Rational Numbers - Exercise 4(D)

Find the additive inverse of :

(i) 9

(ii) -11

(iii) $\dfrac{-8}{13}$

(iv) $\dfrac{5}{-6}$

(v) 0

Answer

(i) 9

Since, 9 + (-9) = 0

∴ Additive inverse of 9 is -9.

(ii) -11

Since, -11 + 11 = 0

∴ Additive inverse of -11 is 11.

(iii) $\dfrac{-8}{13}$

For a rational number $\dfrac{a}{b}$, the additive inverse is $\dfrac{-a}{b}$.

$\dfrac{-8}{13}$ + $\dfrac{8}{13}$ = 0

∴ Additive inverse of $\dfrac{-8}{13}$ is $\dfrac{8}{13}$.

(iv) $\dfrac{5}{-6}$

First, express the number with a positive denominator:

$\dfrac{5}{-6} = \dfrac{5 \times (-1)}{-6 \times (-1)} = \dfrac{-5}{6}$.

Now, find the additive inverse:

$\dfrac{-5}{6}$ + $\dfrac{5}{6}$ = 0

∴ Additive inverse of $\dfrac{5}{-6} = \dfrac{5}{6}$.

(v) 0

Zero is its own additive inverse because 0 + 0 = 0.

∴ Additive inverse of 0 is 0.

Subtract :

(i) $\dfrac{3}{5}$ from $\dfrac{1}{2}$

(ii) $\dfrac{-4}{7}$ from $\dfrac{2}{3}$

(iii) $\dfrac{-5}{6}$ from $\dfrac{-3}{4}$

(iv) $\dfrac{-7}{9}$ from 0

(v) 4 from $\dfrac{-6}{11}$

(vi) $\dfrac{3}{8}$ from $\dfrac{-5}{6}$

Answer

(i) $\dfrac{3}{5}$ from $\dfrac{1}{2}$

We have:

$\phantom{=} \Big(\dfrac{1}{2} - \dfrac{3}{5}\Big) \\[1em] = \dfrac{1}{2} + \Big(\text{additive inverse of } \dfrac{3}{5}\Big) \\[1em] = \dfrac{1}{2} + \dfrac{-3}{5}$

The L.C.M. of 2 and 5 is 10.

Now, expressing each fraction with denominator 10:

$= \dfrac{1 \times 5}{2 \times 5} + \dfrac{-3 \times 2}{5 \times 2} \\[1em] = \dfrac{5}{10} + \dfrac{-6}{10} \\[1em] = \dfrac{5 + (-6)}{10} \\[1em] = \dfrac{-1}{10}$

Hence, the answer is $\dfrac{-1}{10}$

(ii) $\dfrac{-4}{7}$ from $\dfrac{2}{3}$

We have:

$\phantom{=} \Big(\dfrac{2}{3} - \dfrac{-4}{7}\Big) \\[1em] = \dfrac{2}{3} + \Big(\text{additive inverse of } \dfrac{-4}{7}\Big) \\[1em] = \dfrac{2}{3} + \dfrac{4}{7}$

The L.C.M. of 3 and 7 is 21.

Now, expressing each fraction with denominator 21:

$= \dfrac{2 \times 7}{3 \times 7} + \dfrac{4 \times 3}{7 \times 3} \\[1em] = \dfrac{14}{21} + \dfrac{12}{21} \\[1em] = \dfrac{14 + 12}{21} \\[1em] = \dfrac{26}{21}$

Hence, the answer is $\dfrac{26}{21}$

(iii) $\dfrac{-5}{6}$ from $\dfrac{-3}{4}$

We have:

$\phantom{=} \Big(\dfrac{-3}{4} - \dfrac{-5}{6}\Big) \\[1em] = \dfrac{-3}{4} + \Big(\text{additive inverse of } \dfrac{-5}{6}\Big) \\[1em] = \dfrac{-3}{4} + \dfrac{5}{6}$

The L.C.M. of 4 and 6 is 12.

Now, expressing each fraction with denominator 12:

$= \dfrac{-3 \times 3}{4 \times 3} + \dfrac{5 \times 2}{6 \times 2} \\[1em] = \dfrac{-9}{12} + \dfrac{10}{12} \\[1em] = \dfrac{-9 + 10}{12} \\[1em] = \dfrac{1}{12}$

Hence, the answer is $\dfrac{1}{12}$

(iv) $\dfrac{-7}{9}$ from 0

We have:

$\phantom{=} \Big(0 - \dfrac{-7}{9}\Big) \\[1em] = 0 + \Big(\text{additive inverse of } \dfrac{-7}{9}\Big) \\[1em] = 0 + \dfrac{7}{9} \\[1em] = \dfrac{7}{9}$

Hence, the answer is $\dfrac{7}{9}$

(v) 4 from $\dfrac{-6}{11}$

we have:

$\phantom{=} \Big(\dfrac{-6}{11} - 4\Big) \\[1em] = \dfrac{-6}{11} + \Big(\text{additive inverse of } 4\Big) \\[1em] = \dfrac{-6}{11} + \dfrac{-4}{1}$

The L.C.M. of 11 and 1 is 11.

