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Relations and Functions Ncert Solutions Maths class 12th Maths Class 12th Relations and Functions

**51. Let * be a binary operation on the set Q of rational numbers as follows:

(i)** a * b = a − b

(ii) a * b = a² + b²

(iii) a * b = a + ab

(iv) a * b = (a − b)²

(v) a * b = ab/4 (vi) a * b = ab²

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

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Vishal Baghel | Contributor-Level 10

4 months ago

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

It can be observed that:

$\begin{array}{l} \frac{1}{2} . \frac{1}{3} = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6} " a n d " \frac{1}{3} . \frac{1}{2} = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = \frac{- 1}{6} \\ ∴ \frac{1}{2} . \frac{1}{3} \neq \frac{1}{3} . \frac{1}{2} w h e r e, \frac{1}{2}, \frac{1}{3} \in Q \end{array}$

Thus, the operation* is not commutative.

It can also be observed that:

$\begin{array}{l} (\frac{1}{2} . \frac{1}{3}) . \frac{1}{4} = (\frac{1}{2} - \frac{1}{3}) . \frac{1}{4} = \frac{1}{6} . \frac{1}{4} = \frac{1}{6} - \frac{1}{4} = \frac{2 - 3}{12} = \frac{- 1}{12} \\ \frac{1}{2} . (\frac{1}{3} . \frac{1}{4}) = \frac{1}{2} . (\frac{1}{3} - \frac{1}{4}) = \frac{1}{2} . \frac{1}{12} = \frac{1}{2} - \frac{1}{12} = \frac{6 - 1}{12} = \frac{5}{12} \\ ∴ (\frac{1}{2} . \frac{1}{3}) . \frac{1}{4} \neq \frac{1}{2} . (\frac{1}{3} . \frac{1}{4}) w h e r e, \frac{1}{2}, \frac{1}{3}, \frac{1}{4} \in Q \end{array}$

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:

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

Thus, the operation* is commutative.

It can be observed that:

$\begin{array}{l} (1 * 2) * 3 = (12 + 22) * 3 = (1 + 4) * 3 = 5 * 3 = 52 + 32 = 25 + 9 = 34 \\ 1 * (2 * 3) = 1 * (22 + 32) = 1 * (4 + 9) = 1 * 13 = 12 + 132 = 1 + 169 = 170 \\ ∴ (1 * 2) * 3 \neq 1 * (2 * 3), w h e r e, 1, 2, 3 \in Q \end{array}$

Thus, the operation* is not commutative.

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

It can be observed that:

$\begin{array}{l} 1.2 = 1 + 1 \times 2 = 1 + 2 = 3 \\ 2.1 = 2 + 2 \times 1 = 2 + 2 = 4 \\ ∴ 1.2 \neq 2.1, w h e r e, 1, 2 \in Q \end{array}$

Thus, the operation* is not commutative.

It can also be observed that:

$\begin{array}{l} (1.2) . 3 = (1 + 1 \times 2) . 3 = 3.3 = 3 + 3 \times 3 = 3 + 9 = 12 \\ 1 . (2.3) = 1 . (2 + 2 \times 3) = 1 + 1 \times 8 = 9 \\ ∴ (1.2) . 3 \neq 1 . (2.3), w h e r e, 1, 2, 3 \in Q \end{array}$

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:

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

Thus, the operation* is commutative

...more

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

It can be observed that:

Thus, the operation* is not commutative.

It can also be observed that:

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:

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

Thus, the operation* is commutative.

It can be observed that:

Thus, the operation* is not commutative.

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

It can be observed that:

$\begin{array}{l} 1.2 = 1 + 1 \times 2 = 1 + 2 = 3 \\ 2.1 = 2 + 2 \times 1 = 2 + 2 = 4 \\ ∴ 1.2 \neq 2.1, w h e r e, 1, 2 \in Q \end{array}$

Thus, the operation* is not commutative.

It can also be observed that:

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:

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

Thus, the operation* is commutative.

It can be observed that:

$\begin{array}{l} (1.2) . 3 = {(1.2)}^{2} . 3 = {(- 1)}^{2} . 3 = 1.3 = {(1.3)}^{2} = {(- 2)}^{2} = 4 \\ 1 . (2.3) = 1 . {(2.3)}^{2} = 1 . {(- 1)}^{2} = 1.1 = {(1.1)}^{2} = 0 \\ ∴ (1.2) . 3 \neq 1 . (2.3), w h e r e, 1, 2, 3 \in Q \end{array}$

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:

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

Thus, the operation* is commutative.

For $a, b, c \in Q$ , we have:

$\begin{array}{l} (a . b) . c = \frac{a b}{4} . c = \frac{\frac{a b}{4} . c}{4} = \frac{a b c}{16} \\ a . (b . c) = a . b \frac{c}{4} = \frac{a . \frac{b c}{4}}{4} = \frac{a b c}{16} \\ = ∴ (a * b) * c \neq a * (b * c) \end{array}$

Thus, the operation* is associative.

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

It can be observed that:

$\begin{array}{l} \frac{1}{2} . \frac{1}{3} = \frac{1}{2} . {(\frac{1}{3})}^{2} = \frac{1}{2} . \frac{1}{9} = \frac{1}{18} \\ \frac{1}{3} . \frac{1}{2} = \frac{1}{3} . {(\frac{1}{2})}^{2} = \frac{1}{3} . \frac{1}{4} = \frac{1}{12} \\ ∴ \frac{1}{2} . \frac{1}{3} \neq \frac{1}{3} . \frac{1}{2}, w h e r e, \frac{1}{2}, \frac{1}{3} \in Q \end{array}$

Thus, the operation* is not commutative.

It can be observed that:

$\begin{array}{l} (\frac{1}{2} . \frac{1}{3}) . \frac{1}{4} = [\frac{1}{2} . {(\frac{1}{3})}^{2}] . \frac{1}{4} = \frac{1}{18} . \frac{1}{4} = \frac{1}{18} . {(\frac{1}{4})}^{2} = \frac{1}{18 \times 16} \\ \frac{1}{2} . (\frac{1}{3} . \frac{1}{4}) = \frac{1}{2} . [\frac{1}{3} . {(\frac{1}{4})}^{2}] = \frac{1}{2} . \frac{1}{48} = \frac{1}{2} . {(\frac{1}{48})}^{2} = \frac{1}{2 \times {(48)}^{2}} \\ ∴ (\frac{1}{2} . \frac{1}{3}) . \frac{1}{4} \neq \frac{1}{2} . (\frac{1}{3} . \frac{1}{4}), w h e r e, \frac{1}{2}, \frac{1}{3}, 1.4 \in Q \end{array}$

Thus, the operation* is not associative.

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

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**51. Let * be a binary operation on the set Q of rational numbers as follows:

(i)** a * b = a − b

(ii) a * b = a² + b²

(iii) a * b = a + ab

(iv) a * b = (a − b)²

(v) a * b = ab/4 (vi) a * b = ab²

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

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51. Let * be a binary operation on the set Q of rational numbers as follows: (i) a * b = a − b (ii) a * b = a2 + b2 (iii) a * b = a + ab (iv) a * b = (a − b)2 (v) a * b = ab/4 (vi) a * b = ab2 Find which of the binary operations are commutative and which are associative.

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**51. Let * be a binary operation on the set Q of rational numbers as follows:

(i)** a * b = a − b

(ii) a * b = a² + b²

(iii) a * b = a + ab

(iv) a * b = (a − b)²

(v) a * b = ab/4 (vi) a * b = ab²

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