Rational Numbers — Exercise 1.3

Multiply and express the result in lowest form:

$\dfrac{6}{-7} \times \dfrac{14}{30}$

Answer

First express the rational number with a positive denominator, then multiply the numerators together and the denominators together.

$\Rightarrow \dfrac{6}{-7} \times \dfrac{14}{30} \\[1em] \Rightarrow \dfrac{-6}{7} \times \dfrac{14}{30} \\[1em] \Rightarrow \dfrac{(-6) \times 14}{7 \times 30} \\[1em] \Rightarrow \dfrac{-84}{210} \\[1em] \Rightarrow \dfrac{-2}{5}$

Hence, $\mathbf{\dfrac{6}{-7} \times \dfrac{14}{30} = \dfrac{-2}{5}}$

Multiply and express the result in lowest form:

$6\dfrac{2}{3} \times 1\dfrac{2}{7}$

Answer

First convert the mixed fractions into improper fractions, then multiply the numerators together and the denominators together.

$\Rightarrow 6\dfrac{2}{3} \times 1\dfrac{2}{7} \\[1em] \Rightarrow \dfrac{20}{3} \times \dfrac{9}{7} \\[1em] \Rightarrow \dfrac{20 \times 9}{3 \times 7} \\[1em] \Rightarrow \dfrac{180}{21} \\[1em] \Rightarrow \dfrac{60}{7} \\[1em] \Rightarrow 8\dfrac{4}{7}$

Hence, $\mathbf{6\dfrac{2}{3} \times 1\dfrac{2}{7} = 8\dfrac{4}{7}}$

Multiply and express the result in lowest form:

$\dfrac{25}{-9} \times \dfrac{-3}{10}$

Answer

First express the rational number with a positive denominator, then multiply the numerators together and the denominators together.

$\Rightarrow \dfrac{25}{-9} \times \dfrac{-3}{10} \\[1em] \Rightarrow \dfrac{-25}{9} \times \dfrac{-3}{10} \\[1em] \Rightarrow \dfrac{(-25) \times (-3)}{9 \times 10} \\[1em] \Rightarrow \dfrac{75}{90} \\[1em] \Rightarrow \dfrac{5}{6}$

Hence, $\mathbf{\dfrac{25}{-9} \times \dfrac{-3}{10} = \dfrac{5}{6}}$

Verify commutative property of multiplication for the following pair of rational numbers:

$\dfrac{4}{5} \text{ and } \dfrac{-7}{8}$

Answer

To verify the commutative property of multiplication, we have to show that $\dfrac{4}{5} \times \dfrac{-7}{8} = \dfrac{-7}{8} \times \dfrac{4}{5}$

Solving L.H.S.,

$\Rightarrow \dfrac{4}{5} \times \dfrac{-7}{8} \\[1em] \Rightarrow \dfrac{4 \times (-7)}{5 \times 8} \\[1em] \Rightarrow \dfrac{-28}{40} \\[1em] \Rightarrow \dfrac{-7}{10}$

Solving R.H.S.,

$\Rightarrow \dfrac{-7}{8} \times \dfrac{4}{5} \\[1em] \Rightarrow \dfrac{(-7) \times 4}{8 \times 5} \\[1em] \Rightarrow \dfrac{-28}{40} \\[1em] \Rightarrow \dfrac{-7}{10}$

As L.H.S. = R.H.S.

$\Rightarrow \dfrac{4}{5} \times \dfrac{-7}{8} = \dfrac{-7}{8} \times \dfrac{4}{5}$

Hence, the commutative property of multiplication is verified.

Verify commutative property of multiplication for the following pair of rational numbers:

$13\dfrac{1}{3} \text{ and } 1\dfrac{1}{8}$

Answer

First convert the mixed fractions into improper fractions.

$13\dfrac{1}{3} = \dfrac{40}{3} \text{ and } 1\dfrac{1}{8} = \dfrac{9}{8}$

To verify the commutative property of multiplication, we have to show that $\dfrac{40}{3} \times \dfrac{9}{8} = \dfrac{9}{8} \times \dfrac{40}{3}$

Solving L.H.S.,

$\Rightarrow \dfrac{40}{3} \times \dfrac{9}{8} \\[1em] \Rightarrow \dfrac{40 \times 9}{3 \times 8} \\[1em] \Rightarrow \dfrac{360}{24} \\[1em] \Rightarrow 15$

Solving R.H.S.,

$\Rightarrow \dfrac{9}{8} \times \dfrac{40}{3} \\[1em] \Rightarrow \dfrac{9 \times 40}{8 \times 3} \\[1em] \Rightarrow \dfrac{360}{24} \\[1em] \Rightarrow 15$

As L.H.S. = R.H.S.

