Rational Numbers — Exercise 1(B)

Find the additive inverse of:

$\dfrac{9}{13}$

Answer

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

Changing the sign of $\dfrac{9}{13}$, we get $\dfrac{-9}{13}$.

[Check : $\dfrac{9}{13} + \dfrac{-9}{13} = \dfrac{0}{13} = 0$]

Hence, the additive inverse of $\mathbf{\dfrac{9}{13} \text{ is } \dfrac{-9}{13}}$.

Find the additive inverse of:

$\dfrac{16}{7}$

Answer

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

Changing the sign of $\dfrac{16}{7}$, we get $\dfrac{-16}{7}$.

[Check : $\dfrac{16}{7} + \dfrac{-16}{7} = \dfrac{0}{7} = 0$]

Hence, the additive inverse of $\mathbf{\dfrac{16}{7} \text{ is } \dfrac{-16}{7}}$.

Find the additive inverse of:

$\dfrac{-3}{23}$

Answer

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

Changing the sign of $\dfrac{-3}{23}$, we get $\dfrac{3}{23}$.

[Check : $\dfrac{-3}{23} + \dfrac{3}{23} = \dfrac{0}{23} = 0$]

Hence, the additive inverse of $\mathbf{\dfrac{-3}{23} \text{ is } \dfrac{3}{23}}$.

Find the additive inverse of:

$\dfrac{8}{-11}$

Answer

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

First express the rational number with a positive denominator.

$\dfrac{8}{-11} = \dfrac{8 \times (-1)}{-11 \times (-1)} = \dfrac{-8}{11}$

Changing the sign of $\dfrac{-8}{11}$, we get $\dfrac{8}{11}$.

[Check : $\dfrac{-8}{11} + \dfrac{8}{11} = \dfrac{0}{11} = 0$]

Hence, the additive inverse of $\mathbf{\dfrac{8}{-11} \text{ is } \dfrac{8}{11}}$.

Find the additive inverse of:

$\dfrac{-22}{15}$

Answer

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

Changing the sign of $\dfrac{-22}{15}$, we get $\dfrac{22}{15}$.

[Check: $\dfrac{-22}{15} + \dfrac{22}{15} = \dfrac{0}{15} = 0$]

Hence, the additive inverse of $\mathbf{\dfrac{-22}{15} \text{ is } \dfrac{22}{15}}$.

Find the additive inverse of:

$\dfrac{-11}{-9}$

Answer

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

First express the rational number with a positive denominator.

$\dfrac{-11}{-9} = \dfrac{-11 \times (-1)}{-9 \times (-1)} = \dfrac{11}{9}$

Changing the sign of $\dfrac{11}{9}$, we get $\dfrac{-11}{9}$.

[Check : $\dfrac{11}{9} + \dfrac{-11}{9} = \dfrac{0}{9} = 0$]

Hence, the additive inverse of $\mathbf{\dfrac{-11}{-9} \text{ is } \dfrac{-11}{9}}$.

Find the sum:

$\dfrac{-7}{17} + \dfrac{6}{17}$

Answer

Both rational numbers have the same (positive) denominator 17, so add the numerators keeping the denominator the same.

$\dfrac{-7}{17} + \dfrac{6}{17} = \dfrac{(-7) + 6}{17} \\[1em] = \dfrac{-1}{17}$

Hence, $\mathbf{\dfrac{-7}{17} + \dfrac{6}{17} = \dfrac{-1}{17}}$.

Find the sum:

$\dfrac{-5}{12} + \dfrac{7}{-12}$

Answer

First express each rational number with a positive denominator.

$\dfrac{7}{-12} = \dfrac{7 \times (-1)}{-12 \times (-1)} = \dfrac{-7}{12}$

Both rational numbers now have the same (positive) denominator 12, so add the numerators keeping the denominator the same.

$\dfrac{-5}{12} + \dfrac{-7}{12} = \dfrac{(-5) + (-7)}{12} \\[1em] = \dfrac{-12}{12} \\[1em] = -1$

Hence, $\mathbf{\dfrac{-5}{12} + \dfrac{7}{-12} = -1}$.

Find the sum:

$\dfrac{8}{15} + \dfrac{5}{12}$

Answer

The denominators are different, so express the rational numbers with the LCM of their denominators as the common denominator and then add.

