Rational Numbers

A number which is not rational is called an irrational number.

Hence, Option 3 is the correct option.

If x ≠ 0 then value of $\frac{0}{x}$ is a rational number.

Hence, Option 1 is the correct option.

Given,

5x + 7 = 0

⇒ 5x = -7

⇒ x = $\frac{-7}{5}$

As x is the form of $\frac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ x is a rational number.

Hence, Option 3 is the correct option.

Rational number $\frac{p}{q}$ is in standard form, if p and q have no common factor and q ≠ 0

Hence, Option 3 is the correct option.

The addition of two rational number $\frac{a}{b}$ and $\frac{c}{d}$ is commutative, if $\frac{a}{b} + \frac{c}{d}$ = $\frac{c}{d} + \frac{a}{b}$

Hence, Option 3 is the correct option.

Hence, Option 2 is the correct option.

$$\begin{aligned}\frac{-5}{8} + \frac{3}{8} \\[1em] =\frac{-5+3}{8} \\[1em] =\frac{-2}{8}\end{aligned}$$

As $\frac{-2}{8}$ is in the form of $\frac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ $\frac{-2}{8}$ is a rational number.

$$\begin{aligned}\frac{-8}{13} + \frac{-4}{13} \\[1em] =\frac{-8+(-4)}{13} \\[1em] =\frac{-12}{13}\end{aligned}$$

As $\frac{-12}{13}$ is in the form of $\frac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ $\frac{-12}{13}$ is a rational number.

$$\begin{aligned}\frac{6}{11} + \frac{-9}{11} \\[1em] =\frac{6+(-9)}{11} \\[1em] =\frac{-3}{11}\end{aligned}$$

As $\frac{-3}{11}$ is in the form of $\frac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ $\frac{-3}{11}$ is a rational number.

$$\begin{aligned}\frac{5}{-26} + \frac{8}{39} \\[1em] \frac{-5}{26} + \frac{8}{39} \\[1em]\end{aligned}$$

LCM of 26 and 39 is 2 x 3 x 13 = 78

$$\begin{aligned}\frac{-5 \times 3}{26 \times 3} + \frac{8 \times 2}{39 \times 2} \\[1em] =\frac{-15}{78} + \frac{16}{78} \\[1em] =\frac{-15+16}{78} \\[1em] =\frac{1}{78}\end{aligned}$$

As $\frac{1}{78}$ is in the form of $\frac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ $\frac{1}{78}$ is a rational number.

$$\begin{aligned}\frac{5}{-6} + \frac{2}{3} \\[1em] = \frac{-5}{6} + \frac{2}{3} \\[1em]\end{aligned}$$

LCM of 6 and 3 is 2 x 3 = 6

$$\begin{aligned}\frac{-5 \times 1}{6 \times 1} + \frac{2 \times 2}{3 \times 2} \\[1em] =\frac{-5}{6} + \frac{4}{6} \\[1em] =\frac{-5+4}{6} \\[1em] =\frac{-1}{6}\end{aligned}$$

As $\frac{-1}{6}$ is in the form of $\frac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ $\frac{-1}{6}$ is a rational number.

$$\begin{aligned}-2 + \frac{2}{5} \\[1em] = \frac{-2}{1} + \frac{2}{5} \\[1em]\end{aligned}$$

LCM of 1 and 5 is 5

$$\begin{aligned}\frac{-2 \times 5}{1 \times 5} + \frac{2 \times 1}{5 \times 1} \\[1em] =\frac{-10}{5} + \frac{2}{5} \\[1em] =\frac{-10+2}{5} \\[1em] =\frac{-8}{5}\end{aligned}$$

As $\frac{-8}{5}$ is in the form of $\frac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ $\frac{-8}{5}$ is a rational number.

$$\begin{aligned}\frac{9}{-4} + \frac{-3}{8} \\[1em] \frac{-9}{4} + \frac{-3}{8} \\[1em]\end{aligned}$$

LCM of 4 and 8 is 2 x 2 x 2 = 8

$$\begin{aligned}\frac{-9 \times 2}{4 \times 2} + \frac{-3 \times 1}{8 \times 1} \\[1em] =\frac{-18}{8} + \frac{-3}{8} \\[1em] =\frac{-18+(-3)}{8} \\[1em] =\frac{-21}{8}\end{aligned}$$

As $\frac{-21}{8}$ is in the form of $\frac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ $\frac{-21}{8}$ is a rational number.

$$\begin{aligned}\frac{7}{-18} + \frac{8}{27} \\[1em] \frac{-7}{18} + \frac{8}{27} \\[1em]\end{aligned}$$

LCM of 18 and 27 is 2 x 3 x 3 x 3 = 54

$$\begin{aligned}\frac{-7 \times 3}{18 \times 3} + \frac{8 \times 2}{27 \times 2} \\[1em] =\frac{-21}{54} + \frac{16}{54} \\[1em] =\frac{-21+16}{54} \\[1em] =\frac{-5}{54}\end{aligned}$$

As $\frac{-5}{54}$ is in the form of $\frac{p}{q}$ where p and q both are integers and q ≠ 0,

∴ $\frac{-5}{54}$ is a rational number.

LCM of 9 and 6 is 2 x 3 x 3 = 18 $$\begin{aligned}\frac{5 \times 2}{9 \times 2} + \frac{-7 \times 3}{6 \times 3} \\[1em] =\frac{10}{18} + \frac{-21}{18} \\[1em] =\frac{10 + (-21)}{18} \\[1em] =\frac{-11}{18}\end{aligned}$$

∴ $\frac{5}{9} + \frac{-7}{6}$ = $\frac{-11}{18}$

$$\begin{aligned}\frac{4}{1} + \frac{-3}{5} \\[1em]\end{aligned}$$

LCM of 1 and 5 is 5

$$\begin{aligned}\frac{4 \times 5}{1 \times 5} + \frac{-3 \times 1}{5 \times 1} \\[1em] = \frac{20}{5} + \frac{-3}{5} \\[1em] = \frac{20 + (-3)}{5} \\[1em] = \frac{17}{5} \\[1em] = 3\frac{2}{5}\end{aligned}$$

∴ $4 + \frac{3}{-5} = 3\frac{2}{5}$

$$\begin{aligned}\frac{-1}{15} + \frac{-5}{12} \\[1em]\end{aligned}$$

LCM of 15 and 12 is 2 x 2 x 3 x 5 = 60

$$\begin{aligned}\frac{-1 \times 4}{15 \times 4} + \frac{-5 \times 5}{12 \times 5} \\[1em] =\frac{-4}{60} + \frac{-25}{60} \\[1em] =\frac{-4 + (-25)}{60} \\[1em] =\frac{-29}{60}\end{aligned}$$

∴ $\frac{1}{-15} + \frac{5}{-12} = \frac{-29}{60}$

$$\begin{aligned}\frac{5}{9} + \frac{-3}{4} \\[1em]\end{aligned}$$

LCM of 9 and 4 is 2 x 2 x 3 x 3 = 36

$$\begin{aligned}\frac{5 \times 4}{9 \times 4} + \frac{-3 \times 9}{4 \times 9} \\[1em] =\frac{20}{36} + \frac{-27}{36} \\[1em] =\frac{20 + (-27)}{36} \\[1em] =\frac{-7}{36}\end{aligned}$$

∴ $\frac{5}{9} + \frac{3}{-4}$ = $\frac{-7}{36}$

$$\begin{aligned}\frac{-8}{9} + \frac{-5}{12} \\[1em]\end{aligned}$$

LCM of 9 and 12 is 2 x 2 x 3 x 3 = 36

$$\begin{aligned}\frac{-8 \times 4}{9 \times 4} + \frac{-5 \times 3}{12 \times 3} \\[1em] =\frac{-32}{36} + \frac{-15}{36} \\[1em] =\frac{-32 + (-15)}{36} \\[1em] =\frac{-47}{36}\end{aligned}$$

∴ $\frac{-8}{9}$ + $\frac{-5}{12}$ = $\frac{-47}{36}$

$\frac{0}{1} + \frac{-2}{7}$

LCM of 1 and 7 is 7

$$\begin{aligned}\frac{0 \times 7}{1 \times 7} + \frac{-2 \times 1}{7 \times 1} \\[1em] =\frac{0}{7} + \frac{-2}{7} \\[1em] =\frac{0 + (-2)}{7} \\[1em] =\frac{-2}{7}\end{aligned}$$

∴ 0 + $\frac{-2}{7}$ = $\frac{-2}{7}$

$\frac{-5}{11} + \frac{0}{1}$

LCM of 11 and 1 is 11

$$\begin{aligned}\frac{-5 \times 1}{11 \times 1} + \frac{0 \times 11}{1 \times 11} \\[1em] =\frac{-5}{11} + \frac{0}{11} \\[1em] =\frac{-5 + 0}{11} \\[1em] =\frac{-5}{11}\end{aligned}$$

∴ $\frac{5}{-11}$ + 0 = $\frac{-5}{11}$

$\frac{2}{1} + \frac{-3}{5}$

LCM of 1 and 5 is 5

$$\begin{aligned}\frac{2 \times 5}{1 \times 5} + \frac{-3 \times 1}{5 \times 1} \\[1em] =\frac{10}{5} + \frac{-3}{5} \\[1em] =\frac{10 + (-3)}{5} \\[1em] =\frac{7}{5}\end{aligned}$$

∴ 2 + $\frac{-3}{5}$ = $\frac{7}{5}$

$\frac{-4}{9} + \frac{1}{1}$

LCM of 9 and 1 is 3 x 3 = 9

$$\begin{aligned}\frac{-4 \times 1}{9 \times 1} + \frac{1 \times 9}{1 \times 9} \\[1em] =\frac{-4}{9} + \frac{9}{9} \\[1em] =\frac{-4 + 9}{9} \\[1em] =\frac{5}{9}\end{aligned}$$

∴ $\frac{4}{-9}$ + 1 = $\frac{5}{9}$

LCM of 7 and 9 is 3 x 3 x 7 = 63

$$\begin{aligned}\frac{3 \times 9}{7 \times 9} + \frac{-4 \times 7}{9 \times 7} + \frac{-11 \times 9}{7 \times 9} + \frac{7 \times 7}{9 \times 7} \\[1em] = \frac{27}{63} + \frac{-28}{63} + \frac{-99}{63} + \frac{49}{63} \\[1em] = \frac{27 + (-28) + (-99) + 49}{63} \\[1em] = \frac{-51}{63} \\[1em]\end{aligned}$$

∴ $\frac{3}{7} + \frac{-4}{9} + \frac{-11}{7} + \frac{7}{9} = \frac{-51}{63}$

LCM of 3 and 5 is 3 x 5 = 15

$$\begin{aligned}\frac{2 \times 5}{3 \times 5} + \frac{-4 \times 3}{5 \times 3} + \frac{1 \times 5}{3 \times 5} + \frac{2 \times 3}{5 \times 3} \\[1em] = \frac{10}{15} + \frac{-12}{15} + \frac{5}{15} + \frac{6}{15} \\[1em] = \frac{10 + (-12) + 5 + 6}{15} \\[1em] = \frac{9}{15} \\[1em]\end{aligned}$$

∴ $\frac{2}{3} + \frac{-4}{5} + \frac{1}{3} + \frac{2}{5} = \frac{9}{15}$

$\frac{4}{7} + \frac{0}{1} + \frac{-8}{9} + \frac{-13}{7} + \frac{17}{9}$

LCM of 7 ,1 and 9 is 3 x 3 x 7 = 63

$$\begin{aligned}\frac{4 \times 9}{7 \times 9} + \frac{0 \times 63}{1 \times 63} + \frac{-8 \times 7}{9 \times 7} + \frac{-13 \times 9}{7 \times 9} + \frac{17 \times 7}{9 \times 7} \\[1em] = \frac{36}{63} + \frac{0}{63} + \frac{-56}{63} + \frac{-117}{63} + \frac{119}{63} \\[1em] = \frac{36 + 0 + (-56) + (-117) + 119}{63} \\[1em] = \frac{-18}{63} \\[1em]\end{aligned}$$

∴ $\frac{4}{7} + 0 + \frac{-8}{9} + \frac{-13}{7} + \frac{17}{9} = \frac{-18}{63}$

$$\begin{aligned}\Big(\frac{3}{8} + \frac{-5}{8}\Big) + \Big(\frac{-5}{12} + \frac{3}{12}\Big) + \Big(\frac{3}{7} + \frac{-2}{7}\Big) \\[1em] = \frac{-2}{8} + \frac{-2}{12} + \frac{1}{7} \\[1em] = \frac{-1}{4} + \frac{-1}{6} + \frac{1}{7} \\[1em]\end{aligned}$$

