Rational Numbers — Check Your Progress

Complete the following table:

Answer

Addition and multiplication are commutative for all these systems of numbers, while subtraction and division are not commutative for any of them.

For example, $5 - 3 \neq 3 - 5 \text{ and } 6 \div 2 \neq 2 \div 6$

Hence, the completed table is

Complete the following table:

Answer

Addition and multiplication are associative for all these systems of numbers, while subtraction and division are not associative for any of them.

For example, $(9 - 5) - 2 \neq 9 - (5 - 2) \text{ and } (12 \div 6) \div 2 \neq 12 \div (6 \div 2)$

Hence, the completed table is

Evaluate the following:

$4\dfrac{2}{5} + 3\dfrac{7}{8}$

Answer

First convert the mixed fractions into improper fractions and then add.

$\Rightarrow 4\dfrac{2}{5} + 3\dfrac{7}{8} \\[1em] \Rightarrow \dfrac{22}{5} + \dfrac{31}{8} \\[1em] [\text{ LCM of } 5 \text{ and } 8 = 40] \\[1em] \Rightarrow \dfrac{22 \times 8 + 31 \times 5}{40} \\[1em] \Rightarrow \dfrac{176 + 155}{40} \\[1em] \Rightarrow \dfrac{331}{40} = 8\dfrac{11}{40}$

Hence, $\mathbf{4\dfrac{2}{5} + 3\dfrac{7}{8} = 8\dfrac{11}{40}}$

Evaluate the following:

$7\dfrac{3}{4} - 4\dfrac{3}{5}$

Answer

First convert the mixed fractions into improper fractions and then subtract.

$\Rightarrow 7\dfrac{3}{4} - 4\dfrac{3}{5} \\[1em] \Rightarrow \dfrac{31}{4} - \dfrac{23}{5} \\[1em] [\text{ LCM of } 4 \text{ and } 5 = 20] \\[1em] \Rightarrow \dfrac{31 \times 5 - 23 \times 4}{20} \\[1em] \Rightarrow \dfrac{155 - 92}{20} \\[1em] \Rightarrow \dfrac{63}{20} = 3\dfrac{3}{20}$

Hence, $\mathbf{7\dfrac{3}{4} - 4\dfrac{3}{5} = 3\dfrac{3}{20}}$

Evaluate the following:

$6\dfrac{8}{9} - \left(\dfrac{-7}{3}\right)$

Answer

Subtracting a negative rational number is the same as adding its additive inverse.

$\Rightarrow 6\dfrac{8}{9} - \left(\dfrac{-7}{3}\right) \\[1em] \Rightarrow \dfrac{62}{9} + \dfrac{7}{3} \\[1em] [\text{ LCM of } 9 \text{ and } 3 = 9] \\[1em] \Rightarrow \dfrac{62 + 7 \times 3}{9} \\[1em] \Rightarrow \dfrac{62 + 21}{9} \\[1em] \Rightarrow \dfrac{83}{9} = 9\dfrac{2}{9}$

Hence, $\mathbf{6\dfrac{8}{9} - \left(\dfrac{-7}{3}\right) = 9\dfrac{2}{9}}$

Evaluate the following:

$\left(-16\dfrac{2}{3}\right) \times 14\dfrac{1}{2}$

Answer

First convert the mixed fractions into improper fractions and then multiply.

$\Rightarrow \left(-16\dfrac{2}{3}\right) \times 14\dfrac{1}{2} \\[1em] \Rightarrow \dfrac{-50}{3} \times \dfrac{29}{2} \\[1em] \Rightarrow \dfrac{(-50) \times 29}{3 \times 2} \\[1em] \Rightarrow \dfrac{-1450}{6} \\[1em] \Rightarrow \dfrac{-725}{3} = -241\dfrac{2}{3}$

Hence, $\mathbf{\left(-16\dfrac{2}{3}\right) \times 14\dfrac{1}{2} = -241\dfrac{2}{3}}$

Evaluate the following:

