrm{a} * \mathrm{b}= \) L.C.M. of \(a\) and \( b \).
Then, \( 5 * 7=\mathrm{LCM} \) of 5 and \( 7=35 \)
\( 20 * 16=\mathrm{LCM} \) of 20 and \( 16=80 \).
(ii) It is given that the binary operation on N given by \( \mathrm{a} * \mathrm{b}= \) L.C.M. of a and b.
We know that LCM of \( a \) and \( b=\mathrm{LCM} \) of \( b \) and \( a, a, b \in N \).
\(
\Rightarrow \mathrm{a} * \mathrm{b}=\mathrm{b} * \mathrm{a}
\)
Therefore, the operation \(*\) is commutative.
(iii) It is given that the binary operation on N given by \( \mathrm{a} * \mathrm{b}= \) L.C.M. of a and b.
For \( a, b, c \in N \)
\( (\mathrm{a} * \mathrm{b}) * \mathrm{c}=(\mathrm{LCM} \) of a and b\( ) * \mathrm{c}=\mathrm{LCM} \) of \( \mathrm{a}, \mathrm{b} \) and c
\( a*\left(b* c\right)=a *( \mathrm{LCM} \) of \( b \) and \( c)= \) LCM of \( a, b \) and \( c \)
\(
\Rightarrow(\mathrm{a} * \mathrm{b}) * \mathrm{c}=\mathrm{a} *(\mathrm{b} * \mathrm{c})
\)
Therefore, the operation \(*\) is associative.
(iv) It is given that the binary operation on N given by \( \mathrm{a} * \mathrm{b}= \) L.C.M. of a and b.
We know that LCM of a and \( 1=\mathrm{a}=\mathrm{LCM} \) of 1 and \( 1, \mathrm{a} \in \mathrm{N} \) \( \Rightarrow \mathrm{a}* 1=\mathrm{a}=1* \mathrm{a}, \mathrm{a} \in \mathrm{N} \)
Therefore, 1 is the identity of \(*\) in N.
(v) It is given that the binary operation on N given by \( \mathrm{a} * \mathrm{b}= \) L.C.M. of \( a \) and \( b \).
An element a in N is invertible w.r.t. the operation \(*\) if there exists an element b in N,
Such that \( a * b=e=b * a \)
Now, if \( \mathrm{e}=1 \)
\( \Rightarrow \) LCM of a and \( \mathrm{b}=1=\mathrm{LCM} \) of b and a
\( \Rightarrow \) This is only possible when \( \mathrm{a}=\mathrm{b}=1 \)
Therefore, 1 is the only invertible element of N w.r.t. the operation \(*\).
7. Is \( * \) defined on the set \( \{1,2,3,4,5\} \) by \( \mathrm{a} * \mathrm{b}= \) L.C.M. of a and b a binary operation? Justify your answer.
Answer
The operation \(*\) on the set \( \mathrm{A}=\{1,2,3,4,5\} \) is defined as
\( a * b=\mathrm{LCM} \) of \( a \) and \( b \).
\( 2 * 3=\mathrm{LCM} \) of 2 and \( 3=6 \).
But 6 does not belong to the given set.
Therefore, the given operation \(*\) is not a binary operation.
8. Let \( * \) be the binary operation on N defined by \( \mathrm{a} * \mathrm{b}= \) H.C.F. of a and b . Is \( * \) commutative? Is \( * \) associative? Does there exist identity for this binary operation on N ?
Answer
It is given that the binary operation on N defined by \( \mathrm{a} * \mathrm{b}= \) H.C.F. of a and b.
We know that HCF of \( a \) and \( b= \) HCF of \( b \) and \( a, a, b \in N \).
\(
\Rightarrow \mathrm{a}* \mathrm{b}=\mathrm{b} * \mathrm{a}
\)
\( \Rightarrow \) The operation \( * \) is commutative.
