Class 12th

(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 observed that:

$\begin{array}{l}\left(1*2\right)*3=\left(12+22\right)*3=\left(1+4\right)*3=5*3=52+32=25+9=34\\ 1*\left(2*3\right)=1*\left(22+32\right)=1*\left(4+9\right)=1*13=12+132=1+169=170\\ \therefore \left(1*2\right)*3\ne 1*\left(2*3\right),where,1,2,3\in Q\end{array}$

Thus, the operation* is not commutative.

(iii) On Q the operation* is defined as a*b=a+ab.

It can be observed that:

$\begin{array}{l}1.2=1+1*2=1+2=3\\ 2.1=2+2*1=2+2=4\\ \therefore 1.2\ne 2.1,where,1,2\in Q\end{array}$

Thus, the operation* is not commutative.

It can also be observed that:

$\begin{array}{l}\left(1.2\right).3=\left(1+1*2\right).3=3.3=3+3*3=3+9=12\\ 1.\left(2.3\right)=1.\left(2+2*3\right)=1+1*8=9\\ \therefore \left(1.2\right).3\ne 1.\left(2.3\right),where,1,2,3\in Q\end{array}$

Thus, the operation* is not associative.

(iv) On Q the operation* is defined as a*b=(a-b)²

For $a,b\in Q$ , we have:

$\begin{array}{l}a*b={\left(a-b\right)}^2\\ b*a={\left(b-a\right)}^2={\left[-\left(a-b\right)\right]}^2={\left(a-b\right)}^2\\ \therefore a*b=b*a\end{array}$

Thus, the operation* is commutative.

It can be observed that:

$\begin{array}{l}\left(1.2\right).3={\left(1.2\right)}^2.3={\left(-1\right)}^2.3=1.3={\left(1.3\right)}^2={\left(-2\right)}^2=4\\ 1.\left(2.3\right)=1.{\left(2.3\right)}^2=1.{\left(-1\right)}^2=1.1={\left(1.1\right)}^2=0\\ \therefore \left(1.2\right).3\ne 1.\left(2.3\right),where,1,2,3\in Q\end{array}$

Thus, the operation* is not associative.

(v) On Q the operation* is defined as a.b=ab/4

For $a,b\in Q$ , we have:

$\begin{array}{l}a.b=\frac{ab}{4}=b\frac{a}{4}=b.a\\ \therefore a*b=b*a\end{array}$

Thus, the operation* is commutative.

For $a,b,c\in Q$ , we have:

$\begin{array}{l}\left(a.b\right).c=\frac{ab}{4}.c=\frac{\frac{ab}{4}.c}{4}=\frac{abc}{16}\\ a.\left(b.c\right)=a.b\frac{c}{4}=\frac{a.\frac{bc}{4}}{4}=\frac{abc}{16}\\ =\therefore \left(a*b\right)*c\ne a*\left(b*c\right)\end{array}$

Thus, the operation* is associative.

(vi) On Q the operation* is defined as a*b=ab²

It can be observed that:

$\begin{array}{l}\frac{1}{2}.\frac{1}{3}=\frac{1}{2}.{\left(\frac{1}{3}\right)}^2=\frac{1}{2}.\frac{1}{9}=\frac{1}{18}\\ \frac{1}{3}.\frac{1}{2}=\frac{1}{3}.{\left(\frac{1}{2}\right)}^2=\frac{1}{3}.\frac{1}{4}=\frac{1}{12}\\ \therefore \frac{1}{2}.\frac{1}{3}\ne \frac{1}{3}.\frac{1}{2},where,\frac{1}{2},\frac{1}{3}\in Q\end{array}$

Thus, the operation* is not commutative.

It can be observed that:

$\begin{array}{l}\left(\frac{1}{2}.\frac{1}{3}\right).\frac{1}{4}=\left[\frac{1}{2}.{\left(\frac{1}{3}\right)}^2\right].\frac{1}{4}=\frac{1}{18}.\frac{1}{4}=\frac{1}{18}.{\left(\frac{1}{4}\right)}^2=\frac{1}{18*16}\\ \frac{1}{2}.\left(\frac{1}{3}.\frac{1}{4}\right)=\frac{1}{2}.\left[\frac{1}{3}.{\left(\frac{1}{4}\right)}^2\right]=\frac{1}{2}.\frac{1}{48}=\frac{1}{2}.{\left(\frac{1}{48}\right)}^2=\frac{1}{2*{\left(48\right)}^2}\\ \therefore \left(\frac{1}{2}.\frac{1}{3}\right).\frac{1}{4}\ne \frac{1}{2}.\left(\frac{1}{3}.\frac{1}{4}\right),where,\frac{1}{2},\frac{1}{3},1.4\in Q\end{array}$

Thus, the operation* is not associative.

Hence, the operations defined (ii), (iv), (v) are commutative and the operation defined in (v) is associative.

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.

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

It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 − 2= −1 ∉ Z⁺.

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

It is seen that for each a, b ∈ Z⁺, there is a unique element ab in Z⁺.

This means that * carries each pair (a, b) to a unique element a * b = ab in Z⁺.

Therefore, * is a binary operation.

(iii) On R, * is defined by a * b = ab².

It is seen that for each a, b ∈ R, there is a unique element ab² in R. 

This means that * carries each pair (a, b) to a unique element a * b = ab² in R.

Therefore, * is a binary operation.

(iv) On Z⁺, * is defined by a * b = |a − b|.

It is seen that for each a, b ∈ Z+, there is a unique element |a − b| in Z⁺. 

This means that * carries each pair (a, b) to a unique element a * b = |a − b| in Z⁺.

Therefore, * is a binary operation.

(v) On Z⁺, * is defined by a * b = a.

* carries each pair (a, b) to a unique element a * b = a in Z⁺.

Therefore, * is a binary operation.