$\Rightarrow 13\dfrac{1}{3} \times 1\dfrac{1}{8} = 1\dfrac{1}{8} \times 13\dfrac{1}{3}$

Hence, the commutative property of multiplication is verified.

Verify commutative property of multiplication for the following pair of rational numbers:

$\dfrac{-7}{-20} \text{ and } \dfrac{5}{-14}$

Answer

First express each rational number with a positive denominator.

$\dfrac{-7}{-20} = \dfrac{7}{20} \text{ and } \dfrac{5}{-14} = \dfrac{-5}{14}$

To verify the commutative property of multiplication, we have to show that $\dfrac{7}{20} \times \dfrac{-5}{14} = \dfrac{-5}{14} \times \dfrac{7}{20}$

Solving L.H.S.,

$\Rightarrow \dfrac{7}{20} \times \dfrac{-5}{14} \\[1em] \Rightarrow \dfrac{7 \times (-5)}{20 \times 14} \\[1em] \Rightarrow \dfrac{-35}{280} \\[1em] \Rightarrow \dfrac{-1}{8}$

Solving R.H.S.,

$\Rightarrow \dfrac{-5}{14} \times \dfrac{7}{20} \\[1em] \Rightarrow \dfrac{(-5) \times 7}{14 \times 20} \\[1em] \Rightarrow \dfrac{-35}{280} \\[1em] \Rightarrow \dfrac{-1}{8}$

As L.H.S. = R.H.S.

$\Rightarrow \dfrac{-7}{-20} \times \dfrac{5}{-14} = \dfrac{5}{-14} \times \dfrac{-7}{-20}$

Hence, the commutative property of multiplication is verified.

Verify the following and name the property also:

$\dfrac{3}{5} \times \left(\dfrac{-4}{7} \times \dfrac{-8}{9}\right) = \left(\dfrac{3}{5} \times \dfrac{-4}{7}\right) \times \dfrac{-8}{9}$

Answer

Solve the bracket first on each side and then multiply.

Solving L.H.S.,

$\Rightarrow \dfrac{3}{5} \times \left(\dfrac{-4}{7} \times \dfrac{-8}{9}\right) \\[1em] \Rightarrow \dfrac{3}{5} \times \dfrac{(-4) \times (-8)}{7 \times 9} \\[1em] \Rightarrow \dfrac{3}{5} \times \dfrac{32}{63} \\[1em] \Rightarrow \dfrac{3 \times 32}{5 \times 63} \\[1em] \Rightarrow \dfrac{96}{315} \\[1em] \Rightarrow \dfrac{32}{105}$

Solving R.H.S.,

$\Rightarrow \left(\dfrac{3}{5} \times \dfrac{-4}{7}\right) \times \dfrac{-8}{9} \\[1em] \Rightarrow \dfrac{3 \times (-4)}{5 \times 7} \times \dfrac{-8}{9} \\[1em] \Rightarrow \dfrac{-12}{35} \times \dfrac{-8}{9} \\[1em] \Rightarrow \dfrac{(-12) \times (-8)}{35 \times 9} \\[1em] \Rightarrow \dfrac{96}{315} \\[1em] \Rightarrow \dfrac{32}{105}$

As L.H.S. = R.H.S.

$\Rightarrow \dfrac{3}{5} \times \left(\dfrac{-4}{7} \times \dfrac{-8}{9}\right) = \left(\dfrac{3}{5} \times \dfrac{-4}{7}\right) \times \dfrac{-8}{9}$

Hence, verified and the property used is associative property of multiplication of rational numbers.

Verify the following and name the property also:

$\dfrac{5}{9} \times \left(\dfrac{-3}{2} + \dfrac{7}{5}\right) = \dfrac{5}{9} \times \dfrac{-3}{2} + \dfrac{5}{9} \times \dfrac{7}{5}$

Answer

Solve the bracket first on the left side and then multiply, and on the right side multiply first and then add.