For \( \mathrm{a}, \mathrm{b} \mathrm{c} \in \mathrm{N} \), we get,
\( (a * b) * c=( \)HCF of \( a \) and \( b) * c= \) HCF of \( a, b \) and \( c \)
\( a*\left(b* c\right)=a *( \)HCF of \( b \) and \( c)= \) HCF of \( a, b \) and \( c \)
\(
\Rightarrow(\mathrm{a} * \mathrm{b}) * \mathrm{c}=\mathrm{a} *(\mathrm{b} * \mathrm{c})
\)
\( \Rightarrow \) The operation \( * \) is associative.
Now, an element \( \mathrm{e} \in\mathrm{N} \) will be the identity for the operation.
Now, if \( \mathrm{a} * \mathrm{e}=\mathrm{a}=\mathrm{e} * \mathrm{a}, \mathrm{a} \in \mathrm{N} \).
But, this is not true for any \( a\in\mathrm{N} \).
Therefore, the operation \(*\) does not have any identity in N.
9 A. Let \(*\) be a binary operation on the set Q of rational numbers as follows: \( \mathrm{a} * \mathrm{b}=\mathrm{a}-\mathrm{b} \)
Find which of the binary operations are commutative and which are associative.
Find which of the binary operations are commutative and which are associative.
Answer
It is given that \(*\) be a binary operation on the set Q of rational numbers as
\(
a * b=a-b
\)
We can have observed that for \( 2,3,4 \in \mathrm{Q} \)
\( 2 * 3=2-3=-1 \) and \( 3 * 2=3-2=1 \)
\(
2 * 3 \neq 3 * 2
\)
\( \Rightarrow \) the operation \(*\) is not commutative.
Also, \( (2 * 3) * 4=(-1) * 4=-1-4=-5 \) and
\(2 *(3 * 4)=2 *(-1)=2-(-1)=3\)
\(\Rightarrow(2 * 3) * 4 \neq 2 *(3 * 4)
\)
Therefore, the operation \(*\) is not associative.
9 B. Let \(*\) be a binary operation on the set Q of rational numbers as follows: \( a * b=a^{2}+b^{2} \)
Find which of the binary operations are commutative and which are associative.
Find which of the binary operations are commutative and which are associative.
Answer
It is given that \(*\) be a binary operation on the set Q of rational numbers is defined as
\(
a * b=a^{2}+b^{2}
\)
For \( \mathrm{a}, \mathrm{b} \in \mathrm{Q} \), we get,
\( a* b=a^{2}+b^{2}=b^{2}+a^{2}=b* a \)
\( \Rightarrow \) the operation is commutative.
We can see that \( (1 * 2) * 3 \neq 1 *(2 * 3) \), where \( 1,2,3 \in \mathrm{Q} \)
Therefore, the operation \(*\) is not associative.
9 C. Let \(*\) be a binary operation on the set Q of rational numbers as follows:
\(
a * b=a+a b
\)
Find which of the binary operations are commutative and which are associative.
\(
a * b=a+a b
\)
Find which of the binary operations are commutative and which are associative.
Answer
It is given that \(*\) be a binary operation on the set Q of rational numbers is defined as
\(
a * b=a+a b
\)
We can see that \( 1 * 2=1+1 \times 2=1+2=3 \)
and \( 2 * 1=2+2 \times 1=2+2=4 \)
\( \Rightarrow 1 * 2 \neq 2 * 1: \) where \( 1,2 \in \mathrm{Q} \).
\( \Rightarrow \) the operation \(*\) is not commutative.
Also, we can see that \( (1 * 2) * 3 \neq 1 *(2 * 3) \), where \( 1,2,3 \in \mathrm{Q} \)
Therefore, the operation \(*\) is not associative.
9 D. Let \(*\) be a binary operation on the set Q of rational numbers as follows: \( \mathrm{a} * \mathrm{b}=(\mathrm{a}-\mathrm{b})^{2} \)
Find which of the binary operations are commutative and which are associative.
Find which of the binary operations are commutative and which are associative.