LCM of 4 ,6 and 7 is 2 x 2 x 3 x 7 = 84

$$\begin{aligned}\frac{-1 \times 21}{4 \times 21} + \frac{-1 \times 14}{6 \times 14} + \frac{1 \times 12}{7 \times 12} \\[1em] = \frac{-21}{84} + \frac{-14}{84} + \frac{12}{84} \\[1em] = \frac{(-21) + (-14) + 12}{84} \\[1em] = \frac{-23}{84} \\[1em]\end{aligned}$$

∴ $\frac{3}{8} + \frac{-5}{12} + \frac{3}{7} + \frac{3}{12} + \frac{-5}{8} + \frac{-2}{7} = \frac{-23}{84}$

To prove:

$\frac{-8}{7} + \frac{5}{14} = \frac{5}{14} + \frac{-8}{7}$

Taking LHS:

$\frac{-8}{7} + \frac{5}{14}$

LCM of 7 and 14 is 2 x 7 = 14

$$\begin{aligned}= \frac{-8 \times 2}{7 \times 2} + \frac{5 \times 1}{14 \times 1} \\[1em] = \frac{-16}{14} + \frac{5}{14} \\[1em] = \frac{-16 + 5}{14} \\[1em] = \frac{-11}{14} \\[1em]\end{aligned}$$

Taking RHS:

$\frac{5}{14} + \frac{-8}{7}$

LCM of 14 and 7 is 2 x 7 = 14

$$\begin{aligned}= \frac{5 \times 1}{14 \times 1} + \frac{-8 \times 2}{7 \times 2} \\[1em] = \frac{5}{14} + \frac{-16}{14} \\[1em] = \frac{5 + (-16)}{14} \\[1em] = \frac{-11}{14} \\[1em]\end{aligned}$$

∴ LHS = RHS

Hence, $\frac{-8}{7} + \frac{5}{14} = \frac{5}{14} + \frac{-8}{7}$

So, the commutative property for the addition of the rational number is verified.

To prove:

$\frac{5}{9} + \frac{5}{-12} = \frac{5}{-12} + \frac{5}{9}$

Taking LHS: $$\begin{aligned}\frac{5}{9} + \frac{5}{-12} \\[1em] = \frac{5}{9} + \frac{-5}{12} \\[1em]\end{aligned}$$

LCM of 9 and 12 is 2 x 2 x 3 x 3 = 36

$$\begin{aligned}= \frac{5 \times 4}{9 \times 4} + \frac{-5 \times 3}{12 \times 3} \\[1em] = \frac{20}{36} + \frac{-15}{36} \\[1em] = \frac{20 + (-15)}{36} \\[1em] = \frac{5}{36} \\[1em]\end{aligned}$$

Taking RHS: $$\begin{aligned}\frac{5}{-12} + \frac{5}{9} \\[1em] = \frac{-5}{12} + \frac{5}{9} \\[1em]\end{aligned}$$

LCM of 12 and 9 is 2 x 2 x 3 x 3 = 36

$$\begin{aligned}= \frac{-5 \times 3}{12 \times 3} + \frac{5 \times 4}{9 \times 4} \\[1em] = \frac{-15}{36} + \frac{20}{36} \\[1em] = \frac{(-15) + 20}{36} \\[1em] = \frac{5}{36} \\[1em]\end{aligned}$$

∴ LHS = RHS

Hence, $\frac{5}{9} + \frac{5}{-12} = \frac{5}{-12} + \frac{5}{9}$

So, the commutative property for the addition of the rational number is verified.

To prove:

$\frac{-4}{5} + \frac{-13}{-15} = \frac{-13}{-15} + \frac{-4}{5}$

Taking LHS: $$\begin{aligned}\frac{-4}{5} + \frac{-13}{-15} \\[1em] = \frac{-4}{5} + \frac{13}{15} \\[1em]\end{aligned}$$

LCM of 5 and 15 is 3 x 5 = 15

$$\begin{aligned}= \frac{-4 \times 3}{5 \times 3} + \frac{13 \times 1}{15 \times 1} \\[1em] = \frac{-12}{15} + \frac{13}{15} \\[1em] = \frac{-12 + 13}{15} \\[1em] = \frac{1}{15} \\[1em]\end{aligned}$$

Taking RHS: $$\begin{aligned}\frac{-13}{-15} + \frac{-4}{5} \\[1em] = \frac{13}{15} + \frac{-4}{5} \\[1em]\end{aligned}$$

LCM of 15 and 5 is 3 x 5 = 15

$$\begin{aligned}= \frac{13 \times 1}{15 \times 1} + \frac{-4 \times 3}{5 \times 3} \\[1em] = \frac{13}{15} + \frac{-12}{15} \\[1em] = \frac{13 + (-12)}{15} \\[1em] = \frac{1}{15} \\[1em]\end{aligned}$$

∴ LHS = RHS

Hence, $\frac{-4}{5} + \frac{-13}{-15} = \frac{-13}{-15} + \frac{-4}{5}$

So, the commutative property for the addition of the rational number is verified.

To prove:

$\frac{2}{-5} + \frac{11}{-15} = \frac{11}{-15} + \frac{2}{-5}$

Taking LHS: $$\begin{aligned}\frac{2}{-5} + \frac{11}{-15} \\[1em] = \frac{-2}{5} + \frac{-11}{15} \\[1em]\end{aligned}$$

LCM of 5 and 15 is 3 x 5 = 15

$$\begin{aligned}= \frac{-2 \times 3}{5 \times 3} + \frac{-11 \times 1}{15 \times 1} \\[1em] = \frac{-6}{15} + \frac{-11}{15} \\[1em] = \frac{-6 + (-11)}{15} \\[1em] = \frac{-17}{15} \\[1em]\end{aligned}$$

Taking RHS: $$\begin{aligned}\frac{11}{-15} + \frac{2}{-5} \\[1em] = \frac{-11}{15} + \frac{-2}{5} \\[1em]\end{aligned}$$

LCM of 15 and 5 is 3 x 5 = 15

$$\begin{aligned}= \frac{-11 \times 1}{15 \times 1} + \frac{-2 \times 3}{5 \times 3} \\[1em] = \frac{-11}{15} + \frac{-6}{15} \\[1em] = \frac{(-11) + (-6)}{15} \\[1em] = \frac{-17}{15} \\[1em]\end{aligned}$$

∴ LHS = RHS

Hence, $\frac{2}{-5} + \frac{11}{-15} = \frac{11}{-15} + \frac{2}{-5}$

So, the commutative property for the addition of the rational number is verified.

To prove:

$3 + \frac{-2}{7} = \frac{-2}{7} + 3$

Taking LHS: $$\begin{aligned}3 + \frac{-2}{7} \\[1em] = \frac{3}{1} + \frac{-2}{7} \\[1em]\end{aligned}$$

LCM of 1 and 7 is 7

$$\begin{aligned}= \frac{3 \times 7}{1 \times 7} + \frac{-2 \times 1}{7 \times 1} \\[1em] = \frac{21}{7} + \frac{-2}{7} \\[1em] = \frac{21 + (-2)}{7} \\[1em] = \frac{19}{7} \\[1em]\end{aligned}$$

Taking RHS: $$\begin{aligned}\frac{-2}{7} + 3 \\[1em] = \frac{-2}{7} + \frac{3}{1} \\[1em]\end{aligned}$$

LCM of 7 and 1 is 7

$$\begin{aligned}= \frac{-2 \times 1}{7 \times 1} + \frac{3 \times 7}{1 \times 7} \\[1em] = \frac{-2}{7} + \frac{21}{7} \\[1em] = \frac{(-2) + 21}{7} \\[1em] = \frac{19}{7} \\[1em]\end{aligned}$$

∴ LHS = RHS

Hence, $3 + \frac{-2}{7} = \frac{-2}{7} + 3$

So, the commutative property for the addition of the rational number is verified.

To prove:

$-2 + \frac{3}{-5} = \frac{3}{-5} + -2$

Taking LHS: $$\begin{aligned}-2 + \frac{3}{-5} \\[1em] = \frac{-2}{1} + \frac{-3}{5}\end{aligned}$$

LCM of 1 and 5 is 5

$$\begin{aligned}= \frac{-2 \times 5}{1 \times 5} + \frac{-3 \times 1}{5 \times 1} \\[1em] = \frac{-10}{5} + \frac{-3}{5} \\[1em] = \frac{-10 + (-3)}{5} \\[1em] = \frac{-13}{5}\end{aligned}$$

Taking RHS: $$\begin{aligned}\frac{3}{-5} + -2 \\[1em] = \frac{-3}{5} + \frac{-2}{1}\end{aligned}$$

LCM of 5 and 1 is 5

$$\begin{aligned}= \frac{-3 \times 1}{5 \times 1} + \frac{-2 \times 5}{1 \times 5} \\[1em] = \frac{-3}{5} + \frac{-10}{5} \\[1em] = \frac{(-3) + (-10)}{5} \\[1em] = \frac{-13}{5}\end{aligned}$$

∴ LHS = RHS

Hence, $-2 + \frac{3}{-5} = \frac{3}{-5} + -2$

So, the commutative property for the addition of the rational number is verified.

To prove:

$\Big(\frac{1}{2} + \frac{2}{3}\Big) + \frac{-1}{6} = \frac{1}{2} + \Big(\frac{2}{3} + \frac{-1}{6}\Big)$

Taking LHS:

$$\begin{aligned}\Big(\frac{1}{2} + \frac{2}{3}\Big) + \frac{-1}{6} \\[1em]\end{aligned}$$ LCM of 2 and 3 is 2 x 3 = 6

$$\begin{aligned}= \Big(\frac{1 \times 3}{2 \times 3} + \frac{2 \times 2}{3 \times 2}\Big) + \frac{-1}{6} \\[1em] = \Big(\frac{3}{6} + \frac{4}{6}\Big) + \frac{-1}{6} \\[1em] = \Big(\frac{3 + 4}{6}\Big) + \frac{-1}{6} \\[1em] = \frac{7}{6}+ \frac{-1}{6} \\[1em] = \frac{7-1}{6} \\[1em] = \frac{6}{6} \\[1em] = 1\end{aligned}$$

Taking RHS: $\frac{1}{2} + \Big(\frac{2}{3} + \frac{-1}{6}\Big)$

LCM of 3 and 6 is 2 x 3 = 6 $$\begin{aligned}\frac{1}{2} + \Big(\frac{2 \times 2}{3 \times 2} + \frac{-1 \times 1}{6 \times 1}\Big) \\[1em] = \frac{1}{2} + \Big(\frac{4}{6} + \frac{-1}{6}\Big) \\[1em] = \frac{1}{2} + \Big( \frac{4 +(-1)}{6}\Big) \\[1em] = \frac{1}{2} + \frac{3}{6} \\[1em]\end{aligned}$$

LCM of 2 and 6 is 2 x 3 = 6

$$\begin{aligned}= \frac{1 \times 3}{2 \times 3} + \frac{3 \times 1}{6 \times 1} \\[1em] = \frac{3}{6} + \frac{3}{6} \\[1em] = \frac{3 + 3}{6} \\[1em] = \frac{6}{6} \\[1em] = 1\end{aligned}$$

∴ LHS = RHS

$\Big(\frac{1}{2} + \frac{2}{3}\Big) + \frac{-1}{6} = \frac{1}{2} + \Big(\frac{2}{3} + \frac{-1}{6}\Big)$

So, the associative property for the addition of the rational number is verified.