Solving L.H.S.,

$\Rightarrow \dfrac{5}{9} \times \left(\dfrac{-3}{2} + \dfrac{7}{5}\right) \\[1em] [\text{ LCM of } 2 \text{ and } 5 = 10] \\[1em] \Rightarrow \dfrac{5}{9} \times \dfrac{(-3) \times 5 + 7 \times 2}{10} \\[1em] \Rightarrow \dfrac{5}{9} \times \dfrac{-15 + 14}{10} \\[1em] \Rightarrow \dfrac{5}{9} \times \dfrac{-1}{10} \\[1em] \Rightarrow \dfrac{-5}{90} \\[1em] \Rightarrow \dfrac{-1}{18}$

Solving R.H.S.,

$\Rightarrow \dfrac{5}{9} \times \dfrac{-3}{2} + \dfrac{5}{9} \times \dfrac{7}{5} \\[1em] \Rightarrow \dfrac{5 \times (-3)}{9 \times 2} + \dfrac{5 \times 7}{9 \times 5} \\[1em] \Rightarrow \dfrac{-15}{18} + \dfrac{35}{45} \\[1em] \Rightarrow \dfrac{-5}{6} + \dfrac{7}{9} \\[1em] [\text{ LCM of } 6 \text{ and } 9 = 18] \\[1em] \Rightarrow \dfrac{(-5) \times 3 + 7 \times 2}{18} \\[1em] \Rightarrow \dfrac{-15 + 14}{18} \\[1em] \Rightarrow \dfrac{-1}{18}$

As L.H.S. = R.H.S.

$\Rightarrow \dfrac{5}{9} \times \left(\dfrac{-3}{2} + \dfrac{7}{5}\right) = \dfrac{5}{9} \times \dfrac{-3}{2} + \dfrac{5}{9} \times \dfrac{7}{5}$

Hence, verified and the property used is distributive property of multiplication over addition of rational numbers.

Find the multiplicative inverse of the following:

$12$

Answer

To find the multiplicative inverse (or reciprocal) of a non-zero rational number, interchange its numerator and denominator.

$\Rightarrow 12 = \dfrac{12}{1} \\[1em] \Rightarrow \text{Multiplicative inverse} = \dfrac{1}{12}$

Hence, the multiplicative inverse of 12 is $\mathbf{\dfrac{1}{12}}$

Find the multiplicative inverse of the following:

$\dfrac{2}{3}$

Answer

To find the multiplicative inverse (or reciprocal) of a non-zero rational number, interchange its numerator and denominator.

Interchanging the numerator and the denominator of $\dfrac{2}{3}$, we get $\dfrac{3}{2}$

Hence, the multiplicative inverse of $\mathbf{\dfrac{2}{3} \text{ is } \dfrac{3}{2}}$

Find the multiplicative inverse of the following:

$\dfrac{-4}{7}$

Answer

To find the multiplicative inverse (or reciprocal) of a non-zero rational number, interchange its numerator and denominator.

$\Rightarrow \text{Multiplicative inverse of } \dfrac{-4}{7} = \dfrac{7}{-4} \\[1em] \Rightarrow \dfrac{-7}{4}$

Hence, the multiplicative inverse of $\mathbf{\dfrac{-4}{7} \text{ is } \dfrac{-7}{4}}$

Find the multiplicative inverse of the following:

$\dfrac{-3}{8} \times \left(\dfrac{-7}{13}\right)$

Answer

First find the product, then interchange its numerator and denominator.

$\Rightarrow \dfrac{-3}{8} \times \left(\dfrac{-7}{13}\right) \\[1em] \Rightarrow \dfrac{(-3) \times (-7)}{8 \times 13} \\[1em] \Rightarrow \dfrac{21}{104}$

Interchanging the numerator and the denominator of $\dfrac{21}{104}$, we get $\dfrac{104}{21} = 4\dfrac{20}{21}$

Hence, the multiplicative inverse of $\mathbf{\dfrac{-3}{8} \times \left(\dfrac{-7}{13}\right) \text{ is } 4\dfrac{20}{21}}$

Using the appropriate properties of operations of rational numbers, evaluate the following:

$\dfrac{2}{5} \times \dfrac{-3}{7} - \dfrac{1}{14} - \dfrac{3}{7} \times \dfrac{3}{5}$

Answer

Rearrange the terms using commutativity of multiplication so that the common factor $\dfrac{3}{7}$ can be taken out using distributivity.