$\dfrac{-7}{8} \div 15\dfrac{3}{4}$

Answer

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

$\Rightarrow \dfrac{-7}{8} \div 15\dfrac{3}{4} \\[1em] \Rightarrow \dfrac{-7}{8} \div \dfrac{63}{4} \\[1em] \Rightarrow \dfrac{-7}{8} \times \dfrac{4}{63} \\[1em] \Rightarrow \dfrac{(-7) \times 4}{8 \times 63} \\[1em] \Rightarrow \dfrac{-28}{504} = \dfrac{-1}{18}$

Hence, $\mathbf{\dfrac{-7}{8} \div 15\dfrac{3}{4} = \dfrac{-1}{18}}$

Evaluate the following:

$\dfrac{-12}{13} \times 1$

Answer

As 1 is the multiplicative identity of rational numbers, the product remains unchanged.

$\Rightarrow \dfrac{-12}{13} \times 1 \\[1em] \Rightarrow \dfrac{-12}{13}$

Hence, $\mathbf{\dfrac{-12}{13} \times 1 = \dfrac{-12}{13}}$

Evaluate the following:

$\dfrac{4}{9} \times 0$

Answer

When a rational number is multiplied by zero, the product is always zero.

$\Rightarrow \dfrac{4}{9} \times 0 \\[1em] \Rightarrow 0$

Hence, $\mathbf{\dfrac{4}{9} \times 0 = 0}$

Evaluate the following:

$0 \div \left(-4\dfrac{7}{10}\right)$

Answer

Zero divided by any non-zero rational number is always zero.

$\Rightarrow 0 \div \left(-4\dfrac{7}{10}\right) \\[1em] \Rightarrow 0 \div \left(\dfrac{-47}{10}\right) \\[1em] \Rightarrow 0 \times \dfrac{10}{-47} \\[1em] \Rightarrow 0$

Hence, $\mathbf{0 \div \left(-4\dfrac{7}{10}\right) = 0}$

What number should be added to $\dfrac{-4}{11}$ to get $\dfrac{-3}{8}$?

Answer

Let the required number be $x$.

According to the given condition,

$\Rightarrow \dfrac{-4}{11} + x = \dfrac{-3}{8} \\[1em] \Rightarrow x = \dfrac{-3}{8} - \left(\dfrac{-4}{11}\right) \\[1em] \Rightarrow x = \dfrac{-3}{8} + \dfrac{4}{11} \\[1em] [\text{ LCM of } 8 \text{ and } 11 = 88] \\[1em] \Rightarrow x = \dfrac{(-3) \times 11 + 4 \times 8}{88} \\[1em] \Rightarrow x = \dfrac{-33 + 32}{88} \\[1em] \Rightarrow x = \dfrac{-1}{88}$

Hence, the required number is $\mathbf{\dfrac{-1}{88}}$

What rational number should be subtracted from the sum of $\dfrac{3}{14} \text{ and } \dfrac{-4}{7}$ to get $\dfrac{13}{21}$?

Answer

First find the sum of the two given rational numbers.

$\Rightarrow \dfrac{3}{14} + \dfrac{-4}{7} \\[1em] [\text{ LCM of } 14 \text{ and } 7 = 14] \\[1em] \Rightarrow \dfrac{3 + (-4) \times 2}{14} \\[1em] \Rightarrow \dfrac{3 - 8}{14} \\[1em] \Rightarrow \dfrac{-5}{14}$

Let the required rational number be $x$.

According to the given condition,

$\Rightarrow \dfrac{-5}{14} - x = \dfrac{13}{21} \\[1em] \Rightarrow x = \dfrac{-5}{14} - \dfrac{13}{21} \\[1em] [\text{ LCM of } 14 \text{ and } 21 = 42] \\[1em] \Rightarrow x = \dfrac{(-5) \times 3 - 13 \times 2}{42} \\[1em] \Rightarrow x = \dfrac{-15 - 26}{42} \\[1em] \Rightarrow x = \dfrac{-41}{42}$

Hence, the required rational number is $\mathbf{\dfrac{-41}{42}}$

If the product of two rational numbers is $\dfrac{25}{42}$ and one of them is $-2\dfrac{6}{7}$, find the other.