To prove:

$\Big(\frac{-2}{5} + \frac{4}{15}\Big) + \frac{-7}{10} = \frac{-2}{5} + \Big(\frac{4}{15} + \frac{-7}{10}\Big)$

Taking LHS:

$$\begin{aligned}\Big(\frac{-2}{5} + \frac{4}{15}\Big) + \frac{-7}{10} \\[1em]\end{aligned}$$ LCM of 5 and 15 is 3 x 5 = 15

$$\begin{aligned}= \Big(\frac{-2 \times 3}{5 \times 3} + \frac{4 \times 1}{15 \times 1}\Big) + \frac{-7}{10} \\[1em] = \Big(\frac{-6}{15} + \frac{4}{15}\Big) + \frac{-7}{10} \\[1em] = \Big(\frac{-6 + 4}{15}\Big) + \frac{-7}{10} \\[1em] = \frac{-2}{15}+ \frac{-7}{10} \\[1em]\end{aligned}$$ LCM of 15 and 10 is 2 x 3 x 5 = 30

$$\begin{aligned}= \frac{-2 \times 2}{15 \times 2} + \frac{-7 \times 3}{10 \times 3} \\[1em] = \frac{-4}{30} + \frac{-21}{30} \\[1em] = \frac{-4 + (-21)}{30} \\[1em] = \frac{-25}{30} \\[1em] = \frac{-5}{6} \\[1em]\end{aligned}$$

Taking RHS: $\frac{-2}{5} + \Big(\frac{4}{15} + \frac{-7}{10}\Big)$

LCM of 15 and 10 is 2 x 3 x 5 = 30 $$\begin{aligned}\frac{-2}{5} + \Big(\frac{4 \times 2}{15 \times 2} + \frac{-7 \times 3}{10 \times 3}\Big) \\[1em] = \frac{-2}{5} + \Big(\frac{8}{30} + \frac{-21}{30}\Big) \\[1em] = \frac{-2}{5} + \Big( \frac{8 +(-21)}{30}\Big) \\[1em] = \frac{-2}{5} + \frac{-13}{30} \\[1em]\end{aligned}$$

LCM of 5 and 30 is 2 x 3 x 5 = 30

$$\begin{aligned}= \frac{-2 \times 6}{5 \times 6} + \frac{-13 \times 1}{30 \times 1} \\[1em] = \frac{-12}{30} + \frac{-13}{30} \\[1em] = \frac{-12 + (-13)}{30} \\[1em] = \frac{-25}{30} \\[1em] = \frac{-5}{6} \\[1em]\end{aligned}$$

∴ LHS = RHS

$\Big(\frac{-2}{5} + \frac{4}{15}\Big) + \frac{-7}{10} = \frac{-2}{5} + \Big(\frac{4}{15} + \frac{-7}{10}\Big)$

So, the associative property for the addition of the rational number is verified.

To prove: $\Big(\frac{-7}{9} + \frac{2}{-3}\Big) + \frac{-5}{18} = \frac{-7}{9} + \Big(\frac{2}{-3} + \frac{-5}{18}\Big)$

Taking LHS:

$$\begin{aligned}\Big(\frac{-7}{9} + \frac{2}{-3}\Big) + \frac{-5}{18} \\[1em] = \Big(\frac{-7}{9} + \frac{-2}{3}\Big) + \frac{-5}{18} \\[1em]\end{aligned}$$ LCM of 9 and 3 is 3 x 3 = 9

$$\begin{aligned}= \Big(\frac{-7 \times 1}{9 \times 1} + \frac{-2 \times 3}{3 \times 3}\Big) + \frac{-5}{18} \\[1em] = \Big(\frac{-7}{9} + \frac{-6}{9}\Big) + \frac{-5}{18} \\[1em] = \Big(\frac{-7 + (-6)}{9}\Big) + \frac{-5}{18} \\[1em] = \frac{-13}{9}+ \frac{-5}{18} \\[1em]\end{aligned}$$

LCM of 9 and 18 is 2 x 9 = 18

$$\begin{aligned}= \frac{-13 \times 2}{9 \times 2} + \frac{-5 \times 1}{18 \times 1} \\[1em] = \frac{-26}{18} + \frac{-5}{18} \\[1em] = \frac{-26 + (-5)}{18} \\[1em] = \frac{-31}{18} \\[1em]\end{aligned}$$

Taking RHS:

$$\begin{aligned}\frac{-7}{9} + \Big(\frac{2}{-3} + \frac{-5}{18}\Big) \\[1em] = \frac{-7}{9} + \Big(\frac{-2}{3} + \frac{-5}{18}\Big) \\[1em]\end{aligned}$$

LCM of 3 and 18 is 2 x 3 x 9 = 18

$$\begin{aligned}\frac{-7}{9} + \Big(\frac{-2 \times 6}{3 \times 6} + \frac{-5 \times 1}{18 \times1}\Big) \\[1em] = \frac{-7}{9} + \Big(\frac{-12}{18} + \frac{-5}{18}\Big) \\[1em] = \frac{-7}{9} + \Big( \frac{-12 +(-5)}{18}\Big) \\[1em] = \frac{-7}{9} + \frac{-17}{18} \\[1em]\end{aligned}$$

LCM of 9 and 18 is 2 x 3 x 3 = 18

$$\begin{aligned}= \frac{-7 \times 2}{9 \times 2} + \frac{-17 \times 1}{18 \times 1} \\[1em] = \frac{-14}{18} + \frac{-17}{18} \\[1em] = \frac{-14 + (-17)}{18} \\[1em] = \frac{-31}{18} \\[1em]\end{aligned}$$

∴ LHS = RHS

$\Big(\frac{-7}{9} + \frac{2}{-3}\Big) + \frac{-5}{18} = \frac{-7}{9} + \Big(\frac{2}{-3} + \frac{-5}{18}\Big)$

So, the associative property for the addition of the rational number is verified.

To prove: $\Big(-1 + \frac{5}{6}\Big) + \frac{-2}{3} = -1 + \Big(\frac{5}{6} + \frac{-2}{3}\Big)$

Taking LHS:

$$\begin{aligned}\Big(-1 + \frac{5}{6}\Big) + \frac{-2}{3} \\[1em] = \Big(\frac{-1}{1} + \frac{5}{6}\Big) + \frac{-2}{3} \\[1em]\end{aligned}$$ LCM of 1 and 6 is 2 x 3 = 6

$$\begin{aligned}= \Big(\frac{-1 \times 6}{1 \times 6} + \frac{5 \times 1}{6 \times 1}\Big) + \frac{-2}{3} \\[1em] = \Big(\frac{-6}{6} + \frac{5}{6}\Big) + \frac{-2}{3} \\[1em] = \Big(\frac{-6 + 5}{6}\Big) + \frac{-2}{3} \\[1em] = \frac{-1}{6}+ \frac{-2}{3} \\[1em]\end{aligned}$$

LCM of 6 and 3 is 2 x 3 = 6

$$\begin{aligned}= \frac{-1 \times 1}{6 \times 1} + \frac{-2 \times 2}{3 \times 2} \\[1em] = \frac{-1}{6} + \frac{-4}{6} \\[1em] = \frac{-1 + (-4)}{6} \\[1em] = \frac{-5}{6} \\[1em]\end{aligned}$$

Taking RHS: $$\begin{aligned}-1 + \Big(\frac{5}{6} + \frac{-2}{3}\Big) \\[1em] = \frac{-1}{1} + \Big(\frac{5}{6} + \frac{-2}{3}\Big) \\[1em]\end{aligned}$$

LCM of 6 and 3 is 2 x 3 = 6 $$\begin{aligned}\frac{-1}{1} + \Big(\frac{5 \times 1}{6 \times 1} + \frac{-2 \times 2}{3 \times 2}\Big) \\[1em] = \frac{-1}{1} + \Big(\frac{5}{6} + \frac{-4}{6}\Big) \\[1em] = \frac{-1}{1} + \Big( \frac{5 +(-4)}{6}\Big) \\[1em] = \frac{-1}{1} + \frac{1}{6} \\[1em]\end{aligned}$$

LCM of 1 and 6 is 2 x 3 = 6

$$\begin{aligned}= \frac{-1 \times 6}{1 \times 6} + \frac{1 \times 1}{6 \times 1} \\[1em] = \frac{-6}{6} + \frac{1}{6} \\[1em] = \frac{-6 + 1}{6} \\[1em] = \frac{-5}{6} \\[1em]\end{aligned}$$

∴ LHS = RHS

$\Big(-1 + \frac{5}{6}\Big) + \frac{-2}{3} = -1 + \Big(\frac{5}{6} + \frac{-2}{3}\Big)$

So, the associative property for the addition of the rational number is verified.

Additive inverse of $\frac{-3}{8}$ = $-\Big(\frac{-3}{8}\Big)$

= $\frac{3}{8}$

$\frac{4}{-9}$ =$\frac{-4}{9}$

Additive inverse of $\frac{-4}{9}$ = $-\Big(\frac{-4}{9}\Big)$

= $\frac{4}{9}$

$\frac{-4}{-13}$ =$\frac{4}{13}$

Additive inverse of $\frac{4}{13}$ = $-\Big(\frac{4}{13}\Big)$

= $-\frac{4}{13}$

0 =$\frac{0}{1}$

Additive inverse of $\frac{0}{1}$ = $-\Big(\frac{0}{1}\Big)$

= 0

-2 =$\frac{-2}{1}$

Additive inverse of $\frac{-2}{1}$ = $-\Big(\frac{-2}{1}\Big)$

= $\frac{2}{1} = 2$

1 =$\frac{1}{1}$

Additive inverse of $\frac{1}{1}$ = $-\Big(\frac{1}{1}\Big)$

= - $\frac{1}{1} = -1$

(i) Additive inverse of $\frac{-5}{-12} = -\frac{5}{12}$.

(ii) $\frac{-5}{-12}$ + its additive inverse = 0

(iii) If $\frac{a}{b}$ is the additive inverse of $\frac{-c}{d}$, then $\frac{-c}{d}$ is the additive inverse of $\frac{a}{b}$.
And, so $\frac{a}{b} + \frac{-c}{d} = \frac{-c}{d} + \frac{a}{b} = 0$.

Explanation

(i) $\frac{-5}{-12} = \frac{5}{12}$
Additive inverse of $\frac{5}{12} = -\frac{5}{12}$

(ii) $\frac{5}{12} + \Big(-\frac{5}{12}\Big)$

$$\begin{aligned}= \frac{5 - 5}{12} \\[1em] = 0\end{aligned}$$

(iii) The sum of number and its additive inverse = Additive identity.

(i) False.

Reason:

$\frac{7 + 5}{9 + 5} = \frac{12}{14}$ ≠ $\frac{7}{9}$

(ii) False

Reason:

$\frac{7 - 5}{9 - 5} = \frac{2}{4}$ ≠ $\frac{7}{9}$

(iii) True

Reason:

$\frac{7 \times 5}{9 \times 5} = \frac{7 \times \cancel{5}}{9 \times \cancel{5}} = \frac{7}{9}$

(iv) True

Reason:

$$\begin{aligned}\frac{7 ÷ 5}{9 ÷ 5} \\[1em] = \frac{7 \times \frac{1}{5}}{9 \times \frac{1}{5}} \\[1em] = \frac{7 \times \cancel{\frac{1}{5}}}{9 \times \cancel{\frac{1}{5}}} \\[1em] = \frac{7}{9}\end{aligned}$$

(v) False

Reason:

$\frac{-5}{-12} = \frac{5}{12}$ is a positive rational number.

(vi) False

Reason:

We need to check if $\frac{-13}{25}$ is smaller than $\frac{-25}{13}$

LCM of 25 and 13 is 325

$\frac{-13}{25} = \frac{-13 \times 13}{25 \times 13} = \frac{-169}{325}$

$\frac{-25}{13} = \frac{-25 \times 25}{13 \times 25} = \frac{-625}{325}$

And, $\frac{-169}{325}$ > $\frac{-625}{325}$

Hence, $\frac{-13}{25}$ > $\frac{-25}{13}$

The weight of an empty fruit basket = $2\frac{1}{3}$ kg

The weight of grapes = $5\frac{5}{6}$ kg

The weight of mangoes = $8\frac{3}{8}$ kg

Total weight of basket with fruits = Weight of empty fruit basket + weight of grapes + weight of mangoes

$$\begin{aligned}= 2\frac{1}{3} \text{ kg} + 5 \frac{5}{6} \text{ kg} + 8 \frac{3}{8} \text{ kg} \\[1em] =\frac{7}{3} \text{ kg} + \frac{35}{6} \text{ kg} + \frac{67}{8} \text{ kg} \\[1em]\end{aligned}$$

LCM of 3, 6 and 8 is 2 x 2 x 2 x 3 = 24

$$\begin{aligned}= \frac{7 \times 8}{3 \times 8} + \frac{35 \times 4}{6 \times 4} + \frac{67 \times 3}{8 \times 3} \text{ kg}\\[1em] = \frac{56}{24} + \frac{140}{24} + \frac{201}{24} \text{ kg}\\[1em] = \frac{56 + 140 + 201}{24} \text{ kg}\\[1em] = \frac{397}{24} \text{ kg}\\[1em] = 16\frac{13}{24} \text{ kg}\end{aligned}$$

Total weight of basket with fruits = $16\frac{13}{24}$ kg.

