Maths

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, we have:

(a * b) * c = (L.C.M of a and b) * c = LCM of a, b, and c

a * (b * c) = a * (LCM of b and c) = L.C.M of a, b, and c

∴ (a * b) * c = a * (b * c) 

Thus, the operation * is associative.

(iv) It is known that:

L.C.M. of a and 1 = a = L.C.M. 1 and a &mnForE; a ∈ N 

⇒ a * 1 = a = 1 * a &mnForE; a ∈ N

Thus, 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.

Here, e = 1

This means that: 

L.C.M of a and b = 1 = L.C.M of b and a

This case is possible only when a and b are equal to 1.

Thus, 1 is the only invertible element of N with respect to the operation *.

 (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 − 3) = 1 * −1 = 1 − (−1) = 2

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Z

Hence, the operation * is not associative.

(ii) On Q, * is defined by a * b = ab + 1.

It is known that:

ab = ba & mn For E; a, b ∈ Q

⇒ ab + 1 = ba + 1 & mn For E; a, b ∈ Q

⇒ a * b = a * b &mn For E; a, b ∈ Q

Therefore, the operation * is commutative.

It can be observed that:

(1 * 2) * 3 = (1 * 2 + 1) * 3 = 3 * 3 = 3 * 3 + 1 = 10

1 * (2 * 3) = 1 * (2 * 3 + 1) = 1 * 7 = 1 * 7 + 1 = 8

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Q

Therefore, the operation * is not associative.

(iii) On Q, * is defined by a * b = ab/2

It is known that:

ab = ba &mnForE; a, b ∈ Q

⇒ab/2 = ba/2 &mnForE; a, b ∈ Q

⇒ a * b = b * a &mnForE; a, b ∈ Q

Therefore, the operation * is commutative.

For all a, b, c ∈ Q, we have:

$\begin{array}{l}\left(a\ast b\right)\ast c=\left(\frac{ab}{2}\right)\ast c=\frac{\left(\frac{ab}{2}\right)c}{2}=\frac{abc}{4}\\ a\ast \left(b\ast c\right)=a\ast \left(\frac{bc}{2}\right)=\frac{a\left(\frac{bc}{2}\right)}{2}=\frac{abc}{4}\\ \therefore \left(a\ast b\right)\ast c=a\ast \left(b\ast c\right)\end{array}$

Therefore, the operation * is associative.

(iv) On Z⁺, * is defined by a * b = 2^ab.

It is known that:

ab = ba &mnForE; a, b ∈ Z⁺

⇒ 2^ab = 2^ba &mnForE; a, b ∈ Z⁺

⇒ a * b = b * a &mnForE; a, b ∈ Z⁺

Therefore, the operation * is commutative.

It can be observed that:

$\begin{array}{l}\left(1\ast 2\right)\ast 3=2^{\left(1*2\right)}\ast 3=4\ast 3=2^{4*3}=2^{12}\\ 1\ast \left(2\ast 3\right)=1\ast 2^{2*3}=1\ast 2^6=1\ast 64=2^{64}\end{array}$

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Z+

Therefore, the operation * is not associative.

(v) On Z⁺, * is defined by a * b = a^b.

It can be observed that:

 1 x2 = 1²=1 and 2x 1 =2¹ = 2

∴ 1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ Z⁺

Therefore, the operation * is not commutative.

It can also be observed that:

$\begin{array}{l}\left(2\ast 3\right)\ast 4=2^3\ast 4=8\ast 4=8^4={\left(2^3\right)}^4=2^{12}\\ 2\ast \left(3\ast 4\right)=2\ast 3^4=2\ast 81=2^{81}\end{array}$

∴(2 * 3) * 4 ≠ 2 * (3 * 4) ; where 2, 3, 4 ∈ Z+

Therefore, the operation * is not associative.

(vi) On R, * − {−1} is defined by a.b = a/b+1

It can be observed that 1 x 2 = 1/2+1 = 1/2 and 2 x 1 = 2/1+1 = 2/2

∴1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ R − {−1}

Therefore, the operation * is not commutative.

It can also be observed that:

$\begin{array}{l}\left(1\ast 2\right)\ast 3=\frac{1}{3}\ast 3=\frac{\frac{1}{3}}{3+1}=\frac{1}{12}\\ 1\ast \left(2\ast 3\right)=1\ast \frac{2}{3+1}=1\ast \frac{2}{4}=1\ast \frac{1}{2}=\frac{1}{\frac{1}{2}+1}=\frac{1}{\frac{3}{2}}=\frac{2}{3}\end{array}$

∴ (1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ R − {−1}

Therefore, the operation * is not associative.