LCM of 15 and 12 = 60

$\dfrac{8}{15} + \dfrac{5}{12} = \dfrac{8 \times 4 + 5 \times 5}{60} \\[1em] = \dfrac{32 + 25}{60} \\[1em] = \dfrac{57}{60} \\[1em] = \dfrac{19}{20}$

Hence, $\mathbf{\dfrac{8}{15} + \dfrac{5}{12} = \dfrac{19}{20}}$.

Find the sum:

$\dfrac{-11}{18} + \dfrac{5}{-12}$

Answer

First express each rational number with a positive denominator.

$\dfrac{5}{-12} = \dfrac{5 \times (-1)}{-12 \times (-1)} = \dfrac{-5}{12}$

The denominators are different, so express the rational numbers with the LCM of their denominators as the common denominator and then add.

LCM of 18 and 12 = 36

$\dfrac{-11}{18} + \dfrac{-5}{12} = \dfrac{(-11) \times 2 + (-5) \times 3}{36} \\[1em] = \dfrac{-22 - 15}{36} \\[1em] = \dfrac{-37}{36} \\[1em] = -1\dfrac{1}{36}$

Hence, $\mathbf{\dfrac{-11}{18} + \dfrac{5}{-12} = -1\dfrac{1}{36}}$.

Find the sum:

$\dfrac{-11}{6} + \dfrac{-3}{4} + \dfrac{5}{8} + \dfrac{-7}{3}$

Answer

All the denominators are positive but different, so express the rational numbers with the LCM of their denominators as the common denominator and then add.

LCM of 6, 4, 8 and 3 = 24

$\dfrac{-11}{6} + \dfrac{-3}{4} + \dfrac{5}{8} + \dfrac{-7}{3} = \dfrac{(-11) \times 4 + (-3) \times 6 + 5 \times 3 + (-7) \times 8}{24} \\[1em] = \dfrac{-44 - 18 + 15 - 56}{24} \\[1em] = \dfrac{15 - 118}{24} \\[1em] = \dfrac{-103}{24} \\[1em] = -4\dfrac{7}{24}$

Hence, $\mathbf{\dfrac{-11}{6} + \dfrac{-3}{4} + \dfrac{5}{8} + \dfrac{-7}{3} = -4\dfrac{7}{24}}$.

Find the sum:

$\dfrac{4}{7} + \dfrac{2}{-3} + \dfrac{5}{21} + \dfrac{-8}{9}$

Answer

First express each rational number with a positive denominator.

$\dfrac{2}{-3} = \dfrac{2 \times (-1)}{-3 \times (-1)} = \dfrac{-2}{3}$

The denominators are different, so express the rational numbers with the LCM of their denominators as the common denominator and then add.

LCM of 7, 3, 21 and 9 = 63

$\dfrac{4}{7} + \dfrac{-2}{3} + \dfrac{5}{21} + \dfrac{-8}{9} = \dfrac{4 \times 9 + (-2) \times 21 + 5 \times 3 + (-8) \times 7}{63} \\[1em] = \dfrac{36 - 42 + 15 - 56}{63} \\[1em] = \dfrac{51 - 98}{63} \\[1em] = \dfrac{-47}{63}$

Hence, $\mathbf{\dfrac{4}{7} + \dfrac{2}{-3} + \dfrac{5}{21} + \dfrac{-8}{9} = \dfrac{-47}{63}}$.

Subtract:

$\dfrac{2}{3}$ from $\dfrac{5}{6}$

Answer

To subtract one rational number from another, add the additive inverse of the rational number to be subtracted.

Subtracting $\dfrac{2}{3}$ from $\dfrac{5}{6}$,

LCM of 6 and 3 = 6

$\dfrac{5}{6} - \dfrac{2}{3} = \dfrac{5 - 2 \times 2}{6} \\[1em] = \dfrac{5 - 4}{6} \\[1em] = \dfrac{1}{6}$

Hence, the required difference is $\mathbf{\dfrac{1}{6}}$.

Subtract:

$\dfrac{-2}{5}$ from $\dfrac{-5}{7}$

Answer

To subtract one rational number from another, add the additive inverse of the rational number to be subtracted.

Subtracting $\dfrac{-2}{5}$ from $\dfrac{-5}{7}$,

LCM of 7 and 5 = 35

$\dfrac{-5}{7} - \left(\dfrac{-2}{5}\right) = \dfrac{-5}{7} + \dfrac{2}{5} \\[1em] = \dfrac{(-5) \times 5 + 2 \times 7}{35} \\[1em] = \dfrac{-25 + 14}{35} \\[1em] = \dfrac{-11}{35}$

Hence, the required difference is $\mathbf{\dfrac{-11}{35}}$.