$\Rightarrow \dfrac{2}{5} \times \dfrac{-3}{7} - \dfrac{1}{14} - \dfrac{3}{7} \times \dfrac{3}{5} \\[1em] \Rightarrow \dfrac{-3}{7} \times \dfrac{2}{5} - \dfrac{3}{7} \times \dfrac{3}{5} - \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{3}{7} \times \left(\dfrac{-2}{5} - \dfrac{3}{5}\right) - \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{3}{7} \times \dfrac{-5}{5} - \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{3}{7} \times (-1) - \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{-3}{7} - \dfrac{1}{14} \\[1em] [\text{ LCM of } 7 \text{ and } 14 = 14] \\[1em] \Rightarrow \dfrac{(-3) \times 2 - 1}{14} \\[1em] \Rightarrow \dfrac{-6 - 1}{14} \\[1em] \Rightarrow \dfrac{-7}{14} \\[1em] \Rightarrow \dfrac{-1}{2}$

Hence, $\mathbf{\dfrac{2}{5} \times \dfrac{-3}{7} - \dfrac{1}{14} - \dfrac{3}{7} \times \dfrac{3}{5} = \dfrac{-1}{2}}$

Using the appropriate properties of operations of rational numbers, evaluate the following:

$\dfrac{8}{9} \times \dfrac{4}{5} + \dfrac{5}{6} - \dfrac{9}{5} \times \dfrac{8}{9}$

Answer

Rearrange the terms using commutativity of multiplication so that the common factor $\dfrac{8}{9}$ can be taken out using distributivity.

$\Rightarrow \dfrac{8}{9} \times \dfrac{4}{5} + \dfrac{5}{6} - \dfrac{9}{5} \times \dfrac{8}{9} \\[1em] \Rightarrow \dfrac{8}{9} \times \dfrac{4}{5} - \dfrac{8}{9} \times \dfrac{9}{5} + \dfrac{5}{6} \\[1em] \Rightarrow \dfrac{8}{9} \times \left(\dfrac{4}{5} - \dfrac{9}{5}\right) + \dfrac{5}{6} \\[1em] \Rightarrow \dfrac{8}{9} \times \dfrac{-5}{5} + \dfrac{5}{6} \\[1em] \Rightarrow \dfrac{8}{9} \times (-1) + \dfrac{5}{6} \\[1em] \Rightarrow \dfrac{-8}{9} + \dfrac{5}{6} \\[1em] [\text{ LCM of } 9 \text{ and } 6 = 18] \\[1em] \Rightarrow \dfrac{(-8) \times 2 + 5 \times 3}{18} \\[1em] \Rightarrow \dfrac{-16 + 15}{18} \\[1em] \Rightarrow \dfrac{-1}{18}$

Hence, $\mathbf{\dfrac{8}{9} \times \dfrac{4}{5} + \dfrac{5}{6} - \dfrac{9}{5} \times \dfrac{8}{9} = \dfrac{-1}{18}}$

Using the appropriate properties of operations of rational numbers, evaluate the following:

$\dfrac{-3}{7} \times \dfrac{14}{15} \times \dfrac{7}{12} \times \left(\dfrac{-30}{35}\right)$

Answer

Rearrange and group the rational numbers using commutativity and associativity of multiplication so that the products can be reduced easily.

$\Rightarrow \dfrac{-3}{7} \times \dfrac{14}{15} \times \dfrac{7}{12} \times \left(\dfrac{-30}{35}\right) \\[1em] \Rightarrow \left(\dfrac{-3}{7} \times \dfrac{14}{15}\right) \times \left(\dfrac{7}{12} \times \dfrac{-30}{35}\right) \\[1em] \Rightarrow \dfrac{(-3) \times 14}{7 \times 15} \times \dfrac{7 \times (-30)}{12 \times 35} \\[1em] \Rightarrow \dfrac{-42}{105} \times \dfrac{-210}{420} \\[1em] \Rightarrow \dfrac{-2}{5} \times \dfrac{-1}{2} \\[1em] \Rightarrow \dfrac{(-2) \times (-1)}{5 \times 2} \\[1em] \Rightarrow \dfrac{2}{10} \\[1em] \Rightarrow \dfrac{1}{5}$

Hence, $\mathbf{\dfrac{-3}{7} \times \dfrac{14}{15} \times \dfrac{7}{12} \times \left(\dfrac{-30}{35}\right) = \dfrac{1}{5}}$

If $p = \dfrac{-8}{27}, q = \dfrac{3}{4} \text{ and } r = \dfrac{-12}{15}$, then verify that