Answer
It is given that \(*\) be a binary operation on the set Q of rational numbers is defined as
\(
a * b=(a-b)^{2}
\)
For \( \mathrm{a}, \mathrm{b} \in \mathrm{Q} \), we have,
\(
a * b=(a-b)^{2}\)
\(b * a=(b-a)^{2}=[-(a-b)]^{2}=(a-b)^{2}\)
\(\Rightarrow a * b=b * a
\)
\( \Rightarrow \) the operation \( * \) is commutative.
Also, we can see that \( (1 * 2) * 3 \neq 1 *(2 * 3) \), where \( 1,2,3 \in \mathrm{Q} \) Therefore, the operation \(*\) is not associative.
9 E. Let \( * \) be a binary operation on the set Q of rational numbers as follows: \( \mathrm{a} * \mathrm{b}=\frac{a b}{4} \)
Find which of the binary operations are commutative and which are associative.
Find which of the binary operations are commutative and which are associative.
Answer
It is given that \(*\) be a binary operation on the set Q of rational numbers is defined as
\(
\mathrm{a} * \mathrm{b}=\frac{a b}{4}
\)
For \( a, b \in Q \), we get,
\(
\mathrm{a} * \mathrm{b}=\frac{a b}{4}=\frac{b a}{4}=\mathrm{b} * \mathrm{a}
\)
\( \Rightarrow \) the operation is commutative.
For, \( \mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{Q} \), we get,
\(
(\mathrm{a} * \mathrm{b}) * \mathrm{c}=\frac{a b}{4} * \mathrm{c}=\frac{\frac{a b}{4} \cdot c}{4}=\frac{a b c}{16}\)
\(\mathrm{a} *(\mathrm{b} * \mathrm{c})=\mathrm{a} * \frac{b c}{4}=\frac{a \cdot \frac{b c}{4}}{4}=\frac{a b c}{16}\)
\(\Rightarrow(\mathrm{a} * \mathrm{b}) * \mathrm{c}=\mathrm{a} *(\mathrm{b} * \mathrm{c})
\)
Therefore, the operation \(*\) is associative.
class 12 maths ncert solutions chapter 1 ex 1.4 || exercise 1.4 class 12 maths ncert solutions || ex 1.4 class 12 maths ncert solutions || relations and functions class 12 ncert solutions
9 F. Let \(*\) be a binary operation on the set Q of rational numbers as follows: \( a * b=a b^{2} \)
Find which of the binary operations are commutative and which are associative.
Find which of the binary operations are commutative and which are associative.
Answer
It is given that \(*\) be a binary operation on the set Q of rational numbers is defined as
\(
a * b=a b^{2}
\)
We can see that for \( 2,3 \in \mathrm{Q} \)
\( 2 * 3=2.3^{2}=18 \) and \( 3 * 2=3.2^{2}=12 \)
\(
\Rightarrow 2 * 3 \neq 3 * 2
\)
Therefore, the operation \(*\) is not commutative.
Also, we can see that \( (1 * 2) * 3 \neq 1 *(2 * 3) \), where \( 1,2,3 \in \mathrm{Q} \)
Therefore, the operation \(*\) is not associative.
10. Find which of the operations given above has identity.
Answer
(i) An element \( e \in Q \) will be the identity element for the operation \(*\) if \( \mathrm{a} * \mathrm{e}=\mathrm{a}=\mathrm{e}* \mathrm{a}, \forall \mathrm{a} \in \mathrm{Q} \)
\(
a* b=a-b
\)
This operation is not commutative,
Therefore, it does not have identity element.
(ii) An element \( \mathrm{e} \in \mathrm{Q} \) will be the identity element for the operation \(*\) if \( \mathrm{a}* \mathrm{e}=\mathrm{a}=\mathrm{e}* \mathrm{a}, \forall \mathrm{a} \in \mathrm{Q} \)
\(
a* b=a^{2}+b^{2}
\)
If \( a * e=a \), then \( a^{2}+e^{2}=a \)
For \( \mathrm{a}=-2,(-2)^{4}+\mathrm{e}^{2} \neq-2 \).
Therefore, there is no identity element.