Now, expressing each fraction with denominator 11:

$= \dfrac{-6}{11} + \dfrac{-4 \times 11}{1 \times 11} \\[1em] = \dfrac{-6}{11} + \dfrac{-44}{11} \\[1em] = \dfrac{-6 + (-44)}{11} \\[1em] = \dfrac{-50}{11}$

Hence, the answer is $\dfrac{-50}{11}$

(vi) $\dfrac{3}{8}$ from $\dfrac{-5}{6}$

we have:

$\phantom{=} \Big(\dfrac{-5}{6} - \dfrac{3}{8}\Big) \\[1em] = \dfrac{-5}{6} + \Big(\text{additive inverse of } \dfrac{3}{8}\Big) \\[1em] = \dfrac{-5}{6} + \dfrac{-3}{8}$

The L.C.M. of 6 and 8 is 24.

Now, expressing each fraction with denominator 24:

$= \dfrac{-5 \times 4}{6 \times 4} + \dfrac{-3 \times 3}{8 \times 3} \\[1em] = \dfrac{-20}{24} + \dfrac{-9}{24} \\[1em] = \dfrac{-20 + (-9)}{24} \\[1em] = \dfrac{-29}{24}$

Hence, the answer is $\dfrac{-29}{24}$

Evaluate :

(i) $\dfrac{5}{6} - \dfrac{7}{8}$

(ii) $\dfrac{5}{12} - \dfrac{17}{18}$

(iii) $\dfrac{11}{15} - \dfrac{13}{20}$

(iv) $\dfrac{-5}{9} - \dfrac{-2}{3}$

(v) $\dfrac{6}{11} - \dfrac{-3}{4}$

(vi) $\dfrac{-2}{3} - \dfrac{3}{4}$

Answer

(i) $\dfrac{5}{6} - \dfrac{7}{8}$

We have:

$\phantom{=} \Big(\dfrac{5}{6} - \dfrac{7}{8}\Big) \\[1em] = \dfrac{5}{6} + \Big(\text{additive inverse of } \dfrac{7}{8}\Big) \\[1em] = \dfrac{5}{6} + \dfrac{-7}{8}$

L.C.M. of 6 and 8 is 24.

Now, expressing each fraction with denominator 24:

$= \dfrac{5 \times 4}{6 \times 4} + \dfrac{-7 \times 3}{8 \times 3} \\[1em] = \dfrac{20}{24} + \dfrac{-21}{24} \\[1em] = \dfrac{20 + (-21)}{24} \\[1em] = \dfrac{-1}{24}$

Hence, the answer is $\dfrac{-1}{24}$

(ii) $\dfrac{5}{12} - \dfrac{17}{18}$

We have:

$\phantom{=} \Big(\dfrac{5}{12} - \dfrac{17}{18}\Big) \\[1em] = \dfrac{5}{12} + \Big(\text{additive inverse of } \dfrac{17}{18}\Big) \\[1em] = \dfrac{5}{12} + \dfrac{-17}{18}$

L.C.M. of 12 and 18 is 36.

Now, expressing each fraction with denominator 36:

$= \dfrac{5 \times 3}{12 \times 3} + \dfrac{-17 \times 2}{18 \times 2} \\[1em] = \dfrac{15}{36} + \dfrac{-34}{36} \\[1em] = \dfrac{15 + (-34)}{36} \\[1em] = \dfrac{-19}{36}$

Hence, the answer is $\dfrac{-19}{36}$

(iii) $\dfrac{11}{15} - \dfrac{13}{20}$

we have:

$\phantom{=} \Big(\dfrac{11}{15} - \dfrac{13}{20}\Big) \\[1em] = \dfrac{11}{15} + \Big(\text{additive inverse of } \dfrac{13}{20}\Big) \\[1em] = \dfrac{11}{15} + \dfrac{-13}{20}$

L.C.M. of 15 and 20 is 60.

