Name the property of multiplication of rational numbers illustrated by the following statements:

(i) $$\dfrac{-5}{16}× \dfrac{8}{15} = \dfrac{8}{15} × \dfrac{-5}{16}$$

(ii) $$\dfrac{-17}{5}×9 = 9 × \dfrac{-17}{5}$$

(iii) $$\dfrac{7}{4} × (\dfrac{-8}{3}+ \dfrac{-13}{12}) = \dfrac{7}{4} × \dfrac{-8}{3} + \dfrac{7}{4} × \dfrac{-13}{12}$$

(iv) $$\dfrac{-5}{9} × (\dfrac{4}{15} × \dfrac{-9}{8}) = ( \dfrac{-5}{9} × \dfrac{4}{15}) × \dfrac{-9}{8}$$

(v) $$\dfrac{ 13}{-17}× 1 = \dfrac{ 13}{-17} = 1 ×\dfrac{ 13}{-17}$$

(vi) $$\dfrac { -11}{16 }× \dfrac{16}{-11 }= 1$$

(vii) $$\dfrac{ 2}{13} × 0 = 0 = 0 ×\dfrac{ 2}{13}$$

(viii) $$\dfrac{-3}{2} × \dfrac{5}{4} + \dfrac{-3}{2} × \dfrac{-7}{6} = \dfrac{-3}{2} ×\dfrac{5}{4} + \dfrac{-7}{6}$$

Asked by Aaryan | 1 year ago |  107

##### Solution :-

(i) $$\dfrac{-5}{16}× \dfrac{8}{15} = \dfrac{8}{15} × \dfrac{-5}{16}$$

According to commutative law, $$\dfrac{ a}{b} × \dfrac{ c}{d} = \dfrac{ c}{d} × \dfrac{ a}{b}$$

The above rational number satisfies commutative property.

(ii) $$\dfrac{-17}{5}×9 = 9 × \dfrac{-17}{5}$$

According to commutative law, $$\dfrac{ a}{b} × \dfrac{ c}{d} = \dfrac{ c}{d} × \dfrac{ a}{b}$$

The above rational number satisfies commutative property.

(iii) $$\dfrac{7}{4} × (\dfrac{-8}{3}+ \dfrac{-13}{12}) = \dfrac{7}{4} × \dfrac{-8}{3} + \dfrac{7}{4} × \dfrac{-13}{12}$$

According to given rational number,

$$\dfrac{ a}{b}× ( \dfrac{ c}{d} + \dfrac{ e}{f}) = ( \dfrac{ a}{b} × \dfrac{ c}{d}) + ( \dfrac{ a}{b}× \dfrac{e}{f})$$

Distributivity of multiplication over addition satisfies.

(iv) $$\dfrac{-5}{9} × (\dfrac{4}{15} × \dfrac{-9}{8}) = ( \dfrac{-5}{9} × \dfrac{4}{15}) × \dfrac{-9}{8}$$

According to associative law,

$$\dfrac{ a}{b}× ( \dfrac{ c}{d} + \dfrac{ e}{f}) = ( \dfrac{ a}{b} × \dfrac{ c}{d}) + ( \dfrac{ a}{b}× \dfrac{e}{f})$$

The above rational number satisfies associativity of multiplication.

(v) $$\dfrac{ 13}{-17}× 1 = \dfrac{ 13}{-17} = 1 ×\dfrac{ 13}{-17}$$

Existence of identity for multiplication satisfies for the given rational number.

(vi) $$\dfrac { -11}{16 }× \dfrac{16}{-11 }= 1$$

Existence of multiplication inverse satisfies for the given rational number.

(vii) $$\dfrac{ 2}{13} × 0 = 0 = 0 ×\dfrac{ 2}{13}$$

By using $$\dfrac{ a}{b}× 0 = 0 × \dfrac{ a}{b}$$

Multiplication of zero satisfies for the given rational number.

(viii) $$\dfrac{-3}{2} × \dfrac{5}{4} + \dfrac{-3}{2} × \dfrac{-7}{6} = \dfrac{-3}{2} ×\dfrac{5}{4} + \dfrac{-7}{6}$$

According to distributive law,

$$\dfrac{ a}{b}× ( \dfrac{ c}{d} + \dfrac{ e}{f}) = ( \dfrac{ a}{b} × \dfrac{ c}{d}) + ( \dfrac{ a}{b}× \dfrac{e}{f})$$

The above rational number satisfies distributive law.

Answered by Sakshi | 1 year ago

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