(i) $p × (q × r) = (p × q) × r$

(ii) $p × (q - r) = p × q - p × r$

Answer

Here $r = \dfrac{-12}{15} = \dfrac{-4}{5}$

(i) To verify the associative property of multiplication, we have to show that $p \times (q \times r) = (p \times q) \times r$

Solving L.H.S.,

$\Rightarrow q \times r = \dfrac{3}{4} \times \dfrac{-4}{5} \\[1em] \Rightarrow \dfrac{3 \times (-4)}{4 \times 5} = \dfrac{-12}{20} = \dfrac{-3}{5} \\[1em] \therefore p \times (q \times r) = \dfrac{-8}{27} \times \dfrac{-3}{5} \\[1em] \Rightarrow \dfrac{(-8) \times (-3)}{27 \times 5} \\[1em] \Rightarrow \dfrac{24}{135} \\[1em] \Rightarrow \dfrac{8}{45}$

Solving R.H.S.,

$\Rightarrow p \times q = \dfrac{-8}{27} \times \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{(-8) \times 3}{27 \times 4} = \dfrac{-24}{108} = \dfrac{-2}{9} \\[1em] \therefore (p \times q) \times r = \dfrac{-2}{9} \times \dfrac{-4}{5} \\[1em] \Rightarrow \dfrac{(-2) \times (-4)}{9 \times 5} \\[1em] \Rightarrow \dfrac{8}{45}$

As L.H.S. = R.H.S.

Hence, proved that $\bm{p \times (q \times r) = (p \times q) \times r}$.

(ii) To verify the distributive property of multiplication over subtraction, we have to show that $p \times (q - r) = p \times q - p \times r$

Solving L.H.S.,

$\Rightarrow q - r = \dfrac{3}{4} - \left(\dfrac{-4}{5}\right) \\[1em] \Rightarrow \dfrac{3}{4} + \dfrac{4}{5} \\[1em] \Rightarrow \dfrac{3 \times 5 + 4 \times 4}{20} \\[1em] \Rightarrow \dfrac{15 + 16}{20} \\[1em] \Rightarrow \dfrac{31}{20} \\[1em] \therefore p \times (q - r) = \dfrac{-8}{27} \times \dfrac{31}{20} \\[1em] \Rightarrow \dfrac{(-8) \times 31}{27 \times 20} \\[1em] \Rightarrow \dfrac{-248}{540} \\[1em] \Rightarrow \dfrac{-62}{135}$

Solving R.H.S.,

$\Rightarrow p \times q = \dfrac{-2}{9} \text{ (found in part (i))} \\[1em] \Rightarrow p \times r = \dfrac{-8}{27} \times \dfrac{-4}{5} = \dfrac{(-8) \times (-4)}{27 \times 5} = \dfrac{32}{135} \\[1em] \therefore p \times q - p \times r = \dfrac{-2}{9} - \dfrac{32}{135} \\[1em] [\text{ LCM of } 9 \text{ and } 135 = 135] \\[1em] \Rightarrow \dfrac{(-2) \times 15 - 32}{135} \\[1em] \Rightarrow \dfrac{-30 - 32}{135} \\[1em] \Rightarrow \dfrac{-62}{135}$

As L.H.S. = R.H.S.

Hence, proved that $\bm{p \times (q - r) = p \times q - p \times r}$.

Fill in the following blanks:

(i) The reciprocal of 0 is ..... .

(ii) The numbers ..... and ..... are their own reciprocals.

(iii) If y be the reciprocal of x, then the reciprocal of y² in terms of x will be ..... .

(iv) The product of a non-zero rational number and its reciprocal is ..... .

(v) The reciprocal of a negative rational number is ..... .

Answer

(i) There is no rational number which when multiplied by 0 gives 1, so 0 has no reciprocal.

Hence, the reciprocal of 0 is not defined.

(ii) As $1 \times 1 = 1 \text{ and } (-1) \times (-1) = 1$, each of these numbers is its own reciprocal.

Hence, the numbers 1 and -1 are their own reciprocals.

(iii) Since $y$ is the reciprocal of x,

$\Rightarrow y = \dfrac{1}{x} \\[1em] \Rightarrow y^2 = \dfrac{1}{x^2} \\[1em] \Rightarrow \text{Reciprocal of } y^2 = x^2$

Hence, the reciprocal of y² in terms of x will be x².

(iv) For every non-zero rational number $\dfrac{p}{q}$, we have $\dfrac{p}{q} \times \dfrac{q}{p} = 1$

Hence, the product of a non-zero rational number and its reciprocal is 1.

