Rational Numbers — Exercise 1.4

Find the value of the following:

$\dfrac{-3}{7} \div 4$

Answer

To divide one rational number by another, multiply the dividend by the reciprocal of the divisor.

$\Rightarrow \dfrac{-3}{7} \div 4 \\[1em] \Rightarrow \dfrac{-3}{7} \div \dfrac{4}{1} \\[1em] \Rightarrow \dfrac{-3}{7} \times \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{(-3) \times 1}{7 \times 4} \\[1em] \Rightarrow \dfrac{-3}{28}$

Hence, $\mathbf{\dfrac{-3}{7} \div 4 = \dfrac{-3}{28}}$

Find the value of the following:

$4\dfrac{5}{8} \div \left(\dfrac{-4}{9}\right)$

Answer

First convert the mixed fraction into an improper fraction, then multiply the dividend by the reciprocal of the divisor.

$\Rightarrow 4\dfrac{5}{8} \div \left(\dfrac{-4}{9}\right) \\[1em] \Rightarrow \dfrac{37}{8} \div \left(\dfrac{-4}{9}\right) \\[1em] \Rightarrow \dfrac{37}{8} \times \dfrac{9}{-4} \\[1em] \Rightarrow \dfrac{37 \times 9}{8 \times (-4)} \\[1em] \Rightarrow \dfrac{333}{-32} = \dfrac{-333}{32} \\[1em] \Rightarrow -10\dfrac{13}{32}$

Hence, $\mathbf{4\dfrac{5}{8} \div \left(\dfrac{-4}{9}\right) = -10\dfrac{13}{32}}$

Find the value of the following:

$\dfrac{-8}{9} \div \dfrac{-3}{5}$

Answer

To divide one rational number by another, multiply the dividend by the reciprocal of the divisor.

$\Rightarrow \dfrac{-8}{9} \div \dfrac{-3}{5} \\[1em] \Rightarrow \dfrac{-8}{9} \times \dfrac{5}{-3} \\[1em] \Rightarrow \dfrac{(-8) \times 5}{9 \times (-3)} \\[1em] \Rightarrow \dfrac{-40}{-27} = \dfrac{40}{27} \\[1em] \Rightarrow 1\dfrac{13}{27}$

Hence, $\mathbf{\dfrac{-8}{9} \div \dfrac{-3}{5} = 1\dfrac{13}{27}}$

State whether the following statements are true or false:

(i) $\dfrac{-9}{13} \div \dfrac{2}{7}$ is a rational number

(ii) $\dfrac{-3}{4} \div \left(\dfrac{5}{9} \div \dfrac{-4}{11}\right) = \left(\dfrac{-3}{4} \div \dfrac{5}{9}\right) \div \dfrac{-4}{11}$

(iii) $\dfrac{13}{14} \div \dfrac{-5}{7} \neq \dfrac{-5}{7} \div \dfrac{13}{14}$

(iv) $\left(-7 \div \dfrac{4}{5}\right) \div \dfrac{-9}{10} \neq -7 \div \left(\dfrac{4}{5} \div \dfrac{-9}{10}\right)$

Answer

(i) If we exclude 0, the remaining rational numbers are closed under division. Here the divisor $\dfrac{2}{7}$ is a non-zero rational number.

$\Rightarrow \dfrac{-9}{13} \div \dfrac{2}{7} \\[1em] \Rightarrow \dfrac{-9}{13} \times \dfrac{7}{2} \\[1em] \Rightarrow \dfrac{-63}{26}$

As $\dfrac{-63}{26}$ is a rational number, the given statement is true.

Hence, the given statement is true.

(ii) For any three non-zero rational numbers

$\dfrac{p}{q}, \dfrac{r}{s} \text{ and } \dfrac{t}{u}$, we have $\dfrac{p}{q} \div \left(\dfrac{r}{s} \div \dfrac{t}{u}\right) \neq \left(\dfrac{p}{q} \div \dfrac{r}{s}\right) \div \dfrac{t}{u}$

So, division of rational numbers is not associative.