Answer

Let the other rational number be $x$.

As the product of the two rational numbers is $\dfrac{25}{42}$,

$\Rightarrow -2\dfrac{6}{7} \times x = \dfrac{25}{42} \\[1em] \Rightarrow \dfrac{-20}{7} \times x = \dfrac{25}{42} \\[1em] \Rightarrow x = \dfrac{25}{42} \div \dfrac{-20}{7} \\[1em] \Rightarrow x = \dfrac{25}{42} \times \dfrac{7}{-20} \\[1em] \Rightarrow x = \dfrac{25 \times 7}{42 \times (-20)} \\[1em] \Rightarrow x = \dfrac{175}{-840} \\[1em] \Rightarrow x = \dfrac{-5}{24}$

Hence, the other rational number is $\mathbf{\dfrac{-5}{24}}$

Divide the sum of $\dfrac{4}{13} \text{ and } \dfrac{-3}{2}$ by their product.

Answer

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

Finding the sum,

$\Rightarrow \dfrac{4}{13} + \dfrac{-3}{2} \\[1em] [\text{ LCM of } 13 \text{ and } 2 = 26] \\[1em] \Rightarrow \dfrac{4 \times 2 + (-3) \times 13}{26} \\[1em] \Rightarrow \dfrac{8 - 39}{26} \\[1em] \Rightarrow \dfrac{-31}{26}$

Finding the product,

$\Rightarrow \dfrac{4}{13} \times \dfrac{-3}{2} \\[1em] \Rightarrow \dfrac{4 \times (-3)}{13 \times 2} \\[1em] \Rightarrow \dfrac{-12}{26} = \dfrac{-6}{13}$

Dividing the sum by the product,

$\Rightarrow \dfrac{-31}{26} \div \dfrac{-6}{13} \\[1em] \Rightarrow \dfrac{-31}{26} \times \dfrac{13}{-6} \\[1em] \Rightarrow \dfrac{(-31) \times 13}{26 \times (-6)} \\[1em] \Rightarrow \dfrac{-403}{-156} \\[1em] \Rightarrow \dfrac{31}{12} = 2\dfrac{7}{12}$

Hence, the required quotient is $\mathbf{2\dfrac{7}{12}}$

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

$\dfrac{3}{13} \times \dfrac{4}{5} - \dfrac{7}{5} - \dfrac{4}{5} \times \dfrac{5}{13}$

Answer

Rearrange the terms using commutativity of multiplication so that the common factor $\dfrac{4}{5}$ can be taken out using distributivity.

$\Rightarrow \dfrac{3}{13} \times \dfrac{4}{5} - \dfrac{7}{5} - \dfrac{4}{5} \times \dfrac{5}{13} \\[1em] \Rightarrow \dfrac{4}{5} \times \dfrac{3}{13} - \dfrac{4}{5} \times \dfrac{5}{13} - \dfrac{7}{5} \\[1em] \Rightarrow \dfrac{4}{5} \times \left(\dfrac{3}{13} - \dfrac{5}{13}\right) - \dfrac{7}{5} \\[1em] \Rightarrow \dfrac{4}{5} \times \dfrac{-2}{13} - \dfrac{7}{5} \\[1em] \Rightarrow \dfrac{-8}{65} - \dfrac{7}{5} \\[1em] [\text{ LCM of } 65 \text{ and } 5 = 65] \\[1em] \Rightarrow \dfrac{-8 - 7 \times 13}{65} \\[1em] \Rightarrow \dfrac{-8 - 91}{65} \\[1em] \Rightarrow \dfrac{-99}{65} = -1\dfrac{34}{65}$

Hence, $\mathbf{\dfrac{3}{13} \times \dfrac{4}{5} - \dfrac{7}{5} - \dfrac{4}{5} \times \dfrac{5}{13} = -1\dfrac{34}{65}}$

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

$\dfrac{5}{7} + \dfrac{2}{11} + \dfrac{-8}{7} + \dfrac{6}{11}$

Answer

Using commutativity and associativity of addition, group together the rational numbers having the same denominator.

