NCERT Solutions for CBSE 12th Mathematics, Chapter 1: Relations & Functions are available here. In this article, you will get solutions of exercise 1.4 (from question number 8 to question number 13). Most of the questions given in this exercise are related to binary operations. These questions are important CBSE Class 12 Maths board exam 2018 and other competitive exams.

*NCERT Solutions for CBSE Class 12th Maths, Chapter 1: Relations and Functions (Exercise 1.4) are given below*

**Question 7:** Is ∗ defined on the set {1, 2, 3, 4, 5} by *a* ∗ *b* = L.C.M. of *a* and *b* a binary operation? Justify your answer.

**Solution 7:**

Set A = {1, 2, 3, 4, 5}

*a* * *b* = L.C.M. of *a* and *b*

Then, the operation table for the given operation * can be represented as:

* |
1 |
2 |
3 |
4 |
5 |

1 |
1 |
2 |
3 |
4 |
5 |

2 |
2 |
2 |
6 |
4 |
10 |

3 |
3 |
6 |
3 |
12 |
15 |

4 |
4 |
4 |
12 |
4 |
20 |

5 |
5 |
10 |
15 |
20 |
5 |

From the table,

3 * 2 = 2 * 3 = 6

5 * 2 = 2 * 5 = 10

3 * 4 = 4 * 3 = 12

3 * 5 = 5 * 3 = 15

4 * 5 = 5 * 4 = 20

Thus, the given operation * is not a binary operation.

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**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 8:**

*a *** **b* = H.C.F. of *a* and *b*

H.C.F. of *a* and *b* = H.C.F. of *b* and *a*

Thus, *a* * *b* = *b *** **a*

Hence, the operation * is commutative.

For *a*, *b*, *c *∈ N,

*a **(*b *** **c*)= *a **(H.C.F. of *b* and *c*) = H.C.F. of *a*, *b*, and *c*

(*a *** **b*)** **c* = (H.C.F. of *a* and *b*) * *c* = H.C.F. of *a*, *b*, and *c*

Thus, (*a *** **b*) * *c = a* * (*b *** **c*)

Hence, the operation * is associative

Now,

An element *e *∈ N will be the identity for the operation *

If *a *** **e* = *a* = e** **a*

But this is not true for any *a *∈ N

Thus, the operation * does not have any identity in N

**Question 9:** Let ∗ be a binary operation on the set 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* = (*a* – *b*)^{2}

(*v*) *a* ∗ *b* = *ab*/4

(*vi*) *a* ∗* b* = *ab*^{2}

**Solution 9:**

(*i*) *a *** **b = a – b*

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

For *a, b* ∈ Q,

*a* * *b* = *a*^{2} + *b*^{2} = *b*^{2} + *a*^{2} = *b* * *a*

∴*a *** **b = b *** **a*

Thus, the operation * is commutative

We can see that,

(1 * 2) * 3 = (1^{2 }+ 2^{2}) * 3 = (1 + 4)*4 = 5*4 = 5^{2 }+ 4^{2 }= 41

1 * (2*3) = 1 * (2^{2} + 3^{2}) = 1 * (4 + 9) = 1*13 = 1^{2 }+ 13^{2} = 170

(1 * 2) * 3 ≠ 1 * (2 * 3)

Thus, the operation * is not associative

(*iii*) *a *** **b = a + ab*

We can observe that,

1 * 2 = 1 + 1 × 2 = 1 + 2 =3

2 * 1 = 2 + 2 × 1 = 2 + 2 = 4

1 * 2 ≠ 2 * 1; where 1, 2 ϵ *Q*

Thus, the operation * is not commutative

Also,

(1 * 2) * 3 = (1 + 1 × 2) * 3 = 3 * 3 = 3 + 3 × 3 = 3 + 9 = 12

1 * (2 * 3) = 1 * (2 + 2 × 3) = 1 * 8 = 1 + 1 × 8 = 9

(1 * 2) * 3 ≠ 1 * (2 * 3); where 1, 2, 3 ϵ *Q*

Thus, the operation * is not associative.