Let $x$ be the other number. $$\begin{aligned}2\frac{3}{4} + x = 8 \\[1em] ⇒\frac{11}{4} + x = 8 \\[1em] ⇒ x = 8 - \frac{11}{4} \\[1em] ⇒ x = \frac{8}{1} - \frac{11}{4}\end{aligned}$$

LCM of 1 and 4 is 2 x 2 = 4

$$\begin{aligned}⇒ x = \frac{8 \times 4}{1 \times 4} - \frac{11 \times 1}{4 \times 1} \\[1em] ⇒ x = \frac{32}{4} - \frac{11}{4} \\[1em] ⇒ x = \frac{32 - 11}{4} \\[1em] ⇒ x = \frac{21}{4}\\[1em] ⇒ x = 5\frac{1}{4}\end{aligned}$$

The sum of two rational numbers is 8, if one of them is $2\frac{3}{5}$, the other number is $5\frac{1}{4}$.

Hence, option 3 is correct option.

$\frac{a}{b} - \Big(\frac{c}{d} - \frac{e}{f}\Big) = \frac{a}{b} - \frac{c}{d} + \frac{e}{f}$

Hence, option 1 is correct option.

Let $x$ be the other number. $$\begin{aligned}4\frac{1}{2} + x = -6 \\[1em] ⇒\frac{9}{2} + x = -6 \\[1em] ⇒ x = -6 - \frac{9}{2} \\[1em] ⇒ x = \frac{-6}{1} - \frac{9}{2}\end{aligned}$$

LCM of 1 and 2 is 2

$$\begin{aligned}⇒ x = -\frac{6 \times 2}{1 \times 2} - \frac{9\times 1}{2 \times 1} \\[1em] ⇒ x = -\frac{12}{2} - \frac{9}{2} \\[1em] ⇒ x = \frac{-12 - 9}{2} \\[1em] ⇒ x = \frac{-21}{2} \\[1em] ⇒ x = -10\frac{1}{2}\end{aligned}$$

The sum of two rational numbers is -6, if one of them is $4\frac{1}{2}$, the other number is -$10\frac{1}{2}$.

Hence, option 4 is correct option.

Let $x$ be subtracted from $5\frac{2}{3}$.

$$\begin{aligned}5\frac{2}{3} - x = -1\frac{2}{3}\\[1em] ⇒\frac{17}{3} - x = -\frac{5}{3}\\[1em] ⇒ x = \frac{17}{3} + \frac{5}{3}\\[1em] ⇒ x = \frac{17 + 5}{3}\\[1em] ⇒ x = \frac{22}{3}\\[1em] ⇒ x = 7\frac{1}{3}\end{aligned}$$

The number subtracted from $5\frac{2}{3}$ to get -$1\frac{2}{3}$ is $7\frac{1}{3}$.

Hence, option 4 is correct option.

Let $x$ be added to $5\frac{2}{3}$. $$\begin{aligned}5\frac{2}{3} + x = -1\frac{2}{3}\\[1em] ⇒\frac{17}{3} + x = -\frac{5}{3}\\[1em] ⇒ x = \frac{-5}{3} - \frac{17}{3}\\[1em] ⇒ x = \frac{-5 - 17}{3}\\[1em] ⇒ x = \frac{-22}{3}\\[1em] ⇒ x = -7\frac{1}{3}\end{aligned}$$

The number added to $5\frac{2}{3}$ to get -1 $\frac{2}{3}$ is -$7\frac{1}{3}$.

Hence, option 2 is correct option.

$\frac{2}{3} - \frac{4}{5}$

LCM of 3 and 5 is 3 x 5 = 15

$$\begin{aligned}\frac{2 \times 5}{3 \times 5} - \frac{4 \times 3}{5 \times 3}\\[1em] = \frac{10}{15} - \frac{12}{15}\\[1em] = \frac{10 - 12}{15}\\[1em] = \frac{-2}{15}\end{aligned}$$

$$\begin{aligned}\frac{-4}{9} - \frac{2}{-3}\\[1em] =\frac{-4}{9} - \frac{-2}{3}\end{aligned}$$

LCM of 9 and 3 is 3 x 3 = 9

$$\begin{aligned}\frac{-4 \times 1}{9 \times 1} - \frac{-2 \times 3}{3 \times 3}\\[1em] = \frac{-4}{9} - \frac{-6}{9}\\[1em] = \frac{-4 - (-6)}{9}\\[1em] = \frac{-4 + 6}{9}\\[1em] = \frac{2}{9}\end{aligned}$$

$$\begin{aligned}-1 - \frac{4}{9} \\[1em] =\frac{-1}{1} - \frac{4}{9} \\[1em]\end{aligned}$$

LCM of 1 and 9 is 3 x 3 = 9

$$\begin{aligned}\frac{-1 \times 9}{1 \times 9} - \frac{4 \times 1}{9 \times 1}\\[1em] = \frac{-9}{9} - \frac{4}{9}\\[1em] = \frac{-9 - 4}{9}\\[1em] = \frac{-13}{9}\\[1em] = -1\frac{4}{9}\end{aligned}$$

$$\begin{aligned}\frac{-2}{7} - \frac{3}{-14}\\[1em] =\frac{-2}{7} - \frac{-3}{14}\end{aligned}$$

LCM of 7 and 14 is 2 x 14 = 14

$$\begin{aligned}\frac{-2 \times 2}{7 \times 2} - \frac{-3 \times 1}{14 \times 1}\\[1em] = \frac{-4}{14} - \frac{-3}{14}\\[1em] = \frac{-4 - (-3)}{14}\\[1em] = \frac{-4 + 3}{14}\\[1em] = \frac{-1}{14}\end{aligned}$$

$$\begin{aligned}\frac{-5}{18} - \frac{-2}{9}\\[1em]\end{aligned}$$

LCM of 18 and 9 is 2 x 3 x 3 = 18

$$\begin{aligned}\frac{-5 \times 1}{18 \times 1} - \frac{-2 \times 2}{9 \times 2}\\[1em] = \frac{-5}{18} - \frac{-4}{18}\\[1em] = \frac{-5 - (-4)}{18}\\[1em] = \frac{-5 + 4}{18}\\[1em] = \frac{-1}{18}\end{aligned}$$

$\frac{5}{21} - \frac{-13}{42}$

LCM of 21 and 42 is 2 x 3 x 7 = 42

$$\begin{aligned}\frac{5 \times 2}{21 \times 2} - \frac{-13 \times 1}{42 \times 1}\\[1em] = \frac{10}{42} - \frac{-13}{42}\\[1em] = \frac{10 - (-13)}{42}\\[1em] = \frac{10 + 13}{42}\\[1em] = \frac{23}{42}\end{aligned}$$

$$\begin{aligned}\frac{-3}{8} - \frac{5}{8}\\[1em] = \frac{-3 - 5}{8}\\[1em] = \frac{-8}{8}\\[1em] = -1\end{aligned}$$

$$\begin{aligned}\frac{4}{11} - \frac{-8}{11}\\[1em] = \frac{4 - (-8)}{11}\\[1em] = \frac{4 + 8}{11}\\[1em] = \frac{12}{11}\\[1em] = 1 \frac{1}{11}\end{aligned}$$

$$\begin{aligned}\frac{-5}{9} - \frac{4}{9}\\[1em] = \frac{-5 - 4}{9}\\[1em] = \frac{-9}{9} = - 1\end{aligned}$$

$\frac{-3}{8} - \frac{1}{4}$

LCM of 8 and 4 is 2 x 2 x 2 = 8

$$\begin{aligned}\frac{-3 \times 1}{8 \times 1} - \frac{1 \times 2}{4 \times 2}\\[1em] = \frac{-3}{8} - \frac{2}{8}\\[1em] = \frac{-3 - 2}{8}\\[1em] = -\frac{5}{8}\end{aligned}$$

$\frac{-13}{16} - \frac{-5}{8}$

LCM of 16 and 8 is 2 x 2 x 2 x 2 = 16

$$\begin{aligned}\frac{-13 \times 1}{16 \times 1} - \frac{-5 \times 2}{8 \times 2}\\[1em] = \frac{-13}{16} - \frac{-10}{16}\\[1em] = \frac{-13 - (-10)}{16}\\[1em] = \frac{-13 + 10}{16}\\[1em] = \frac{-3}{16}\end{aligned}$$

$\frac{5}{33} - \frac{-9}{22}$

LCM of 33 and 22 is 2 x 3 x 11 = 66

$$\begin{aligned}\frac{5 \times 2}{33 \times 2} - \frac{-9 \times 3}{22 \times 3}\\[1em] = \frac{10}{66} - \frac{-27}{66}\\[1em] = \frac{10 - (-27)}{66}\\[1em] = \frac{10 + 27}{66}\\[1em] = \frac{37}{66}\end{aligned}$$

Let $x$ be the other number. $$\begin{aligned}\frac{2}{5} + x = \frac{9}{20} \\[1em] ⇒ x = \frac{9}{20} - \frac{2}{5}\end{aligned}$$

LCM of 20 and 5 is 2 x 2 x 5 = 20

$$\begin{aligned}⇒ x = \frac{9 \times 1}{20 \times 1} - \frac{2 \times 4}{5 \times 4}\\[1em] ⇒ x = \frac{9}{20} - \frac{8}{20}\\[1em] ⇒ x = \frac{9 - 8}{20}\\[1em] ⇒ x = \frac{1}{20}\end{aligned}$$

The sum of two rational numbers is $\frac{9}{20}$, if one of them is $\frac{2}{5}$, the other number is $\frac{1}{20}$

Let $x$ be the other number. $$\begin{aligned}\frac{-8}{15} + x = \frac{-2}{3}\\[1em] ⇒ x = \frac{-2}{3} - \frac{-8}{15}\end{aligned}$$

LCM of 3 and 15 is 3 x 5 = 15

$$\begin{aligned}⇒ x = \frac{-2 \times 5}{3 \times 5} - \frac{-8 \times 1}{15 \times 1}\\[1em] ⇒ x = \frac{-10}{15} - \frac{-8}{15}\\[1em] ⇒ x = \frac{-10 - (-8)}{15}\\[1em] ⇒ x = \frac{-10 + 8}{15}\\[1em] ⇒ x = \frac{-2}{15}\end{aligned}$$

The sum of two rational numbers is $\frac{-2}{3}$, if one of them is $\frac{-8}{15}$, the other number is $\frac{-2}{15}$.

Let $x$ be the other number. $$\begin{aligned}\frac{-8}{5} + x = -6 \\[1em] \Rightarrow x = \frac{-6}{1} - \frac{-8}{5}\end{aligned}$$

LCM of 1 and 5 is 5

$$\begin{aligned}\Rightarrow x = \frac{-6 \times 5}{1 \times 5} - \frac{-8 \times 1}{5 \times 1}\\[1em] \Rightarrow x = \frac{-30}{5} - \frac{-8}{5}\\[1em] \Rightarrow x = \frac{-30 - (-8)}{5}\\[1em] \Rightarrow x = \frac{-30 + 8}{5}\\[1em] \Rightarrow x = \frac{-22}{5}\\[1em] \Rightarrow x = -4\frac{2}{5}\end{aligned}$$

The sum of two rational numbers is -6, if one of them is $\frac{-8}{5}$, the other number is $-4\frac{2}{5}$.

Let $x$ be added to $\frac{-7}{8}$.

$$\begin{aligned}\frac{-7}{8} + x = \frac{5}{9}\\[1em] ⇒ x = \frac{5}{9} - \frac{-7}{8}\end{aligned}$$

LCM of 9 and 8 is 2 x 2 x 2 x 3 x 3 = 72

$$\begin{aligned}\Rightarrow x = \frac{5 \times 8}{9 \times 8} - \frac{-7 \times 9}{8 \times 9}\\[1em] \Rightarrow x = \frac{40}{72} - \frac{-63}{72}\\[1em] \Rightarrow x = \frac{40 - (-63)}{72}\\[1em] \Rightarrow x = \frac{40 + 63}{72}\\[1em] \Rightarrow x = \frac{103}{72}\\[1em] \Rightarrow x = 1 \frac{31}{72}\end{aligned}$$

The number added to $\frac{-7}{8}$ to get $\frac{5}{9}$ is $1\frac{31}{72}$.