Now, expressing each fraction with denominator 60:

$= \dfrac{11 \times 4}{15 \times 4} + \dfrac{-13 \times 3}{20 \times 3} \\[1em] = \dfrac{44}{60} + \dfrac{-39}{60}\\[1em] = \dfrac{44 + (-39)}{60} \\[1em] = \dfrac{5}{60}$

Hence, the answer is $\dfrac{5}{60}$

(iv) $\dfrac{-5}{9} - \dfrac{-2}{3}$

We have:

$\phantom{=} \Big(\dfrac{-5}{9} - \dfrac{-2}{3}\Big) \\[1em] = \dfrac{-5}{9} + \Big(\text{additive inverse of } \dfrac{-2}{3}\Big) \\[1em] = \dfrac{-5}{9} + \dfrac{2}{3}$

L.C.M. of 9 and 3 is 9.

Now, expressing each fraction with denominator 9:

$= \dfrac{-5}{9} + \dfrac{2 \times 3}{3 \times 3} \\[1em] = \dfrac{-5}{9} + \dfrac{6}{9} \\[1em] = \dfrac{-5 + 6}{9} \\[1em] = \dfrac{1}{9}$

Hence, the answer is $\dfrac{1}{9}$

(v) $\dfrac{6}{11} - \dfrac{-3}{4}$

We have:

$\phantom{=} \Big(\dfrac{6}{11} - \dfrac{-3}{4}\Big) \\[1em] = \dfrac{6}{11} + \Big(\text{additive inverse of } \dfrac{-3}{4}\Big) \\[1em] = \dfrac{6}{11} + \dfrac{3}{4}$

L.C.M. of 11 and 4 is 44.

Now, expressing each fraction with denominator 44:

$= \dfrac{6 \times 4}{11 \times 4} + \dfrac{3 \times 11}{4 \times 11} \\[1em] = \dfrac{24}{44} + \dfrac{33}{44} \\[1em] = \dfrac{24 + 33}{44} \\[1em] = \dfrac{57}{44}$

Hence, the answer is $\dfrac{57}{44}$

(vi) $\dfrac{-2}{3} - \dfrac{3}{4}$

We have:

$\phantom{=} \Big(\dfrac{-2}{3} - \dfrac{3}{4}\Big) \\[1em] = \dfrac{-2}{3} + \Big(\text{additive inverse of } \dfrac{3}{4}\Big) \\[1em] = \dfrac{-2}{3} + \dfrac{-3}{4}$

L.C.M. of 3 and 4 is 12.

Now, expressing each fraction with denominator 12:

$= \dfrac{-2 \times 4}{3 \times 4} + \dfrac{-3 \times 3}{4 \times 3} \\[1em] = \dfrac{-8}{12} + \dfrac{-9}{12} \\[1em] = \dfrac{-8 + (-9)}{12} \\[1em] = \dfrac{-17}{12}$

Hence, the answer is $\dfrac{-17}{12}$

The sum of two rational numbers is $\dfrac{-5}{8}$. If one of them is $\dfrac{7}{16}$, find the other.

Answer

Given:

Let p and q be two rational numbers.

One rational number = p = $\dfrac{7}{16}$

Other rational number = q = ?

Sum of two rational numbers = (p + q) = $\dfrac{-5}{8}$

q = $\dfrac{-5}{8}$ - p

Substituting the values in above, we get:

$q = \dfrac{-5}{8} - \dfrac{7}{16} \\[1em] = \dfrac{-5}{8} + \Big(\text{additive inverse of } \dfrac{7}{16}\Big) \\[1em] q = \dfrac{-5}{8} + \dfrac{-7}{16}$

L.C.M. of 8 and 16 is 16.

Now, expressing each fraction with denominator 16:

$\dfrac{-5 \times 2}{8 \times 2} + \dfrac{-7 \times 1}{16 \times 1} \\[1em] = \dfrac{-10}{16} + \dfrac{-7}{16} \\[1em] = \dfrac{-10 + (-7)}{16} \\[1em] = \dfrac{-17}{16}$

The other rational number q is $\dfrac{-17}{16}$.

The sum of two rational numbers is -4. If one of them is $\dfrac{-3}{5}$, find the other.

Answer

Let p and q be two rational numbers.

One rational number = p = $\dfrac{-3}{5}$

Other rational number = q = ?

Sum of two rational numbers = (p + q) = -4

q = -4 - p

Substituting the values in above, we get:

$q = -4 - \Big(\dfrac{-3}{5}\Big) \\[1em] = -4 + \Big(\text{additive inverse of } \dfrac{-3}{5}\Big) \\[1em] q = \dfrac{-4}{1} + \dfrac{3}{5}$

L.C.M. of 1 and 5 is 5.

Now, expressing each fraction with denominator 5:

$\dfrac{-4 \times 5}{1 \times 5} + \dfrac{3 \times 1}{5 \times 1} \\[1em] = \dfrac{-20}{5} + \dfrac{3}{5} \\[1em] = \dfrac{-20 + 3}{5} \\[1em] = \dfrac{-17}{5}$

The other rational number q is $\dfrac{-17}{5}$.