Hence, the given statement is false.

(iii) For any two non-zero rational numbers

$\dfrac{p}{q} \text{ and } \dfrac{r}{s}$, we have $\dfrac{p}{q} \div \dfrac{r}{s} \neq \dfrac{r}{s} \div \dfrac{p}{q}$

So, division of rational numbers is not commutative.

Hence, the given statement is true.

(iv) For any three non-zero rational numbers

$\dfrac{p}{q}, \dfrac{r}{s} \text{ and } \dfrac{t}{u}$, we have $\left(\dfrac{p}{q} \div \dfrac{r}{s}\right) \div \dfrac{t}{u} \neq \dfrac{p}{q} \div \left(\dfrac{r}{s} \div \dfrac{t}{u}\right)$

So, division of rational numbers is not associative.

Hence, the given statement is true.

The product of two rational numbers is $\dfrac{-11}{12}$. If one of them is $2\dfrac{4}{9}$, find the other.

Answer

Let the other rational number be $x$.

As the product of the two rational numbers is $\dfrac{-11}{12}$,

$\Rightarrow 2\dfrac{4}{9} \times x = \dfrac{-11}{12} \\[1em] \Rightarrow \dfrac{22}{9} \times x = \dfrac{-11}{12} \\[1em] \Rightarrow x = \dfrac{-11}{12} \div \dfrac{22}{9} \\[1em] \Rightarrow x = \dfrac{-11}{12} \times \dfrac{9}{22} \\[1em] \Rightarrow x = \dfrac{(-11) \times 9}{12 \times 22} \\[1em] \Rightarrow x = \dfrac{-99}{264} \\[1em] \Rightarrow x = \dfrac{-3}{8}$

Hence, the other rational number is $\mathbf{\dfrac{-3}{8}}$

By what rational number should $\dfrac{-7}{12}$ be multiplied to get the product as $\dfrac{5}{14}$?

Answer

Let the required rational number be $x$.

According to the given condition,

$\Rightarrow \dfrac{-7}{12} \times x = \dfrac{5}{14} \\[1em] \Rightarrow x = \dfrac{5}{14} \div \dfrac{-7}{12} \\[1em] \Rightarrow x = \dfrac{5}{14} \times \dfrac{12}{-7} \\[1em] \Rightarrow x = \dfrac{5 \times 12}{14 \times (-7)} \\[1em] \Rightarrow x = \dfrac{60}{-98} \\[1em] \Rightarrow x = \dfrac{-30}{49}$

Hence, the required rational number is $\mathbf{\dfrac{-30}{49}}$.

By what rational number should -3 be divided to get $\dfrac{-9}{13}$?

Answer

Let the required rational number be $x$.

According to the given condition,

$\Rightarrow -3 \div x = \dfrac{-9}{13} \\[1em] \Rightarrow \dfrac{-3}{1} \times \dfrac{1}{x} = \dfrac{-9}{13} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{-9}{13} \div (-3) \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{-9}{13} \times \dfrac{1}{-3} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{-9}{-39} = \dfrac{3}{13} \\[1em] \Rightarrow x = \dfrac{13}{3} = 4\dfrac{1}{3}$

Hence, the required rational number is $\mathbf{\dfrac{13}{3}}$.

Divide the sum of $\dfrac{-13}{8} \text{ and } \dfrac{5}{12}$ by their difference.

Answer

First find the sum and the difference of the given rational numbers.

LCM of 8 and 12 = 24

Finding the sum,

$\Rightarrow \dfrac{-13}{8} + \dfrac{5}{12} \\[1em] \Rightarrow \dfrac{(-13) \times 3 + 5 \times 2}{24} \\[1em] \Rightarrow \dfrac{-39 + 10}{24} \\[1em] \Rightarrow \dfrac{-29}{24}$

Finding the difference,

$\Rightarrow \dfrac{-13}{8} - \dfrac{5}{12} \\[1em] \Rightarrow \dfrac{(-13) \times 3 - 5 \times 2}{24} \\[1em] \Rightarrow \dfrac{-39 - 10}{24} \\[1em] \Rightarrow \dfrac{-49}{24}$