(iii) An element \( e \in Q \) will be the identity element for the operation * if \( \mathrm{a}* \mathrm{e}=\mathrm{a}=\mathrm{e}* \mathrm{a}, \forall \mathrm{a} \in \mathrm{Q} \)
Now, \( a * b=a+a b \)
This is not commutative.
Therefore, there is no identity element.
(iv) An element \( e \in Q \) will be the identity element for the operation \(*\) if \( \mathrm{a}* \mathrm{e}=\mathrm{a}=\mathrm{e}* \mathrm{a}, \forall \mathrm{a} \in \mathrm{Q} \)
\(
a* b=(a-b)^{2}
\)
If \( a* \mathrm{e}=\mathrm{a} \), then \( (\mathrm{a}-\mathrm{e})^{2}=\mathrm{a} \).
A square is always positive, thus for \( \mathrm{a}=-2,(-2-\mathrm{e})^{2} \neq-2 \).
Therefore, there is no identity element.
(v) An element \( \mathrm{e} \in \mathrm{Q} \) will be the identity element for the operation \(*\) if \( \mathrm{a} * \mathrm{e}=\mathrm{a}=\mathrm{e}* \mathrm{a}, \forall \mathrm{a} \in \mathrm{Q} \)
\( \mathrm{a} * \mathrm{b}=\frac{a b}{4} \)
If \( \mathrm{a} * \mathrm{e}=\mathrm{a} \), then \( \frac{a e}{4}=\mathrm{a} \)
Therefore, \( \mathrm{e}=4 \) is the identity element.
\( \mathrm{a} * 4=4 * \mathrm{a}=\frac{4 a}{4}=\mathrm{a} \).
(vi) An element \( e \in\mathrm{Q} \) will be the identity element for the operation \(*\) if \( \mathrm{a} * \mathrm{e}=\mathrm{a}=\mathrm{e} * \mathrm{a}, \forall \mathrm{a} \in \mathrm{Q} \)
Now, \( \mathrm{a} * \mathrm{b}=\mathrm{ab}^{2} \)
This operation is not commutative,
Therefore, there is not have identity element.
11. Let \( \mathrm{A}=\mathrm{N} \times \mathrm{N} \) and \( * \) be the binary operation on A defined by \( (\mathrm{a}, \mathrm{b}) *(\mathrm{c} , d) =(a+c, b+d) \)
Show that \( * \) is commutative and associative. Find the identity element for \( * \) on \( A \), if any.
Show that \( * \) is commutative and associative. Find the identity element for \( * \) on \( A \), if any.
Answer
It is given that \( \mathrm{A}=\mathrm{N} \times \mathrm{N} \) and \( * \) be the binary operation on A defined by \( (\mathrm{a}, \mathrm{b}) *(\mathrm{c}, \mathrm{d})=(\mathrm{a}+\mathrm{c}, \mathrm{b}+\mathrm{d}) \)
Let \( (\mathrm{a}, \mathrm{b}),(\mathrm{c}, \mathrm{d}) \in \mathrm{A} \)
Then, \( a, b, c, d \in N \)
Now, we have,
\(
(a, b) *(c, d)=(a+c, b+d)
\)
\(
(\mathrm{c}, \mathrm{d}) *(\mathrm{a}, \mathrm{d})=(\mathrm{c}+\mathrm{a}, \mathrm{d}+\mathrm{b})=(\mathrm{a}+\mathrm{c}, \mathrm{b}+\mathrm{d})\)
\(\Rightarrow(\mathrm{a}, \mathrm{b}) *(\mathrm{c}, \mathrm{d})=(\mathrm{c}, \mathrm{d}) *(\mathrm{a}, \mathrm{b})
\)
\( \Rightarrow \) The operation \(*\) is commutative.