Let $x$ be added to $\frac{-5}{9}$.

$$\begin{aligned}\frac{-5}{9} + x = \frac{-2}{3}\\[1em] ⇒ x = \frac{-2}{3} - \frac{-5}{9}\end{aligned}$$

LCM of 3 and 9 is 3 x 3 = 9

$$\begin{aligned}\Rightarrow x = \frac{-2 \times 3}{3 \times 3} - \frac{-5 \times 1}{9 \times 1}\\[1em] \Rightarrow x = \frac{-6}{9} - \frac{-5}{9}\\[1em] \Rightarrow x = \frac{-6 - (-5)}{9}\\[1em] \Rightarrow x = \frac{-6 + 5}{9}\\[1em] \Rightarrow x = \frac{-1}{9}\end{aligned}$$

The number added to $\frac{-5}{9}$ to get $\frac{-2}{3}$ is $\frac{-1}{9}$.

Let $x$ be subtracted from $\frac{-5}{6}$.

$$\begin{aligned}\frac{-5}{6} - x = \frac{4}{9}\\[1em] \Rightarrow x = \frac{-5}{6} - \frac{4}{9}\end{aligned}$$

LCM of 6 and 9 is 2 x 3 x 3 = 18

$$\begin{aligned}\Rightarrow x = \frac{-5 \times 3}{6 \times 3} - \frac{4 \times 2}{9 \times 2}\\[1em] \Rightarrow x = \frac{-15}{18} - \frac{8}{18}\\[1em] \Rightarrow x = \frac{-15 - 8}{18}\\[1em] \Rightarrow x = \frac{-23}{18}\\[1em] \Rightarrow x = -1\frac{5}{18}\end{aligned}$$

The number subtracted from $\frac{-5}{6}$ to get $\frac{4}{9}$ is -$1\frac{5}{18}$.

Let $x$ be subtracted from -2.

$$\begin{aligned}-2 - x = \frac{3}{8}\\[1em] \Rightarrow \frac{-2}{1} - x = \frac{3}{8}\\[1em] \Rightarrow x = \frac{-2}{1} - \frac{3}{8}\end{aligned}$$

LCM of 1 and 8 is 2 x 2 x 2 = 8

$$\begin{aligned}\Rightarrow x = \frac{-2 \times 8}{1 \times 8} - \frac{3 \times 1}{8 \times 1}\\[1em] \Rightarrow x = \frac{-16}{8} - \frac{3}{8}\\[1em] \Rightarrow x = \frac{-16 - 3}{8}\\[1em] \Rightarrow x = \frac{-19}{8}\\[1em] \Rightarrow x = -2 \frac{3}{8}\end{aligned}$$

The number subtracted from -2 to get $\frac{3}{8}$ is -$2\frac{3}{8}$.

Let $x$ be added to -2.

$$\begin{aligned}-2 + x = \frac{3}{8}\\[1em] \Rightarrow \frac{-2}{1} + x = \frac{3}{8}\\[1em] \Rightarrow x = \frac{3}{8} - \frac{-2}{1}\end{aligned}$$

LCM of 8 and 1 is 2 x 2 x 2 = 8

$$\begin{aligned}\Rightarrow x = \frac{3 \times 1}{8 \times 1} - \frac{-2 \times 8}{1 \times 8}\\[1em] \Rightarrow x = \frac{3}{8} - \frac{-16}{8}\\[1em] \Rightarrow x = \frac{3 - (-16)}{8}\\[1em] \Rightarrow x = \frac{3 + 16}{8}\\[1em] \Rightarrow x = \frac{19}{8}\\[1em] \Rightarrow x = 2\frac{3}{8}\end{aligned}$$

The number added to -2 to get $\frac{3}{8}$ is $2\frac{3}{8}$.

$$\begin{aligned}\Big(\frac{3}{7} - \frac{-11}{7}\Big) + \Big(\frac{-4}{9} - \frac{7}{9}\Big)\\[1em] =\Big(\frac{3 - (-11)}{7}\Big) + \Big(\frac{-4 - 7}{9}\Big)\\[1em] =\Big(\frac{14}{7}\Big) + \Big(\frac{-11}{9}\Big)\\[1em] =\Big(\frac{2}{1}\Big) + \Big(\frac{-11}{9}\Big)\end{aligned}$$

LCM of 1 and 9 is 3 x 3 = 9

$$\begin{aligned}=\frac{2 \times 9}{1 \times 9} + \frac{-11 \times 1}{9 \times 1}\\[1em] =\frac{18}{9} + \frac{-11}{9}\\[1em] =\frac{18 + (-11)}{9}\\[1em] =\frac{7}{9}\end{aligned}$$

$$\begin{aligned}\Big(\frac{2}{3} - \frac{1}{3}\Big) + \Big(\frac{-4}{5} - \frac{2}{5}\Big)\\[1em] =\Big(\frac{2 - 1}{3}\Big) + \Big(\frac{-4 - 2}{5}\Big)\\[1em] =\Big(\frac{1}{3}\Big) + \Big(\frac{-6}{5}\Big)\end{aligned}$$

LCM of 3 and 5 is 3 x 5 = 15

$$\begin{aligned}=\frac{1 \times 5}{3 \times 5} + \frac{-6 \times 3}{5 \times 3}\\[1em] =\frac{5}{15} + \frac{-18}{15}\\[1em] =\frac{5 + (-18)}{15}\\[1em] =\frac{-13}{15}\end{aligned}$$

$$\begin{aligned}\Big(\frac{4}{7} - \frac{-13}{7}\Big) + \Big(-\frac{-8}{9} + \frac{17}{9}\Big)\\[1em] =\Big(\frac{4 - (-13)}{7}\Big) + \Big(\frac{8 + 17}{9}\Big)\\[1em] =\Big(\frac{4 + 13}{7}\Big) + \Big(\frac{25}{9}\Big)\\[1em] =\Big(\frac{17}{7}\Big) + \Big(\frac{25}{9}\Big)\end{aligned}$$

LCM of 7 and 9 is 3 x 3 x 7 = 63

$$\begin{aligned}=\frac{17 \times 9}{7 \times 9} + \frac{25 \times 7}{9 \times 7}\\[1em] =\frac{153}{63} + \frac{175}{63}\\[1em] =\frac{153 + 175}{63}\\[1em] =\frac{328}{63}\\[1em] =5\frac{13}{63}\end{aligned}$$

Let the number be $x$. $$\begin{aligned}5\frac{2}{3} × x = -1\frac{2}{3}\\[1em] ⇒ \frac{17}{3} × x = -\frac{5}{3}\\[1em] ⇒ x = -\frac{5}{3} ÷ \frac{17}{3}\\[1em] ⇒ x = -\frac{5}{3} × \frac{3}{17}\\[1em] ⇒ x = -\frac{5 \times 3}{3 \times 17}\\[1em] ⇒ x = -\frac{15}{51}\\[1em] ⇒ x = -\frac{5}{17}\end{aligned}$$

Hence, option 3 is the correct option.

We know that, multiplication of rational numbers is distributive over their addition/subtraction.

∴ a x (b - c) = a x b - a x c

Hence, option 3 is the correct option.

Let a positive rational number be $\frac{a}{b}$.

$$\begin{aligned}\frac{a}{b} \times \frac{b}{a}\\[1em] = \frac{a \times b}{b \times a}\\[1em] = \frac{ab}{ab}\\[1em] = 1\end{aligned}$$

Hence, option 3 is the correct option.

Let breadth be b

Length = $4\frac{2}{5}$cm = $\frac{22}{5}$cm.

Area = $7\frac{1}{3}$ cm² =$\frac{22}{3}$ cm².

Area of rectangle = length x breadth $$\begin{aligned}\frac{22}{3} = \frac{22}{5} \times b\\[1em] ⇒ b = \frac{22}{3} ÷ \frac{22}{5}\\[1em] ⇒ b = \frac{22}{3} \times \frac{5}{22}\\[1em] ⇒ b = \frac{22 \times 5}{3 \times 22} \\[1em] ⇒ b = \frac{110}{66}\\[1em] ⇒ b = \frac{5}{3}\\[1em] ⇒ b = 1\frac{2}{3}\end{aligned}$$

Hence, option 2 is the correct option.

Rational number = $\frac{2}{7}$

Additive inverse = $-\frac{2}{7}$

$$\begin{aligned}\frac{2}{7} \times -\frac{2}{7}\\[1em] = -\frac{2 \times 2}{7 \times 7}\\[1em] = -\frac{4}{49}.\end{aligned}$$

Hence, option 2 is the correct option.

$$\begin{aligned}\frac{-14}{5} \times \frac{-6}{7}\\[1em] =\frac{-14 \times (-6)}{5 \times 7}\\[1em] =\frac{84}{35}\\[1em] =\frac{12}{5}\\[1em] =2\frac{2}{5}\end{aligned}$$

Hence, $\frac{-14}{5} \times \frac{-6}{7} = 2\frac{2}{5}$

$$\begin{aligned}\frac{7}{6} \times \frac{-18}{91}\\[1em] =\frac{7 \times (-18)}{6 \times 91}\\[1em] =\frac{-126}{546}\\[1em] =\frac{-3}{13}\end{aligned}$$

Hence, $\frac{7}{6} \times \frac{-18}{91} = \frac{-3}{13}$

$$\begin{aligned}\frac{-125}{72} \times \frac{9}{-5}\\[1em] =\frac{-125 \times 9}{72 \times (-5)}\\[1em] =\frac{-1125}{-360}\\[1em] =\frac{25}{8}\\[1em] =3\frac{1}{8}\end{aligned}$$

Hence, $\frac{-125}{72} \times \frac{9}{-5} = 3\frac{1}{8}$

$$\begin{aligned}\frac{-11}{9} \times \frac{-51}{-44}\\[1em] =\frac{-11 \times (-51)}{9 \times -44}\\[1em] =\frac{561}{-396}\\[1em] =-\frac{17}{12}\\[1em] =-1\frac{5}{12}\end{aligned}$$

Hence, $\frac{-11}{9} \times \frac{-51}{-44} = -1\frac{5}{12}$

$$\begin{aligned}-\frac{16}{5} \times \frac{20}{8}\\[1em] =-\frac{16 \times 20}{5 \times 8}\\[1em] =-\frac{320}{40}\\[1em] =-8\end{aligned}$$

Hence, $-\frac{16}{5} \times \frac{20}{8} = -8$

$$\begin{aligned}\frac{5}{6} \times \frac{8}{9}\\[1em] =\frac{5 \times 8}{6 \times 9}\\[1em] =\frac{40}{54}\\[1em] =\frac{20}{27}\end{aligned}$$

Hence, $\frac{5}{6} \times \frac{8}{9} = \frac{20}{27}$

$$\begin{aligned}\frac{2}{7} \times \frac{-14}{9}\\[1em] =\frac{2 \times (-14)}{7 \times 9}\\[1em] =\frac{-28}{63}\\[1em] =\frac{-4}{9}\end{aligned}$$

Hence, $\frac{2}{7} \times \frac{-14}{9} = \frac{-4}{9}$

$$\begin{aligned}\frac{-7}{8} \times \frac{4}{1}\\[1em] =\frac{-7 \times 4}{8 \times 1}\\[1em] =\frac{-28}{8}\\[1em] =\frac{-7}{2}\\[1em] =-3\frac{1}{2}\end{aligned}$$

Hence, $\frac{-7}{8} \times 4 = -3\frac{1}{2}$

$$\begin{aligned}\frac{36}{-7} \times \frac{-9}{28}\\[1em] =\frac{36 \times (-9)}{-7 \times 28}\\[1em] =\frac{-324}{-196}\\[1em] =\frac{81}{49}\\[1em] =1\frac{32}{49}\end{aligned}$$

Hence, $\frac{36}{-7} \times \frac{-9}{28} = 1\frac{32}{49}$

$$\begin{aligned}\frac{-7}{10} \times \frac{-8}{15}\\[1em] =\frac{-7 \times (-8)}{10 \times 15}\\[1em] =\frac{56}{150}\\[1em] =\frac{28}{75}\end{aligned}$$

Hence, $\frac{-7}{10} \times \frac{-8}{15} = \frac{28}{75}$

$$\begin{aligned}\Big(\frac{2 \times 5}{-3 \times 4}\Big) + \Big(\frac{5 \times 3}{9 \times (-10)}\Big)\\[1em] =\Big(\frac{10}{-12}\Big) + \Big(\frac{15}{-90}\Big)\\[1em] =\Big(\frac{-5}{6}\Big) + \Big(\frac{-1}{6}\Big)\\[1em] =\Big(\frac{-5 + (-1)}{6}\Big)\\[1em] =\Big(\frac{-6}{6}\Big)\\[1em] =-1\end{aligned}$$