$\Rightarrow \dfrac{5}{7} + \dfrac{2}{11} + \dfrac{-8}{7} + \dfrac{6}{11} \\[1em] \Rightarrow \left(\dfrac{5}{7} + \dfrac{-8}{7}\right) + \left(\dfrac{2}{11} + \dfrac{6}{11}\right) \\[1em] \Rightarrow \dfrac{5 + (-8)}{7} + \dfrac{2 + 6}{11} \\[1em] \Rightarrow \dfrac{-3}{7} + \dfrac{8}{11} \\[1em] [\text{ LCM of } 7 \text{ and } 11 = 77] \\[1em] \Rightarrow \dfrac{(-3) \times 11 + 8 \times 7}{77} \\[1em] \Rightarrow \dfrac{-33 + 56}{77} \\[1em] \Rightarrow \dfrac{23}{77}$

Hence, $\mathbf{\dfrac{5}{7} + \dfrac{2}{11} + \dfrac{-8}{7} + \dfrac{6}{11} = \dfrac{23}{77}}$

Find the additive inverse of the following:

$-13\dfrac{7}{8}$

Answer

To find the additive inverse of any rational number, change its sign.

$\Rightarrow -13\dfrac{7}{8} = \dfrac{-111}{8} \\[1em] \Rightarrow \text{Additive inverse} = -\left(\dfrac{-111}{8}\right) \\[1em] \Rightarrow \dfrac{111}{8} = 13\dfrac{7}{8}$

Hence, the additive inverse of $\mathbf{-13\dfrac{7}{8} \text{ is } 13\dfrac{7}{8}}$

Find the additive inverse of the following:

$4\dfrac{3}{6}$

Answer

First reduce the fractional part to its lowest form, then change the sign.

$\Rightarrow 4\dfrac{3}{6} = 4\dfrac{1}{2} = \dfrac{9}{2} \\[1em] \Rightarrow \text{Additive inverse} = -\left(\dfrac{9}{2}\right) \\[1em] \Rightarrow \dfrac{-9}{2} = -4\dfrac{1}{2}$

Hence, the additive inverse of $\mathbf{4\dfrac{3}{6} \text{ is } -4\dfrac{1}{2}}$

Find the multiplicative inverse of the following:

$\dfrac{-23}{46}$

Answer

First reduce the rational number to its lowest form, then interchange its numerator and denominator.

$\Rightarrow \dfrac{-23}{46} = \dfrac{-23 \div 23}{46 \div 23} = \dfrac{-1}{2} \\[1em] \Rightarrow \text{Multiplicative inverse} = \dfrac{2}{-1} = -2 \\[1em] [\text{Check : } \dfrac{-1}{2} \times (-2) = 1]$

Hence, the multiplicative inverse of $\mathbf{\dfrac{-23}{46} \text{ is } -2}$

Find the multiplicative inverse of the following:

$0$

Answer

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

Hence, the multiplicative inverse of 0 does not exist.

Represent the following rational numbers on the number line:

$\dfrac{-3}{11}$

Answer

As $\dfrac{-3}{11}$ is a negative proper fraction, it lies between -1 and 0.

Divide the line segment between -1 and 0 into eleven equal parts. The third point of division to the left of 0 represents $\dfrac{-3}{11}$.

Hence, the point marked on the number line represents $\mathbf{\dfrac{-3}{11}}$

Represent the following rational numbers on the number line:

$\dfrac{5}{7}$

Answer

As $\dfrac{5}{7}$ is a positive proper fraction, it lies between 0 and 1.

Divide the line segment between 0 and 1 into seven equal parts. The fifth point of division from 0 represents $\dfrac{5}{7}$.

Hence, the point marked on the number line represents $\mathbf{\dfrac{5}{7}}$

Insert five rational numbers between $\dfrac{-3}{7} \text{ and } \dfrac{2}{5}$

Answer

The given rational numbers have different denominators, so first write them with a common denominator.