(v) The reciprocal is obtained by interchanging the numerator and the denominator, so the sign of the rational number does not change. For example, the reciprocal of $\dfrac{-3}{5} \text{ is } \dfrac{5}{-3} = \dfrac{-5}{3}$

Hence, the reciprocal of a negative rational number is a negative rational number.

Using distributivity, find

$\bigg\lbrace\dfrac{7}{5} \times \left(\dfrac{-3}{12}\right)\bigg\rbrace + \bigg\lbrace\dfrac{7}{5} \times \dfrac{5}{12}\bigg\rbrace$

Answer

As $\dfrac{7}{5}$ is common to both the products, take it out using distributivity.

$\Rightarrow \bigg\lbrace\dfrac{7}{5} \times \left(\dfrac{-3}{12}\right)\bigg\rbrace + \bigg\lbrace\dfrac{7}{5} \times \dfrac{5}{12}\bigg\rbrace \\[1em] \Rightarrow \dfrac{7}{5} \times \left(\dfrac{-3}{12} + \dfrac{5}{12}\right) \\[1em] \Rightarrow \dfrac{7}{5} \times \dfrac{-3 + 5}{12} \\[1em] \Rightarrow \dfrac{7}{5} \times \dfrac{2}{12} \\[1em] \Rightarrow \dfrac{7}{5} \times \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{7}{30}$

Hence, $\mathbf{\bigg\lbrace\dfrac{7}{5} \times \left(\dfrac{-3}{12}\right)\bigg\rbrace + \bigg\lbrace\dfrac{7}{5} \times \dfrac{5}{12}\bigg\rbrace = \dfrac{7}{30}}$

Using distributivity, find

$\bigg\lbrace\dfrac{9}{16} \times \dfrac{4}{12}\bigg\rbrace + \bigg\lbrace\dfrac{9}{16} \times \left(\dfrac{-3}{9}\right)\bigg\rbrace$

Answer

As $\dfrac{9}{16}$ is common to both the products, take it out using distributivity.

$\Rightarrow \bigg\lbrace\dfrac{9}{16} \times \dfrac{4}{12}\bigg\rbrace + \bigg\lbrace\dfrac{9}{16} \times \left(\dfrac{-3}{9}\right)\bigg\rbrace \\[1em] \Rightarrow \dfrac{9}{16} \times \left(\dfrac{4}{12} + \dfrac{-3}{9}\right) \\[1em] \Rightarrow \dfrac{9}{16} \times \left(\dfrac{1}{3} - \dfrac{1}{3}\right) \\[1em] \Rightarrow \dfrac{9}{16} \times 0 \\[1em] \Rightarrow 0$

Hence, $\mathbf{\bigg\lbrace\dfrac{9}{16} \times \dfrac{4}{12}\bigg\rbrace + \bigg\lbrace\dfrac{9}{16} \times \left(\dfrac{-3}{9}\right)\bigg\rbrace = 0}$

Find the sum of additive inverse and multiplicative inverse of 9.

Answer

To find the additive inverse, change the sign of the number, and to find the multiplicative inverse, interchange its numerator and denominator.

Additive inverse of 9 = -9

Multiplicative inverse of 9 = $\dfrac{1}{9}$

Adding both the inverses,

$\Rightarrow -9 + \dfrac{1}{9} \\[1em] \Rightarrow \dfrac{(-9) \times 9 + 1}{9} \\[1em] \Rightarrow \dfrac{-81 + 1}{9} \\[1em] \Rightarrow \dfrac{-80}{9} \\[1em] \Rightarrow -8\dfrac{8}{9}$

Hence, the required sum is $\mathbf{-8\dfrac{8}{9}}$

Find the product of additive inverse and multiplicative inverse of $-\dfrac{3}{7}$.

Answer

To find the additive inverse, change the sign of the number, and to find the multiplicative inverse, interchange its numerator and denominator.

Additive inverse of $-\dfrac{3}{7} = \dfrac{3}{7}$

Multiplicative inverse of $-\dfrac{3}{7} = \dfrac{-7}{3}$

Multiplying both the inverses,

$\Rightarrow \dfrac{3}{7} \times \dfrac{-7}{3} \\[1em] \Rightarrow \dfrac{3 \times (-7)}{7 \times 3} \\[1em] \Rightarrow \dfrac{-21}{21} \\[1em] \Rightarrow -1$

Hence, the required product is -1.