Hence, $\Big(\frac{2}{-3} \times \frac{5}{4}\Big) + \Big(\frac{5}{9} \times \frac{3}{-10}\Big) = \Big(\frac{-6}{6}\Big)$

$$\begin{aligned}\Big(\frac{2 \times 1}{1 \times 4}\Big) - \Big(\frac{-18 \times -7}{7 \times 15}\Big) \\[1em] =\Big(\frac{2}{4}\Big) - \Big(\frac{126}{105}\Big) \\[1em] =\Big(\frac{1}{2}\Big) - \Big(\frac{6}{5}\Big) \\[1em]\end{aligned}$$ LCM of 2 and 5 is 2 x 5 = 10 $$\begin{aligned}=\Big(\frac{1 \times 5}{2 \times 5}\Big) - \Big(\frac{6 \times 2}{5 \times 2}\Big)\\[1em] =\Big(\frac{5}{10}\Big) - \Big(\frac{12}{10}\Big)\\[1em] =\Big(\frac{5 - 12}{10}\Big)\\[1em] =\Big(\frac{-7}{10}\Big)\end{aligned}$$

Hence, $\Big(2 \times \frac{1}{4}\Big) - \Big(\frac{-18}{7} \times \frac{-7}{15}\Big) = \Big(\frac{-7}{10}\Big)$

$$\begin{aligned}\Big(\frac{-5 \times 2}{1 \times 15}\Big) - \Big(\frac{-6 \times 2}{1 \times 9}\Big)\\[1em] =\Big(\frac{-10}{15}\Big) - \Big(\frac{-12}{9}\Big)\\[1em] =\Big(\frac{-2}{3}\Big) - \Big(\frac{-4}{3}\Big)\\[1em] =\Big(\frac{-2 - (-4)}{3}\Big)\\[1em] =\Big(\frac{-2 + 4}{3})\\[1em] =\Big(\frac{2}{3}\Big)\end{aligned}$$

Hence, $\Big(-5 \times \frac{5}{12}\Big) - \Big(-6 \times \frac{2}{9}\Big) = \Big(\frac{2}{3}\Big)$

$$\begin{aligned}\Big(\frac{8 \times -3}{5 \times 2}\Big) + \Big(\frac{-3 \times 9}{10 \times 16}\Big)\\[1em] =\Big(\frac{-24}{10}\Big) + \Big(\frac{-27}{160}\Big)\\[1em] =\Big(\frac{-12}{5}\Big) + \Big(\frac{-27}{160}\Big)\\[1em]\end{aligned}$$

LCM of 5 and 160 is 2 x 2 x 2 x 2 x 2 x 5 = 160

$$\begin{aligned}=\Big(\frac{-12 \times 32}{5 \times 32}\Big) + \Big(\frac{-27 \times 1}{160 \times 1}\Big)\\[1em] =\Big(\frac{-384}{160}\Big) + \Big(\frac{-27}{160}\Big)\\[1em] =\Big(\frac{-384 + (-27)}{160}\Big)\\[1em] =\Big(\frac{-411}{160}\Big)\\[1em] =-2\frac{91}{160}\end{aligned}$$

Hence, $\Big(\frac{8}{5} \times \frac{-3}{2}\Big) + \Big(\frac{-3}{10} \times \frac{9}{16}\Big) = -2\frac{91}{160}$

$$\begin{aligned}\frac{7}{-5} \times 1\\[1em] =\frac{7 \times 1}{-5 \times 1}\\[1em] =\frac{7}{-5}\end{aligned}$$

Hence, $\frac{7}{-5} \times 1 = \frac{7}{-5}$

$$\begin{aligned}\frac{-3}{-4} \times 1\\[1em] =\frac{-3 \times 1}{-4 \times 1}\\[1em] =\frac{-3}{-4}\\[1em] =\frac{3}{4}\end{aligned}$$

Hence, $\frac{-3}{-4} \times 1 = \frac{3}{4}$

$$\begin{aligned}0 \times 1 \\[1em] = 0\end{aligned}$$

Hence, 0 x 1 = 0

$$\begin{aligned}\frac{-8}{13} \times 1\\[1em] =\frac{-8 \times 1}{13 \times 1}\\[1em] =\frac{-8}{13}\end{aligned}$$

Hence, $\frac{-8}{13} \times 1 = \frac{-8}{13}$

$$\begin{aligned}\frac{-6}{-7} \times 1\\[1em] =\frac{-6 \times 1}{-7 \times 1}\\[1em] =\frac{-6}{-7}\\[1em] =\frac{6}{7}\end{aligned}$$

Hence, $\frac{-6}{-7} \times 1 = \frac{6}{7}$

To prove:

$\frac{-1}{5} \times \frac{2}{9} = \frac{2}{9} \times \frac{-1}{5}$

Taking LHS:

$$\begin{aligned}\frac{-1}{5} \times \frac{2}{9}\\[1em] =\frac{-1 \times 2}{5 \times 9}\\[1em] =\frac{-2}{45}\\[1em]\end{aligned}$$

Taking RHS:

$$\begin{aligned}\frac{2}{9} \times \frac{-1}{5}\\[1em] =\frac{2 \times -1}{9 \times 5}\\[1em] =\frac{-2}{45}\\[1em]\end{aligned}$$

∴ LHS = RHS

$\frac{-1}{5} \times \frac{2}{9} = \frac{2}{9} \times \frac{-1}{5}$

To prove:

$\frac{5}{-3} \times \frac{13}{-11} = \frac{13}{-11} \times \frac{5}{-3}$

Taking LHS:

$$\begin{aligned}\frac{5}{-3} \times \frac{13}{-11}\\[1em] =\frac{5 \times 13}{-3 \times -11}\\[1em] =\frac{65}{33}\\[1em]\end{aligned}$$

Taking RHS:

$$\begin{aligned}\frac{13}{-11} \times \frac{5}{-3}\\[1em] =\frac{13 \times 5}{-11 \times -3}\\[1em] =\frac{65}{33}\\[1em]\end{aligned}$$

∴ LHS = RHS

$\frac{5}{-3} \times \frac{13}{-11} = \frac{13}{-11} \times \frac{5}{-3}$

To prove:

$3 \times \frac{-8}{9} = \frac{-8}{9} \times 3$

Taking LHS:

$$\begin{aligned}3 \times \frac{-8}{9}\\[1em] =\frac{3 \times -8}{1 \times 9}\\[1em] =\frac{-24}{9}\\[1em] =\frac{-8}{3}\end{aligned}$$

Taking RHS:

$$\begin{aligned}\frac{-8}{9} \times 3\\[1em] =\frac{-8 \times 3}{9 \times 1}\\[1em] =\frac{-24}{9}\\[1em] =\frac{-8}{3}\end{aligned}$$

∴ LHS = RHS

$3 \times \frac{-8}{9} = \frac{-8}{9} \times 3$

To prove:

$0 \times \frac{-12}{17} = \frac{-12}{17} \times 0$

Taking LHS:

$$\begin{aligned}0 \times \frac{-12}{17}\\[1em] =\frac{0 \times -12}{1 \times 17}\\[1em] =\frac{0}{17}\\[1em] =0\end{aligned}$$

Taking RHS:

$$\begin{aligned}\frac{-12}{17} \times 0\\[1em] =\frac{-12 \times 0}{17 \times 1}\\[1em] =\frac{0}{17}\\[1em] =0\end{aligned}$$

∴ LHS = RHS

$0 \times \frac{-12}{17} = \frac{-12}{17} \times 0$

The multiplicative inverse of 5 = reciprocal of 5 = $\frac{1}{5}$.

The multiplicative inverse of -3 = reciprocal of -3 = $-\frac{1}{3}$.

The multiplicative inverse of $\frac{5}{11}$ = reciprocal of $\frac{5}{11}$ = $\frac{11}{5} = 2\frac{1}{5}$.

The multiplicative inverse of $\frac{-7}{-8}$ = reciprocal of $\frac{7}{8}$ = $\frac{8}{7} = 1\frac{1}{7}$.

The multiplicative inverse of $\frac{-8}{-7}$ = reciprocal of $\frac{8}{7}$ = $\frac{7}{8}$.

$$\begin{aligned}\frac{3}{5} \times \frac{2}{3}\\[1em] =\frac{3 \times 2}{5 \times 3}\\[1em] =\frac{6}{15}\\[1em] =\frac{2}{5}\end{aligned}$$

The multiplicative inverse of $\frac{2}{5}$ = reciprocal of $\frac{2}{5} = \frac{5}{2} = 2\frac{1}{2}$.

$$\begin{aligned}\frac{-8}{3} \times \frac{13}{-7}\\[1em] =\frac{-8 \times 13}{3 \times -7}\\[1em] =\frac{-104}{-21}\\[1em] =\frac{104}{21}\end{aligned}$$

The multiplicative inverse of $\frac{104}{21}$ = reciprocal of $\frac{104}{21}$ = $\frac{21}{104}$.

$$\begin{aligned}\frac{-3}{5} \times \frac{-1}{13}\\[1em] =\frac{-3 \times (-1)}{5 \times 13}\\[1em] =\frac{3}{65}\end{aligned}$$

The multiplicative inverse of $\frac{3}{65}$ = reciprocal of $\frac{3}{65}$ = $\frac{65}{3} = 21\frac{2}{3}$.

To prove:

$(x + y) \times z = x \times z + y \times z$

Taking LHS:

$$\begin{aligned}(x + y) \times z\\[1em] =\Big(\frac{4}{5} + \frac{-2}{3}\Big) \times -4\end{aligned}$$

LCM of 5 and 3 is 3 x 5 = 15

$$\begin{aligned}=\Big(\frac{4 \times 3}{5 \times 3} + \frac{-2 \times 5}{3 \times 5}\Big) \times -4\\[1em] =\Big(\frac{12}{15} + \frac{-10}{15}\Big) \times -4\\[1em] =\Big(\frac{12 + (-10)}{15}\Big) \times -4\\[1em] =\Big(\frac{2}{15}\Big) \times -4\\[1em] =\Big(\frac{2 \times -4}{15 \times 1}\Big)\\[1em] =\Big(\frac{-8}{15}\Big)\end{aligned}$$

Taking RHS:

$$\begin{aligned}x \times z + y \times z\\[1em] =\frac{4}{5} \times -4 + \frac{-2}{3} \times -4\\[1em] = \frac{4 \times -4}{5 \times 1} + \frac{-2 \times -4}{3 \times 1}\\[1em] =\frac{-16}{5} + \frac{8}{3}\\[1em]\end{aligned}$$

LCM of 5 and 3 is 3 x 5 = 15

$$\begin{aligned}=\frac{-16 \times 3}{5 \times 3} + \frac{8 \times 5}{3 \times 5}\\[1em] =\frac{-48}{15} + \frac{40}{15}\\[1em] =\frac{-48 + 40}{15}\\[1em] =\frac{-8}{15}\end{aligned}$$

∴ LHS = RHS

$(x + y) \times z = x \times z + y \times z$

To prove:

$(x + y) \times z = x \times z + y \times z$

Taking LHS:

$$\begin{aligned}(x + y) \times z\\[1em] =\Big(2 + \frac{4}{5}\Big) \times \frac{3}{-10}\\[1em] =\Big(\frac{2}{1} + \frac{4}{5}\Big) \times \frac{3}{-10}\end{aligned}$$

LCM of 1 and 5 is 5.

$$\begin{aligned}=\Big(\frac{2 \times 5}{1 \times 5} + \frac{4 \times 1}{5 \times 1}\Big) \times \frac{3}{-10}\\[1em] =\Big(\frac{10}{5} + \frac{4}{5}\Big) \times \frac{3}{-10}\\[1em] =\Big(\frac{10 + 4}{5}\Big) \times \frac{3}{-10}\\[1em] =\Big(\frac{14}{5}\Big) \times \frac{3}{-10}\\[1em] =\Big(\frac{14 \times 3}{5 \times -10}\Big)\\[1em] =\Big(\frac{42}{-50}\Big)\\[1em] =\Big(\frac{-21}{25}\Big)\end{aligned}$$

Taking RHS:

$$\begin{aligned}x \times z + y \times z\\[1em] =2 \times \frac{3}{-10} + \frac{4}{5} \times \frac{3}{-10}\\[1em] = \frac{2 \times 3}{1 \times -10} + \frac{4 \times 3}{5 \times -10}\\[1em] =\frac{6}{-10} + \frac{12}{-50}\\[1em] =\frac{-3}{5} + \frac{-6}{25}\end{aligned}$$