$[\text{ LCM of } 7 \text{ and } 5 = 35] \\[1em] \Rightarrow \dfrac{-3}{7} = \dfrac{-3 \times 5}{7 \times 5} = \dfrac{-15}{35} \\[1em] \Rightarrow \dfrac{2}{5} = \dfrac{2 \times 7}{5 \times 7} = \dfrac{14}{35}$

We note that -14, -13, -12, …, 12, 13 are the integers between -15 and 14.

Choosing any five of these integers as numerators over the common denominator 35, say -1, 0, 1, 12 and 13, we get the rational numbers $\dfrac{-1}{35}, \dfrac{0}{35}, \dfrac{1}{35}, \dfrac{12}{35} \text{ and } \dfrac{13}{35}$.

Hence, five rational numbers between $\mathbf{\dfrac{-3}{7} \text{ and } \dfrac{2}{5}}$ are $\mathbf{\dfrac{-1}{35}, 0, \dfrac{1}{35}, \dfrac{12}{35} \text{ and } \dfrac{13}{35}}$

A wedding cake weighed 8 kg. If $\dfrac{2}{5}$ th of its weight was flour, $\dfrac{5}{16}$ th was sugar, $\dfrac{1}{4}$ th was cream and the rest were nuts, find the weight of nuts.

Answer

First find the fraction of the cake which was nuts, then find that fraction of the total weight.

Total weight of the wedding cake = 8 kg

Finding the fraction of the cake which was flour, sugar and cream,

$\Rightarrow \dfrac{2}{5} + \dfrac{5}{16} + \dfrac{1}{4} \\[1em] [\text{ LCM of } 5, 16 \text{ and } 4 = 80] \\[1em] \Rightarrow \dfrac{2 \times 16 + 5 \times 5 + 1 \times 20}{80} \\[1em] \Rightarrow \dfrac{32 + 25 + 20}{80} \\[1em] \Rightarrow \dfrac{77}{80}$

Finding the fraction of the cake which was nuts,

$\Rightarrow 1 - \dfrac{77}{80} \\[1em] \Rightarrow \dfrac{80 - 77}{80} \\[1em] \Rightarrow \dfrac{3}{80}$

Finding the weight of nuts,

$\Rightarrow \dfrac{3}{80} \times 8 \\[1em] \Rightarrow \dfrac{24}{80} \\[1em] \Rightarrow \dfrac{3}{10} \text{ kg} \\[1em] \Rightarrow \dfrac{3}{10} \times 1000 \text{ g} = 300 \text{ g}$

Hence, the weight of nuts is $\mathbf{300 \text{ g}}$

If $p = \dfrac{-4}{9}, q = \dfrac{2}{3} \text{ and } r = -\dfrac{8}{11}$, then verify:

$p + (q + r) = (p + q) + r$

Answer

To verify the associative property of addition, we have to show that $p + (q + r) = (p + q) + r$

Solving L.H.S.,

$\Rightarrow q + r = \dfrac{2}{3} + \left(\dfrac{-8}{11}\right) \\[1em] = \dfrac{2 \times 11 + (-8) \times 3}{33} \\[1em] = \dfrac{22 - 24}{33} \\[1em] = \dfrac{-2}{33}$

Calculating, $p + (q + r)$ now :

$\Rightarrow p + (q + r) = \dfrac{-4}{9} + \dfrac{-2}{33} \\[1em] = \dfrac{(-4) \times 11 + (-2) \times 3}{99} \\[1em] = \dfrac{-44 - 6}{99} \\[1em] = \dfrac{-50}{99}$

Solving R.H.S.,

$\Rightarrow p + q = \dfrac{-4}{9} + \dfrac{2}{3} \\[1em] = \dfrac{-4 + 2 \times 3}{9} \\[1em] = \dfrac{-4 + 6}{9} \\[1em] = \dfrac{2}{9}$

Calculating, $(p + q) + r$ now :

$\Rightarrow (p + q) + r = \dfrac{2}{9} + \left(\dfrac{-8}{11}\right) \\[1em] = \dfrac{2 \times 11 + (-8) \times 9}{99} \\[1em] = \dfrac{22 - 72}{99} \\[1em] = \dfrac{-50}{99}$