Subtract:

$\dfrac{4}{9}$ from $\dfrac{-7}{8}$

Answer

To subtract one rational number from another, add the additive inverse of the rational number to be subtracted.

Subtracting $\dfrac{4}{9}$ from $\dfrac{-7}{8}$,

LCM of 8 and 9 = 72

$\dfrac{-7}{8} - \dfrac{4}{9} = \dfrac{(-7) \times 9 - 4 \times 8}{72} \\[1em] = \dfrac{-63 - 32}{72} \\[1em] = \dfrac{-95}{72} \\[1em] = -1\dfrac{23}{72}$

Hence, the required difference is $\mathbf{-1\dfrac{23}{72}}$.

Subtract:

$\dfrac{-11}{6}$ from $\dfrac{8}{3}$

Answer

To subtract one rational number from another, add the additive inverse of the rational number to be subtracted.

Subtracting $\dfrac{-11}{6}$ from $\dfrac{8}{3}$,

LCM of 3 and 6 = 6

$\dfrac{8}{3} - \left(\dfrac{-11}{6}\right) = \dfrac{8}{3} + \dfrac{11}{6} \\[1em] = \dfrac{8 \times 2 + 11}{6} \\[1em] = \dfrac{16 + 11}{6} \\[1em] = \dfrac{27}{6} = \dfrac{9}{2} \\[1em] = 4\dfrac{1}{2}$

Hence, the required difference is $\mathbf{4\dfrac{1}{2}}$.

The sum of two rational numbers is $\dfrac{-4}{9}$. If one of them is $\dfrac{13}{6}$, then find the other.

[Hint: Other rational number = $\dfrac{-4}{9} - \dfrac{13}{6}$]

Answer

Let the other rational number be $x$.

As the sum of the two rational numbers is $\dfrac{-4}{9}$,

$\Rightarrow \dfrac{13}{6} + x = \dfrac{-4}{9} \\[1em] \Rightarrow x = \dfrac{-4}{9} - \dfrac{13}{6} \\[1em] [\text{ LCM of } 9 \text{ and } 6 = 18] \\[1em] \Rightarrow x = \dfrac{(-4) \times 2 - 13 \times 3}{18} \\[1em] \Rightarrow x = \dfrac{-8 - 39}{18} \\[1em] \Rightarrow x = \dfrac{-47}{18} = -2\dfrac{11}{18}$

Hence, the other rational number is $\mathbf{\dfrac{-47}{18}}$.

What number should be added to $\dfrac{-2}{3}$ to get $\dfrac{-1}{7}$?

Answer

Let the required number be $x$.

According to the given condition,

$\Rightarrow \dfrac{-2}{3} + x = \dfrac{-1}{7} \\[1em] \Rightarrow x = \dfrac{-1}{7} - \left(\dfrac{-2}{3}\right) \\[1em] \Rightarrow x = \dfrac{-1}{7} + \dfrac{2}{3} \\[1em] [\text{ LCM of } 7 \text{ and } 3 = 21] \\[1em] \Rightarrow x = \dfrac{(-1) \times 3 + 2 \times 7}{21} \\[1em] \Rightarrow x = \dfrac{-3 + 14}{21} \\[1em] \Rightarrow x = \dfrac{11}{21}$

Hence, the required number is $\mathbf{\dfrac{11}{21}}$.

What number should be subtracted from $-2$ to get $\dfrac{7}{11}$?

Answer

Let the required number be $x$.

According to the given condition,

$\Rightarrow -2 - x = \dfrac{7}{11} \\[1em] \Rightarrow x = -2 - \dfrac{7}{11} \\[1em] \Rightarrow x = \dfrac{(-2) \times 11 - 7}{11} \\[1em] \Rightarrow x = \dfrac{-22 - 7}{11} \\[1em] \Rightarrow x = \dfrac{-29}{11} = -2\dfrac{7}{11}$

Hence, the required number is $\mathbf{\dfrac{-29}{11}}$.

What number should be added to -1 so as to get $\dfrac{5}{7}$?

Answer

Let the required number be $x$.