LCM of 5 and 25 is 5 x 5 = 25

$$\begin{aligned}=\frac{-3 \times 5}{5 \times 5} + \frac{-6 \times 1}{25 \times 1}\\[1em] =\frac{-15}{25} + \frac{-6}{25}\\[1em] =\frac{-15 + (-6)}{25}\\[1em] =\frac{-21}{25}\end{aligned}$$

∴ LHS = RHS

$(x + y) \times z = x \times z + y \times z$

To prove:

$x \times (y - z) = x \times y - x \times z$

Taking LHS:

$$\begin{aligned}x \times (y - z)\\[1em] =\frac{4}{5} \times \Big(\frac{-7}{4} - 3\Big)\\[1em] =\frac{4}{5} \times \Big(\frac{-7}{4} - \frac{3}{1}\Big)\end{aligned}$$

LCM of 4 and 1 is 2 x 2 = 4.

$$\begin{aligned}=\frac{4}{5} \times \Big(\frac{-7 \times 1}{4 \times 1} - \frac{3 \times 4}{1 \times 4}\Big)\\[1em] =\frac{4}{5} \times \Big(\frac{-7}{4} - \frac{12}{4}\Big)\\[1em] =\frac{4}{5} \times \Big(\frac{-7 - 12}{4}\Big)\\[1em] =\frac{4}{5} \times \Big(\frac{-19}{4}\Big)\\[1em] =\Big(\frac{4 \times -19}{5 \times 4}\Big)\\[1em] =\Big(\frac{-76}{20}\Big)\\[1em] =\Big(\frac{-19}{5}\Big)\end{aligned}$$

Taking RHS:

$$\begin{aligned}x \times y - x \times z\\[1em] =\frac{4}{5} \times \frac{-7}{4} - \frac{4}{5} \times 3\\[1em] = \frac{4 \times -7}{5 \times 4} - \frac{4 \times 3}{5 \times 1}\\[1em] =\frac{-28}{20} - \frac{12}{5}\\[1em] =\frac{-7}{5} - \frac{12}{5}\\[1em] =\frac{-7 - 12}{5}\\[1em] =\frac{-19}{5}\end{aligned}$$

∴ LHS = RHS

$x \times (y - z) = x \times y - x \times z$

To prove:

$x \times (y - z) = x \times y - x \times z$

Taking LHS:

$$\begin{aligned}x \times (y - z)\\[1em] =\frac{3}{4} \times \Big(\frac{8}{9} - (-5)\Big)\\[1em] =\frac{3}{4} \times \Big(\frac{8}{9} - \frac{-5}{1}\Big)\end{aligned}$$

LCM of 9 and 1 is 3 x 3 = 9.

$$\begin{aligned}=\frac{3}{4} \times \Big(\frac{8 \times 1}{9 \times 1} - \frac{-5 \times 9}{1 \times 9}\Big)\\[1em] =\frac{3}{4} \times \Big(\frac{8}{9} - \frac{-45}{9}\Big)\\[1em] =\frac{3}{4} \times \Big(\frac{8 - (-45)}{9}\Big)\\[1em] =\frac{3}{4} \times \Big(\frac{8 + 45}{9}\Big)\\[1em] =\frac{3}{4} \times \Big(\frac{53}{9}\Big)\\[1em] =\Big(\frac{3 \times 53}{4 \times 9}\Big)\\[1em] =\Big(\frac{159}{36}\Big)\\[1em] =\Big(\frac{53}{12}\Big)\end{aligned}$$

Taking RHS:

$$\begin{aligned}x \times y - x \times z\\[1em] =\frac{3}{4} \times \frac{8}{9} - \frac{3}{4} \times -5\\[1em] = \frac{3 \times 8}{4 \times 9} - \frac{3 \times -5}{4 \times 1}\\[1em] =\frac{24}{36} - \frac{-15}{4}\\[1em] =\frac{2}{3} - \frac{-15}{4}\end{aligned}$$

LCM of 3 and 4 is 2 x 2 x 3 = 12

$$\begin{aligned}=\frac{2 \times 4}{3 \times 4} - \frac{-15 \times 3}{4 \times 3}\\[1em] =\frac{8}{12} - \frac{-45}{12}\\[1em] =\frac{8 - (-45)}{12}\\[1em] =\frac{8 + 45}{12}\\[1em] =\frac{53}{12}\end{aligned}$$

∴ LHS = RHS

$x \times (y - z) = x \times y - x \times z$

(i) Commutativity property

Reason

If $\frac{a}{b}$ and $\frac{c}{d}$ are any two rational numbers, then:

$\frac{a}{b} \times \frac{c}{d} = \frac{c}{d} \times \frac{a}{b}$

(ii) Associativity property

Reason

If $\frac{a}{b}, \frac{c}{d}$ and $\frac{e}{f}$ are any three rational numbers, then:

$\frac{a}{b} \times \Big(\frac{c}{d} \times \frac{e}{f}\Big) = \Big(\frac{a}{b} \times \frac{c}{d}\Big) \times \frac{e}{f}$

(iii) Distributivity property

Reason

If $\frac{a}{b}, \frac{c}{d}$ and $\frac{e}{f}$ are any three rational numbers, then:

$\frac{a}{b} \times \Big(\frac{c}{d} + \frac{e}{f}\Big) = \Big(\frac{a}{b} \times \frac{c}{d}\Big) + \Big(\frac{a}{b} \times \frac{e}{f}\Big)$

(iv) Existence of inverse

Reason

The multiplicative inverse of $\frac{a}{b}$ = reciprocal of $\frac{a}{b} = \frac{b}{a}$.

(v) Existence of identity

Reason

For a rational number $\frac{a}{b}$,

$1 \times \frac{a}{b} = \frac{a}{b} \times 1 = \frac{a}{b}$.

(i) The product of two positive rational numbers is always positive.

(ii) The product of two negative rational numbers is always positive.

(iii) If two rational numbers have opposite signs then their product is always negative.

(iv) The reciprocal of a positive rational number is positive and the reciprocal of a negative rational number is negative.

(v) Rational number 0 has no reciprocal.

(vi) The product of a non-zero rational number and its reciprocal is 1.

(vii) The numbers 1 and -1 are their own reciprocal.

(viii) If m is reciprocal of n, then the reciprocal of n is m.

Explanation

(i) Let 2 positive rational numbers be $\frac{a}{b}$ and $\frac{c}{d}$.

Hence,

$$\begin{aligned}\frac{a}{b} \times \frac{c}{d}\\[1em] = \frac{a \times c}{b \times d}\\[1em] = \frac{ac}{bd}\end{aligned}$$

$\frac{ac}{bd}$ is also positive rational number.

(ii) Let 2 negative rational numbers be -$\frac{a}{b}$ and -$\frac{c}{d}$.

Hence,

$$\begin{aligned}-\frac{a}{b} \times -\frac{c}{d}\\[1em] = \frac{-a \times -c}{b \times d}\\[1em] = \frac{ac}{bd}\end{aligned}$$

$\frac{ac}{bd}$ is positive rational number.

(iii) Let 2 rational numbers be $\frac{a}{b}$ and -$\frac{c}{d}$.

Hence,

$$\begin{aligned}\frac{a}{b} \times -\frac{c}{d}\\[1em] = \frac{a \times -c}{b \times d}\\[1em] = \frac{-ac}{bd}\end{aligned}$$

-$\frac{ac}{bd}$ is negative rational number.

(iv) Let the positive rational number be $\frac{a}{b}$.

Reciprocal of $\frac{a}{b} = \frac{b}{a}$

$\frac{b}{a}$ is a positive rational number.

Let the negative rational number be -$\frac{a}{b}$.

Reciprocal of -$\frac{a}{b} = -\frac{b}{a}$

-$\frac{b}{a}$ is a negative rational number.

(v) Reciprocal of $\frac{0}{1} = \frac{1}{0}$

$\frac{1}{0}$ is not defined.

(vi) Let the positive rational number be $\frac{a}{b}$.

Reciprocal of $\frac{a}{b} = \frac{b}{a}$

$\frac{a}{b} \times \frac{b}{a} =\frac{a \times b}{b \times a} =\frac{ab}{ab} =1$

(vii) Reciprocal of $\frac{1}{1} = \frac{1}{1} = 1$.

Reciprocal of $\frac{-1}{1} = \frac{1}{-1} = -1$.

(viii) If reciprocal of $\frac{m}{1} = \frac{n}{1}$

Reciprocal of $\frac{n}{1} = \frac{m}{1}$

Length = 9 cm

Breadth = $10\frac{2}{3}$ cm

Area = length x breadth

$$\begin{aligned}= 9 \times 10\frac{2}{3}\\[1em] = 9 \times \frac{32}{3}\\[1em] = \frac{9 \times 32}{1 \times 3}\\[1em] = \frac{288}{3}\\[1em] = 96 \text{ cm}^2\end{aligned}$$

Area = 96 cm²

Perimeter = 2 x (length + breadth)

$$\begin{aligned}= 2 \times \Big(9 + 10\frac{2}{3}\Big)\\[1em] = 2 \times \Big(\frac{9}{1} + \frac{32}{3}\Big)\end{aligned}$$

LCM of 1 and 3 is 3.

$$\begin{aligned}= 2 \times \Big(\frac{9 \times 3}{1 \times 3} + \frac{32 \times 1}{3 \times 1}\Big)\\[1em] = 2 \times \Big(\frac{27}{3} + \frac{32}{3}\Big)\\[1em] = 2 \times \Big(\frac{27 + 32}{3}\Big)\\[1em] = 2 \times \Big(\frac{59}{3}\Big)\\[1em] = \Big(\frac{59 \times 2}{3 \times 1}\Big)\\[1em] = \Big(\frac{118}{3}\Big)\\[1em] = 39\Big(\frac{1}{3}\Big)\end{aligned}$$

Perimeter = $39\frac{1}{3}$ cm

Hence, area of the rectangular piece of paper is 96 cm² and its perimeter is $39\frac{1}{3}$ cm.

Length = $7\frac{2}{5}$ m = $\frac{37}{5}$ m

Breadth = $4\frac{1}{6}$ = $\frac{25}{6}$m

Area = length x breadth

$$\begin{aligned}= \frac{37}{5} \times \frac{25}{6}\\[1em] = \frac{37 \times 25}{5 \times 6}\\[1em] = \frac{925}{30}\\[1em] = \frac{185}{6}\\[1em] = 30\frac{5}{6} \text{ m}^2\end{aligned}$$

Perimeter = 2 x (length + breadth)

$$\begin{aligned}= 2 \times \Big(\frac{37}{5} + \frac{25}{6}\Big)\\[1em]\end{aligned}$$

LCM of 5 and 6 is 2 x 3 x 5 = 30.

$$\begin{aligned}= 2 \times \Big(\frac{37 \times 6}{5 \times 6} + \frac{25 \times 5}{6 \times 5}\Big)\\[1em] = 2 \times \Big(\frac{222}{30} + \frac{125}{30}\Big)\\[1em] = 2 \times \Big(\frac{222 + 125}{30}\Big)\\[1em] = 2 \times \Big(\frac{347}{30}\Big)\\[1em] = \Big(\frac{347 \times 2}{30 \times 1}\Big)\\[1em] = \Big(\frac{694}{30}\Big)\\[1em] = \Big(\frac{347}{15}\Big)\\[1em] = 23\Big(\frac{2}{15}\Big)\end{aligned}$$

Area = $30\frac{5}{6}$ m² and perimeter = $23\frac{2}{15}$ m

$$\begin{aligned}-\frac{4}{9} ÷ -\frac{2}{3}\\[1em] = -\frac{4}{9} \times -\frac{3}{2}\\[1em] = \frac{4 \times 3}{9 \times 2}\\[1em] = \frac{12}{18}\\[1em] = \frac{2}{3}\end{aligned}$$

$-\frac{4}{9}$ divided by $-\frac{2}{3}$ gives $\frac{2}{3}$

Hence, Option 1 is the correct option.

Let $x$ be the number.

$$\begin{aligned}\frac{1}{2} ÷ x = -\frac{2}{3}\\[1em] ⇒ \frac{1}{2} \times \frac{1}{x} = -\frac{2}{3} \\[1em] ⇒ \frac{1}{2x} = -\frac{2}{3} \\[1em] ⇒ 2x = -\frac{3}{2} \\[1em] ⇒ x = \frac{1}{2} \times -\frac{3}{2}\\[1em] ⇒ x = -\frac{1 \times 3}{2 \times 2}\\[1em] ⇒ x = -\frac{3}{4}\end{aligned}$$

The rational number by which should $\frac{1}{2}$ be divided to get $-\frac{2}{3}$ is $-\frac{3}{4}$

Hence, Option 2 is the correct option.