Dividing the sum by the difference,

$\Rightarrow \dfrac{-29}{24} \div \dfrac{-49}{24} \\[1em] \Rightarrow \dfrac{-29}{24} \times \dfrac{24}{-49} \\[1em] \Rightarrow \dfrac{(-29) \times 24}{24 \times (-49)} \\[1em] \Rightarrow \dfrac{29}{49}$

Hence, the required quotient is $\mathbf{\dfrac{29}{49}}$

Divide the sum of $\dfrac{8}{3}$ and $\dfrac{4}{7}$ by the product of $\dfrac{-3}{7}$ and $\dfrac{14}{9}$.

Answer

First find the sum of $\dfrac{8}{3} \text{ and } \dfrac{4}{7} \text{, and the product of } \dfrac{-3}{7} \text{ and } \dfrac{14}{9}$.

Finding the sum,

$\Rightarrow \dfrac{8}{3} + \dfrac{4}{7} \\[1em] [\text{ LCM of } 3 \text{ and } 7 = 21] \\[1em] \Rightarrow \dfrac{8 \times 7 + 4 \times 3}{21} \\[1em] \Rightarrow \dfrac{56 + 12}{21} \\[1em] \Rightarrow \dfrac{68}{21}$

Finding the product,

$\Rightarrow \dfrac{-3}{7} \times \dfrac{14}{9} \\[1em] \Rightarrow \dfrac{(-3) \times 14}{7 \times 9} \\[1em] \Rightarrow \dfrac{-42}{63} \\[1em] \Rightarrow \dfrac{-2}{3}$

Dividing the sum by the product,

$\Rightarrow \dfrac{68}{21} \div \dfrac{-2}{3} \\[1em] \Rightarrow \dfrac{68}{21} \times \dfrac{3}{-2} \\[1em] \Rightarrow \dfrac{68 \times 3}{21 \times (-2)} \\[1em] \Rightarrow \dfrac{204}{-42} = \dfrac{-34}{7} \\[1em] \Rightarrow -4\dfrac{6}{7}$

Hence, the required quotient is $\mathbf{-4\dfrac{6}{7}}$

If $p = \dfrac{-3}{2}, q = \dfrac{4}{5} \text{ and } r = \dfrac{-7}{12}$, then verify that (p ÷ q) ÷ r ≠ p ÷ (q ÷ r)

Answer

To verify that division of rational numbers is not associative, we have to show that $(p \div q) \div r \neq p \div (q \div r)$

Solving L.H.S.,

$\Rightarrow p \div q = \dfrac{-3}{2} \div \dfrac{4}{5} \\[1em] \Rightarrow \dfrac{-3}{2} \times \dfrac{5}{4} = \dfrac{-15}{8} \\[1em] \therefore (p \div q) \div r = \dfrac{-15}{8} \div \dfrac{-7}{12} \\[1em] \Rightarrow \dfrac{-15}{8} \times \dfrac{12}{-7} \\[1em] \Rightarrow \dfrac{-180}{-56} \\[1em] \Rightarrow \dfrac{45}{14}$

Solving R.H.S.,

$\Rightarrow q \div r = \dfrac{4}{5} \div \dfrac{-7}{12} \\[1em] \Rightarrow \dfrac{4}{5} \times \dfrac{12}{-7} = \dfrac{48}{-35} = \dfrac{-48}{35} \\[1em] \therefore p \div (q \div r) = \dfrac{-3}{2} \div \dfrac{-48}{35} \\[1em] \Rightarrow \dfrac{-3}{2} \times \dfrac{35}{-48} \\[1em] \Rightarrow \dfrac{-105}{-96} \\[1em] \Rightarrow \dfrac{35}{32}$

As $\dfrac{45}{14} \neq \dfrac{35}{32}$

$\Rightarrow (p \div q) \div r \neq p \div (q \div r)$

Hence, proved that $\bm{(p \div q) \div r \neq p \div (q \div r)}$.