Ncert Solutions Maths class 12th

A = N × N

* is a binary operation on A and is defined by:

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

Let (a, b), (c, d) ∈ A

Then, a, b, c, d ∈ N

We have:

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

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

[Addition is commutative in the set of natural numbers]

∴(a, b) * (c, d) = (

Let the identity be I.

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

$a*e=a=e*a,\forall a\in Q.$

We are given

$\begin{array}{l}a*b=ab4\\ \Rightarrow a*e=a\Rightarrow ae4=a\Rightarrow e=4\end{array}$

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

(i) On Q the operation* is defined as a*b=a-b.

It can be observed that:

$\begin{array}{l}\frac{1}{2}.\frac{1}{3}=\frac{1}{2}-\frac{1}{3}=\frac{3-2}{6}=\frac{1}{6}"and"\frac{1}{3}.\frac{1}{2}=\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=\frac{-1}{6}\\ \therefore \frac{1}{2}.\frac{1}{3}\ne \frac{1}{3}.\frac{1}{2}where,\frac{1}{2},\frac{1}{3}\in Q\end{array}$

Thus, the operation* is not commutative.

It can also be observed that:

$\begin{array}{l}\left(\frac{1}{2}.\frac{1}{3}\right).\frac{1}{4}=\left(\frac{1}{2}-\frac{1}{3}\right).\frac{1}{4}=\frac{1}{6}.\frac{1}{4}=\frac{1}{6}-\frac{1}{4}=\frac{2-3}{12}=\frac{-1}{12}\\ \frac{1}{2}.\left(\frac{1}{3}.\frac{1}{4}\right)=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{4}\right)=\frac{1}{2}.\frac{1}{12}=\frac{1}{2}-\frac{1}{12}=\frac{6-1}{12}=\frac{5}{12}\\ \therefore \left(\frac{1}{2}.\frac{1}{3}\right).\frac{1}{4}\ne \frac{1}{2}.\left(\frac{1}{3}.\frac{1}{4}\right)where,\frac{1}{2},\frac{1}{3},\frac{1}{4}\in Q\end{array}$

Thus, the operation* is not associative.

(ii) On Q the operation* is defined as a*b=a²+b.

For $a,b\in Q$ , we have:

$\begin{array}{l}a.b=a^2+b^2=b^2+a^2=b.a\\ \therefore a*b=b*a\end{array}$

Thus, the operation* is commutative.

It can be ob

The binary operation * on N is defined as:

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

It is known that: 

H.C.F. of a and b = H.C.F. of b and a & mn For E; a, b ∈ N. 

∴a * b = b * a

Thus, the operation * is commutative.

For a, b, c ∈ N, we have:

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

a * (b *

The operation * on the set A = {1, 2, 3, 4, 5} is defined as 

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

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

It can be observed from the obtained table that:

3 * 2 = 2 * 3 = 6 ∉ A, 5 * 2 = 2 * 5 = 10 ∉ A, 3 * 4 = 4 * 3 = 12 ∉ A

3 * 5 = 5 * 3 = 1

The binary operation * on N is defined as a * b = L.C.M. of a and b.

(i) 5 * 7 = L.C.M. of 5 and 7 = 35

20 * 16 = L.C.M of 20 and 16 = 80

(ii) It is known that: 

L.C.M of a and b = L.C.M of b and a & mnForE; a, b ∈ N. 

∴a * b = b * a

Thus, the operation * is commutative.

(iii) For a, b, c ∈ N,

The binary operation *′ on the set {1, 2, 3 4, 5} is defined as a *′ b = H.C.F of a and b.

The operation table for the operation *′ can be given as:

We observe that the operation tables for the operations * and *′ are the same.

Thus, the operation *′ is same as the operation*.

(i) (2 * 3) * 4 = 1 * 4 = 1

2 * (3 * 4) = 2 * 1 = 1

(ii) For every a, b ∈ {1, 2, 3, 4, 5}, we have a * b = b * a. Therefore, the operation * is commutative.

(iii) (2 * 3) = 1 and (4 * 5) = 1

∴ (2 * 3) * (4 * 5) = 1 * 1 = 1

The binary operation? on the set {1, 2, 3, 4, 5} is defined as a? b = min {a, b} &mn For E; a, b? {1, 2, 3, 4, 5}.

Thus, the operation table for the given operation? can be given as:

 (i) On Z, * is defined by a * b = a − b.

It can be observed that 1 * 2 = 1 − 2 = 1 and 2 * 1 = 2 − 1 = 1.

∴1 * 2 ≠ 2 * 1; where 1, 2 ∈ Z

Hence, the operation * is not commutative.

Also we have:

(1 * 2) * 3 = (1 − 2) * 3 = −1 * 3 = −1 − 3 = −4

1 * (2 * 3) = 1 * (2 −