The sum of two rational numbers is $\dfrac{-5}{4}$. If one of them is -3, find the other.

Answer

Let p and q be two rational numbers.

One rational number = p = -3

Other rational number = q = ?

Sum of two rational numbers = (p + q) = $\dfrac{-5}{4}$

q = $\dfrac{-5}{4}$ - p

Substituting the values in above, we get:

$q = \dfrac{-5}{4} - (-3) \\[1em] = \dfrac{-5}{4} + \Big(\text{additive inverse of } -3\Big) \\[1em] q = \dfrac{-5}{4} + \dfrac{3}{1}$

L.C.M. of 4 and 1 is 4.

Now, expressing each fraction with denominator 4:

$\dfrac{-5 \times 1}{4 \times 1} + \dfrac{3 \times 4}{1 \times 4} \\[1em] = \dfrac{-5}{4} + \dfrac{12}{4} \\[1em] = \dfrac{-5 + 12}{4} \\[1em] = \dfrac{7}{4}$

The other rational number q is $\dfrac{7}{4}$.

What should be added to $\dfrac{-5}{6}$ to get $\dfrac{-2}{3}$ ?

Answer

Let the required number be x. Then,

$\dfrac{-5}{6} + x = \dfrac{-2}{3} \\[1em] \Rightarrow x = \dfrac{-2}{3} - \left(\dfrac{-5}{6}\right) \\[1em] = \dfrac{-2}{3} + \left(\text{additive inverse of } \dfrac{-5}{6}\right) \\[1em] = \dfrac{-2}{3} + \dfrac{5}{6}$

L.C.M. of denominators 3 and 6 is 6.

Now, expressing each fraction with denominator 6:

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

The required number is $\dfrac{1}{6}$.

What should be added to $\dfrac{2}{5}$ to get -1 ?

Answer

Let the required number be x. Then,

$\dfrac{2}{5} + x = -1 \\[1em] \Rightarrow x = -1 - \dfrac{2}{5} \\[1em] = \dfrac{-1}{1} + \left(\text{additive inverse of } \dfrac{2}{5}\right) \\[1em] = \dfrac{-1}{1} + \dfrac{-2}{5}$

L.C.M. of denominators 1 and 5 is 5.

Now, expressing each fraction with denominator 5:

$\dfrac{-1 \times 5}{1 \times 5} + \dfrac{-2 \times 1}{5 \times 1} \\[1em] = \dfrac{-5}{5} + \dfrac{-2}{5} \\[1em] = \dfrac{-5 + (-2)}{5} \\[1em] = \dfrac{-7}{5}$

The required number is $\dfrac{-7}{5}$.

What should be subtracted from $\dfrac{-3}{4}$ to get $\dfrac{-5}{6}$

Answer

Let the required number be x. Then,

$\dfrac{-3}{4} - x = \dfrac{-5}{6} \\[1em] \Rightarrow \dfrac{-3}{4} = \dfrac{-5}{6} + x \\[1em] \Rightarrow x = \dfrac{-3}{4} - \left(\dfrac{-5}{6}\right) \\[1em] = \dfrac{-3}{4} + \left(\text{additive inverse of } \dfrac{-5}{6}\right) \\[1em] x = \dfrac{-3}{4} + \dfrac{5}{6}$

L.C.M. of denominators 4 and 6 is 12.

Now, expressing each fraction with denominator 12:

$\dfrac{-3 \times 3}{4 \times 3} + \dfrac{5 \times 2}{6 \times 2} \\[1em] = \dfrac{-9}{12} + \dfrac{10}{12} \\[1em] = \dfrac{-9 + 10}{12} \\[1em] = \dfrac{1}{12}$

The required number is $\dfrac{1}{12}$.

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

Answer

Let the required number be x. Then,

$\dfrac{-2}{3} - x = 1 \\[1em] \Rightarrow \dfrac{-2}{3} = 1 + x \\[1em] \Rightarrow x = \dfrac{-2}{3} - 1 \\[1em] = \dfrac{-2}{3} + (\text{additive inverse of } 1) \\[1em] = \dfrac{-2}{3} + \dfrac{-1}{1}$

L.C.M. of denominators 3 and 1 is 3.

Now, expressing each fraction with denominator 3:

$\dfrac{-2 \times 1}{3 \times 1} + \dfrac{-1 \times 3}{1 \times 3} \\[1em] = \dfrac{-2}{3} + \dfrac{-3}{3} \\[1em] = \dfrac{-2 + (-3)}{3} = \dfrac{-5}{3}$

The required number is $\dfrac{-5}{3}$.