(*iv*)

*a *** **b* = (*a − b*)^{2}

For *a*, *b* ∈ Q,

*a *** **b* = (*a − b*)^{2}

*b *** **a* = (*b − a*)^{2} = [− (*a − b*)]^{2} = (*a − b*)^{2}

∴ *a *** **b = b *** **a*

Thus, the operation * is commutative.

We can see that,

(1 * 2)*3 = (1 ‒ 2)^{2} * 3 = (‒1)^{2} * 3 = 1 * 3 = (1 ‒ 3)^{2} = 4

1 * (2 * 3) = 1 * (2 ‒ 3)^{2} = 1 * (‒1)^{2} = 1 * 1 = (1 ‒ 1)^{2} = 0

(1 * 2) * 3 ≠ 1 * (2 * 3) where 1, 2, 3 ϵ *Q*

Thus, the operation * is not associative.

(*v*)

(*vi*)

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

**Question 10:** Show that none of the operations given above has identity.

**Solution 10:**

An element *e *∈ Q will be the identity element for the operation * if

*a *** **e = a = e *** **a*, *a* ∈ Q

But, there is no such element *e *∈ Q with respect to each of the six operations which can satisfy the above condition.

Hence, none of the six operations has identity

**Question 11:** Let A = **N** × **N** and ∗ be the binary operation on A defined by (*a*, *b*) ∗ (*c*, *d*) = (*a* + *c*, *b* + *d*)

Show that ∗ is commutative and associative. Find the identity element for ∗ on A, if any.

**Solution 11:**

A = N × N

(*a, b*) * (*c, d*) = (*a + c, b + d*)

Let (*a, b*), (*c, d*) ∈ A

Then,

*a, b, c, d* ∈ N

Now,

(*a, b*) * (*c, d*) = (*a + c, b + d*)

(*c, d*) * (*a, b*) = (*c + a, d + b*) = (*a + c, b + d*)

Hence, (*a, b*) * (*c, d*) = (*c, d*) * (*a, b*)

Therefore, the operation * is commutative.

Now,

Let (*a, b*), (*c, d*), (*e, f*) ∈A

Then, *a, b, c, d, e*, *f *∈ N

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*)]

[(*a*, *b*) * (*c*, *d*)]*(*e*, *f*) = (*a*, *b*) * [(*c*, *d*)*(*e*, *f*)]

Hence, the operation * is associative

An element *e* = (*e*_{1}, *e*_{2}) ϵ A will be an identity element for the operation * if

*a ** *e* = *a* = *e* * *a* ∀ *a* = (*a*_{1}, *a*_{2}) ϵ *A,*

i.e., (*a*_{1} + *e*_{1}, *a*_{2} + *e*_{2}) = (*a*_{1}, *a*_{2}) = (*e*_{1} + *a*_{1}, *e*_{2} + *a*_{2}) which is not true for any element in A.

Thus, the operation * does not have any identity element.

**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* ∀ *a* ∈ N.

(*ii*) If ∗ is a commutative binary operation on N, then *a* ∗ (*b* ∗ *c*) = (*c* ∗ *b*) ∗ *a*

**Solution 12:**

(*i*)

*a *** **b* = *a + b* *a, b *∈ N

For *b* = *a* = 3,

we have:

3 * 3 = 3 + 3 = 6 ≠ 3

Thus, first statement is false.

(*ii*)

R.H.S. = (*c *** **b*) * *a*

= (*b *** **c*) * *a *[* is commutative]

= *a* * (*b *** **c*) [* is commutative]

= L.H.S.

Hence, *a* * (*b *** **c*) = (*c *** **b*) * *a*

Thus, second statement is true.

**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 13:**

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

We have,

*a *** **b* = *a*^{3} + *b*^{3} [Addition is commutative in N]

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

= *b *** **a*

Hence, the operation * is commutative.

We can observed that,

1 * (2 * 3) = 1 * (2^{2} + 3^{2}) = 1 * (8 + 27) = 1 × 35 = 1^{3} + (35)^{3} = 1 + 42875 = 42876

1 * (2 * 3) = (1^{3} + 2^{3}) * 3 = 9 * 3 = 9^{3} + 3^{3} = 729 + 27 = 756

1 * 2) * 3 ≠ 1 * (2 * 3)

Therefore, the operation * is not associative.

Thus, the operation * is commutative, but not associative.

Hence, the correct answer is B.

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