Now, \( (\mathrm{a}, \mathrm{b}),(\mathrm{c}, \mathrm{d}),(\mathrm{e}, \mathrm{f}) \in \mathrm{A} \)
Then, \( a, b, c, d, e, f \in N \)
Now, we have,
\(((a, b) *(c, d)) *(e, f)=(a+c, b+d) *(e, f)=(a+c+e, b+d+f)\)
\((a, b) *((c, d) *(e, f))=(a, b) *(c+e, d+f)=(a+c+e, b+d+f)
\)
Then, \( ((\mathrm{a}, \mathrm{b}) *(\mathrm{c}, \mathrm{d})) *(\mathrm{e}, \mathrm{f})=(\mathrm{a}, \mathrm{b}) *((\mathrm{c}, \mathrm{d}) *(\mathrm{e}, \mathrm{f})) \)
Therefore, the operation \(*\) is associative.
Now, an element \( e=\left(e_{1}, e_{2}\right) \in \mathrm{A} \) will be an identity for the operation \(*\) if \( a * \mathrm{e}=\mathrm{a}=\mathrm{e} * \mathrm{a} \forall \mathrm{a}=\left(\mathrm{a}_{1}, \mathrm{a}_{2}\right) \in A \),
\(
\left(a_{1}+e_{1}, a_{2}+e_{2}\right)=\left(a_{1}, a_{2}\right)=\left(e_{1}+a_{1}, e_{2}+a_{2}\right)
\)
which is not true for any element in A.?
Therefore, the operation \(*\) does not have any identity element.
12 A. State whether the following statements are true or false. Justify. For an arbitrary binary operation \( * \) on a set \( N, a * a=a \forall a \in N \).
Answer
For an arbitrary binary operation \( * \) on a set \( \mathrm{N}, \mathrm{a} * \mathrm{a}=\mathrm{a} \forall \mathrm{a} \in \mathrm{N} \).
The above statement is false.
Explanation: It is given that an operation \( * \) on a set \( N, a * a=a \forall a \in N \)
Then, in particular, for \( \mathrm{b}=\mathrm{a}=3 \), we get,
\(
3 * 3=3+3=6 \neq 3
\)
12 B. State whether the following statements are true or false. Justify. If \( * \) is a commutative binary operation on N, then \( \mathrm{a} *(\mathrm{b} * \mathrm{c})=(\mathrm{c} * \mathrm{b}) * \mathrm{a} \)
Answer
If \( * \) is a commutative binary operation on N, then \( \mathrm{a} *(\mathrm{b} * \mathrm{c})=(\mathrm{c} * \mathrm{b}) * \mathrm{a} \) The above statement if true.
Explanation: RHS \( =(\mathrm{c} * \mathrm{b}) * \mathrm{a} \)
\( =(\mathrm{b} * \mathrm{c}) * \mathrm{a}(* \) is commutative\( ) \)
\( =\mathrm{a} *(\mathrm{b} * \mathrm{c})(\mathrm{as} * \) is commutative\( ) \)
\( = \) LHS.
Therefore, \( a *(b * c)=(c * b) * a \).
Hence Proved
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?
A. Is \( * \) both associative and commutative?
B. Is \( * \) commutative but not associative?
C. Is \( * \) associative but not commutative?
D. Is \(*\) neither commutative nor associative?
Answer
On N, the operation \( * \) is defined as \( \mathrm{a}* \mathrm{b}=\mathrm{a}^{3}+\mathrm{b}^{3} \)
For, \( \mathrm{a}, \mathrm{b} \in \mathrm{N} \), we get,
\( a* b=a^{3}+b^{3}=b^{3}+a^{3}=b * a \) [Addition is commutative in \( N \)]
\( \Rightarrow \) the operation \( * \) is commutative.
We can have observed that
\( \left(1* 2\right) * 3=\left(1^{3}+2^{3}\right) * 3=9 * 3=9^{3}+3^{3}=729 +27=756 \)
Also, \( 1*\left(2* 3\right)=1 *\left(2^{3}+3^{3}\right)=1 *(8+27)=1 \times 35=1^{3}+35^{3}=1+(35)^{3}= \) \( 1+42875=42876 \).
Therefore, \( (1 * 2) * 3 \neq 1 *(2 * 3) \); where \( 1,2,3 \in \mathrm{N} \)
Therefore, the operation * is not associative.
Therefore, the operation * is commutative, but not associative.
