Ncert Solutions Maths class 12th

 (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.