According to the given condition,

$\Rightarrow -1 + x = \dfrac{5}{7} \\[1em] \Rightarrow x = \dfrac{5}{7} - (-1) \\[1em] \Rightarrow x = \dfrac{5}{7} + 1 \\[1em] \Rightarrow x = \dfrac{5 + 7}{7} \\[1em] \Rightarrow x = \dfrac{12}{7} = 1\dfrac{5}{7}$

Hence, the required number is $\mathbf{\dfrac{12}{7}}$.

What number should be subtracted from $-\dfrac{2}{3}$ to get $-\dfrac{1}{6}$?

[Hint: Required number = $\dfrac{-2}{3} - \left(-\dfrac{1}{6}\right)$]

Answer

Let the required number be $x$.

According to the given condition,

$\Rightarrow \dfrac{-2}{3} - x = \dfrac{-1}{6} \\[1em] \Rightarrow x = \dfrac{-2}{3} - \left(\dfrac{-1}{6}\right) \\[1em] \Rightarrow x = \dfrac{-2}{3} + \dfrac{1}{6} \\[1em] [\text{ LCM of } 3 \text{ and } 6 = 6] \\[1em] \Rightarrow x = \dfrac{(-2) \times 2 + 1}{6} \\[1em] \Rightarrow x = \dfrac{-4 + 1}{6} \\[1em] \Rightarrow x = \dfrac{-3}{6} = \dfrac{-1}{2}$

Hence, the required number is $\mathbf{\dfrac{-1}{2}}$.

State whether each of the following statements is true or false:

(i) The difference of two rational numbers is always a rational number.

(ii) $\dfrac{2}{3} - \dfrac{3}{5} = \dfrac{3}{5} - \dfrac{2}{3}$

(iii) 1 is the additive identity for rational numbers.

(iv) Subtraction is commutative on rational numbers.

(v) There exists a rational number which is its own additive inverse.

(vi) $\dfrac{19}{-5} + \dfrac{-3}{11} = \dfrac{19}{5} + \dfrac{3}{11}$

Answer

(i) Rational numbers are closed under subtraction, i.e. if $\dfrac{a}{b} \text{ and } \dfrac{c}{d}$ are two rational numbers, then $\dfrac{a}{b} - \dfrac{c}{d}$ is also a rational number.

Hence, the given statement is true.

(ii) Solving L.H.S.,

$\dfrac{2}{3} - \dfrac{3}{5} = \dfrac{2 \times 5 - 3 \times 3}{15} = \dfrac{10 - 9}{15} = \dfrac{1}{15}$

Solving R.H.S.,

$\dfrac{3}{5} - \dfrac{2}{3} = \dfrac{3 \times 3 - 2 \times 5}{15} = \dfrac{9 - 10}{15} = \dfrac{-1}{15}$

Clearly, L.H.S. ≠ R.H.S.

Hence, the given statement is false.

(iii) For every rational number $\dfrac{a}{b}$, we have $\dfrac{a}{b} + 0 = \dfrac{a}{b} = 0 + \dfrac{a}{b}$, so 0 is the additive identity and 1 is the multiplicative identity.

Hence, the given statement is false.

(iv) For any two rational numbers $\dfrac{a}{b} \text{ and } \dfrac{c}{d}$, we have $\dfrac{a}{b} - \dfrac{c}{d} ≠ \dfrac{c}{d} - \dfrac{a}{b}$. For example, $\dfrac{2}{3} - \dfrac{3}{5} = \dfrac{1}{15}$ but $\dfrac{3}{5} - \dfrac{2}{3} = \dfrac{-1}{15}$.

Hence, the given statement is false.

(v) The additive inverse of the rational number 0 is 0 itself, as 0 + 0 = 0.

Hence, the given statement is true.

(vi) Solving L.H.S.,

$\Rightarrow \dfrac{19}{-5} + \dfrac{-3}{11} \\[1em] \Rightarrow \dfrac{-19}{5} + \dfrac{-3}{11} \\[1em] \Rightarrow \dfrac{(-19) \times 11 + (-3) \times 5}{55} = \dfrac{-209 - 15}{55} = \dfrac{-224}{55}$

Solving R.H.S.,

$\Rightarrow \dfrac{19}{5} + \dfrac{3}{11} \\[1em] \Rightarrow \dfrac{19 \times 11 + 3 \times 5}{55} = \dfrac{209 + 15}{55} = \dfrac{224}{55}$

Clearly, L.H.S. ≠ R.H.S.

Hence, the given statement is false.