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

$\Rightarrow p + (q + r) = (p + q) + r$

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

If $p = \dfrac{-4}{9}, q = \dfrac{2}{3} \text{ and } r = -\dfrac{8}{11}$, then verify:

$p \times q = q \times p$

Answer

To verify the commutative property of multiplication, we have to show that $p \times q = q \times p$

Solving L.H.S.,

$\Rightarrow p \times q = \dfrac{-4}{9} \times \dfrac{2}{3} \\[1em] = \dfrac{(-4) \times 2}{9 \times 3} \\[1em] = \dfrac{-8}{27}$

Solving R.H.S.,

$\Rightarrow q \times p = \dfrac{2}{3} \times \dfrac{-4}{9} \\[1em] = \dfrac{2 \times (-4)}{3 \times 9} \\[1em] = \dfrac{-8}{27}$

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

$\Rightarrow p \times q = q \times p$

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

If $p = \dfrac{-4}{9}, q = \dfrac{2}{3} \text{ and } r = -\dfrac{8}{11}$, then verify:

$p \times (q + r) = p \times q + p \times r$

Answer

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

Solving L.H.S.,

Here $q + r = \dfrac{-2}{33}$ (found in part (i))

$\Rightarrow p \times (q + r) = \dfrac{-4}{9} \times \dfrac{-2}{33} \\[1em] = \dfrac{(-4) \times (-2)}{9 \times 33} \\[1em] = \dfrac{8}{297}$

Solving R.H.S.,

Here $p \times q = \dfrac{-8}{27}$ (found in part (ii))

$\Rightarrow p \times r = \dfrac{-4}{9} \times \left(\dfrac{-8}{11}\right) \\[1em] = \dfrac{(-4) \times (-8)}{9 \times 11} \\[1em] = \dfrac{32}{99}$

Calculating, $p \times q + p \times r$ now :

$\Rightarrow p \times q + p \times r = \dfrac{-8}{27} + \dfrac{32}{99} \\[1em] [\text{ LCM of } 27 \text{ and } 99 = 297] \\[1em] = \dfrac{(-8) \times 11 + 32 \times 3}{297} \\[1em] = \dfrac{-88 + 96}{297} \\[1em] = \dfrac{8}{297}$

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

$\Rightarrow p \times (q + r) = p \times q + p \times r$

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

If $p = \dfrac{-4}{9}, q = \dfrac{2}{3} \text{ and } r = -\dfrac{8}{11}$, then verify:

$(p + q) \div r = p \div r + q \div r$

Answer

To verify the given result, we have to show that $(p + q) \div r = p \div r + q \div r$

Solving L.H.S.,

Here $p + q = \dfrac{2}{9}$ (found in part (i))

$\Rightarrow (p + q) \div r = \dfrac{2}{9} \div \left(\dfrac{-8}{11}\right) \\[1em] = \dfrac{2}{9} \times \dfrac{11}{-8} \\[1em] = \dfrac{22}{-72} \\[1em] = \dfrac{-11}{36}$

Solving R.H.S.,

$\Rightarrow p \div r = \dfrac{-4}{9} \div \left(\dfrac{-8}{11}\right) \\[1em] = \dfrac{-4}{9} \times \dfrac{11}{-8} \\[1em] = \dfrac{-44}{-72} \\[1em] = \dfrac{11}{18}$

Calculating, $q \div r$ now :

$\Rightarrow q \div r = \dfrac{2}{3} \div \left(\dfrac{-8}{11}\right) \\[1em] = \dfrac{2}{3} \times \dfrac{11}{-8} \\[1em] = \dfrac{22}{-24} \\[1em] = \dfrac{-11}{12}$

Calculating, $p \div r + q \div r$ now :

$\Rightarrow p \div r + q \div r = \dfrac{11}{18} + \left(\dfrac{-11}{12}\right) \\[1em] [\text{ LCM of } 18 \text{ and } 12 = 36] \\[1em] = \dfrac{11 \times 2 + (-11) \times 3}{36} \\[1em] = \dfrac{22 - 33}{36} \\[1em] = \dfrac{-11}{36}$

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

$\Rightarrow (p + q) \div r = p \div r + q \div r$

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