We know that, division of rational numbers is not associative.

∴ a ÷ (b ÷ c) ≠ a ÷ b ÷ c

Hence, Option 4 is the correct option.

Let the number be $x$.

$$\begin{aligned}3\frac{5}{6} \times x = -7\frac{2}{3}\\[1em] ⇒ \frac{23}{6} \times x = -\frac{23}{3}\\[1em] ⇒ x = -\frac{23}{3} ÷ \frac{23}{6}\\[1em] ⇒ x = -\frac{23}{3} \times \frac{6}{23}\\[1em] ⇒ x = -\frac{23 \times 6}{3 \times 23}\\[1em] ⇒ x = -\frac{23 \times 6}{3 \times 23}\\[1em] ⇒ x = -\frac{138}{69}\\[1em] ⇒ x = -2\end{aligned}$$

Hence, Option 4 is the correct option.

$$\begin{aligned}(8 ÷ 3) ÷ (3 ÷ 8)\\[1em] = \frac{8}{3} ÷ \frac{3}{8}\\[1em] = \frac{8}{3} \times \frac{8}{3}\\[1em] = \frac{64}{9}\end{aligned}$$

Hence, Option 1 is the correct option.

$$\begin{aligned}1 ÷ \frac{1}{3}\\[1em] = 1 \times \frac{3}{1}\\[1em] = \frac{1 \times 3}{1 \times 1}\\[1em] = \frac{3}{1} \\[1em] = 3\end{aligned}$$

Hence, $1 ÷ \frac{1}{3} = 3$

$$\begin{aligned}3 ÷ \frac{3}{5}\\[1em] = 3 \times \frac{5}{3}\\[1em] = \frac{3 \times 5}{1 \times 3}\\[1em] = \frac{15}{3}\\[1em] = 5\end{aligned}$$

Hence, $3 ÷ \frac{3}{5} = 5$

$$\begin{aligned}-\frac{5}{12} ÷ \frac{1}{16}\\[1em] = -\frac{5}{12} \times \frac{16}{1}\\[1em] = -\frac{5 \times 16}{12 \times 1}\\[1em] = -\frac{80}{12}\\[1em] = -\frac{20}{3}\\[1em] = -6\frac{2}{3}\end{aligned}$$

Hence, $-\frac{5}{12} ÷ \frac{1}{16} = -6\frac{2}{3}$

$$\begin{aligned}-\frac{21}{16} ÷ \frac{-7}{8}\\[1em] = -\frac{21}{16} \times -\frac{8}{7}\\[1em] = \frac{21 \times 8}{16 \times 7}\\[1em] = \frac{168}{112}\\[1em] = \frac{3}{2}\\[1em] = 1\frac{1}{2}\end{aligned}$$

Hence, $-\frac{21}{16} ÷ \Big(\frac{-7}{8}\Big) = 1\frac{1}{2}$

$$\begin{aligned}0 ÷ \frac{-4}{7}\\[1em] = 0 \times \frac{-7}{4}\\[1em] = \frac{0 \times -7}{1 \times 4}\\[1em] = \frac{0}{4}\\[1em] = 0\end{aligned}$$

Hence, $0 ÷ \Big(\frac{-4}{7}\Big) = 0$

$$\begin{aligned}\frac{8}{-5} ÷ \frac{24}{25}\\[1em] = -\frac{8}{5} \times \frac{25}{24}\\[1em] = -\frac{8 \times 25}{5 \times 24}\\[1em] = -\frac{200}{120}\\[1em] = -\frac{5}{3}\\[1em] = -1\frac{2}{3}\end{aligned}$$

Hence, $\frac{8}{-5} ÷ \frac{24}{25} = -1\frac{2}{3}$

$$\begin{aligned}-\frac{3}{4} ÷ (-9)\\[1em] = -\frac{3}{4} \times \frac{-1}{9}\\[1em] = \frac{3 \times 1}{4 \times 9}\\[1em] = \frac{3}{36}\\[1em] = \frac{1}{12}\end{aligned}$$

Hence, $-\frac{3}{4} ÷ (-9) = \frac{1}{12}$

$$\begin{aligned}\frac{3}{4} ÷ -\frac{5}{12}\\[1em] = \frac{3}{4} \times -\frac{12}{5}\\[1em] = -\frac{3 \times 12}{4 \times 5}\\[1em] = -\frac{36}{20}\\[1em] = -\frac{9}{5}\\[1em] = -1\frac{4}{5}\end{aligned}$$

Hence, $\frac{3}{4} ÷ \Big(-\frac{5}{12}\Big) = -1\frac{4}{5}$

$$\begin{aligned}-5 ÷ -\frac{10}{11}\\[1em] = -5 \times -\frac{11}{10}\\[1em] = \frac{5 \times 11}{1 \times 10}\\[1em] = \frac{55}{10}\\[1em] = \frac{11}{2}\\[1em] = 5\frac{1}{2}\end{aligned}$$

Hence, $-5 ÷ \Big(-\frac{10}{11}\Big) = 5\frac{1}{2}$

$$\begin{aligned}\frac{-7}{11} ÷ \frac{-3}{44}\\[1em] = -\frac{7}{11} \times -\frac{44}{3}\\[1em] = \frac{7 \times 44}{11 \times 3}\\[1em] = \frac{308}{33}\\[1em] = \frac{28}{3}\\[1em] = 9\frac{1}{3}\end{aligned}$$

Hence, $\frac{-7}{11} ÷ \Big(\frac{-3}{44}\Big) = 9\frac{1}{3}$

$$\begin{aligned}3 ÷ \frac{1}{3}\\[1em] = 3 \times \frac{3}{1}\\[1em] = \frac{3 \times 3}{1 \times 1}\\[1em] = \frac{9}{1}\\[1em] = 9\end{aligned}$$

Hence, $3 ÷ \frac{1}{3} = 9$

$$\begin{aligned}-2 ÷ -\frac{1}{2}\\[1em] = -2 \times -\frac{2}{1}\\[1em] = \frac{2 \times 2}{1 \times 1}\\[1em] = \frac{4}{1}\\[1em] = 4\end{aligned}$$

Hence, $-2 ÷ -\frac{1}{2} = 4$

$$\begin{aligned}0 ÷ \frac{7}{-9}\\[1em] = 0 \times \frac{-9}{7}\\[1em] = \frac{0 \times -9}{1 \times 7}\\[1em] = \frac{0}{7}\\[1em] = 0\end{aligned}$$

Hence, $0 ÷ \frac{7}{-9} = 0$

$$\begin{aligned}\frac{-5}{8} ÷ \frac{1}{4}\\[1em] = \frac{-5}{8} \times \frac{4}{1}\\[1em] = \frac{-5 \times 4}{8 \times 1}\\[1em] = \frac{-20}{8}\\[1em] = \frac{-5}{2}\\[1em] =-2\frac{1}{2}\end{aligned}$$

Hence, $\frac{-5}{8} ÷ \frac{1}{4} = -2\frac{1}{2}$

$$\begin{aligned}-\frac{3}{4} ÷ -\frac{9}{16}\\[1em] = -\frac{3}{4} \times -\frac{16}{9}\\[1em] = \frac{3 \times 16}{4 \times 9}\\[1em] = \frac{48}{36}\\[1em] = \frac{4}{3}\\[1em] = 1\frac{1}{3}\end{aligned}$$

Hence, $-\frac{3}{4} ÷ -\frac{9}{16} = 1\frac{1}{3}$

Let the number be $x$.

$$\begin{aligned}\frac{4}{7} \times x = -2\\[1em] \Rightarrow \frac{4}{7} \times x = -\frac{2}{1}\\[1em] \Rightarrow x = -\frac{2}{1} ÷ \frac{4}{7}\\[1em] \Rightarrow x = -\frac{2}{1} \times \frac{7}{4}\\[1em] \Rightarrow x = -\frac{2 \times 7}{1 \times 4}\\[1em] \Rightarrow x = -\frac{14}{4}\\[1em] \Rightarrow x = -\frac{7}{2}\\[1em] \Rightarrow x = -3\frac{1}{2}\end{aligned}$$

The other number is $-3\frac{1}{2}$.

Let the number be $x$.

$$\begin{aligned}\frac{-2}{27} \times x = -\frac{4}{9}\\[1em] \Rightarrow x = -\frac{4}{9} ÷ \frac{-2}{27}\\[1em] \Rightarrow x = -\frac{4}{9} \times \frac{-27}{2}\\[1em] \Rightarrow x = \frac{4 \times 27}{9 \times 2}\\[1em] \Rightarrow x = \frac{108}{18}\\[1em] \Rightarrow x = 6\end{aligned}$$

The other number is 6.

$$\begin{aligned}m \times n = -\frac{25}{9}\\[1em] \Rightarrow \frac{5}{3} \times n = - \frac{25}{9}\\[1em] \Rightarrow n = -\frac{25}{9} ÷ \frac{5}{3}\\[1em] \Rightarrow n = -\frac{25}{9} \times \frac{3}{5}\\[1em] \Rightarrow n = -\frac{25 \times 3}{9 \times 5}\\[1em] \Rightarrow n = -\frac{75}{45}\\[1em] \Rightarrow n = -\frac{5}{3}\\[1em] \Rightarrow n = -1\frac{2}{3}\end{aligned}$$

if $m = \frac{5}{3}$, then $n = -1\frac{2}{3}.$

$$\begin{aligned}m \times n = -\frac{25}{9}\\[1em] \Rightarrow m \times -\frac{10}{9} = - \frac{25}{9}\\[1em] \Rightarrow m = -\frac{25}{9} ÷ -\frac{10}{9}\\[1em] \Rightarrow m = \frac{25}{9} \times \frac{9}{10}\\[1em] \Rightarrow m = \frac{25 \times 9}{9 \times 10}\\[1em] \Rightarrow m = \frac{225}{90}\\[1em] \Rightarrow m = \frac{5}{2}\\[1em] \Rightarrow m = 2\frac{1}{2}\end{aligned}$$

if $n = -\frac{10}{9}$, then $n = 2\frac{1}{2}.$

Let the number be $x$

$$\begin{aligned}-\frac{3}{4} \times x = -\frac{9}{16}\\[1em] \Rightarrow x = -\frac{9}{16} ÷ -\frac{3}{4}\\[1em] \Rightarrow x = \frac{9}{16} \times \frac{4}{3}\\[1em] \Rightarrow x = \frac{9 \times 4}{16 \times 3}\\[1em] \Rightarrow x = \frac{36}{48}\\[1em] \Rightarrow x = \frac{3}{4}\end{aligned}$$

$-\frac{3}{4}$ must be multiplied by $\frac{3}{4}$ so that the product is $-\frac{9}{16}$.

Let the number be $x$

$$\begin{aligned}-\frac{8}{13} \times x = 16\\[1em] \Rightarrow x = 16 ÷ -\frac{8}{13}\\[1em] \Rightarrow x = -\frac{16}{1} \times \frac{13}{8}\\[1em] \Rightarrow x = -\frac{16 \times 13}{1 \times 8}\\[1em] \Rightarrow x = -\frac{208}{8}\\[1em] \Rightarrow x = -26\end{aligned}$$

$-\frac{8}{13}$ must be multiplied by -26 so that the product is 16

Let the cost of one litre of milk be ₹$x$.

$$\begin{aligned}3\frac{1}{2} \times x = 49\\[1em] \Rightarrow \frac{7}{2} \times x = 49\\[1em] \Rightarrow x = 49 ÷ \frac{7}{2}\\[1em] \Rightarrow x = 49 \times \frac{2}{7}\\[1em] \Rightarrow x = \frac{49 \times 2}{1 \times 7}\\[1em] \Rightarrow x = \frac{98}{7}\\[1em] \Rightarrow x = 14\end{aligned}$$

The cost of one litre of milk = ₹14.

Let the cost of 1 metre of cloth be ₹ $x$.

$$\begin{aligned}3\frac{2}{5} \times x = 88\frac{1}{2}\\[1em] \Rightarrow \frac{17}{5} \times x = \frac{177}{2}\\[1em] \Rightarrow x = \frac{177}{2} ÷ \frac{17}{5}\\[1em] \Rightarrow x = \frac{177}{2} \times \frac{5}{17}\\[1em] \Rightarrow x = \frac{177 \times 5}{2 \times 17}\\[1em] \Rightarrow x = \frac{885}{34}\\[1em] \Rightarrow x = 26\frac{1}{34}\end{aligned}$$

Hence, The cost of 1 meter of cloth is ₹ $26\frac{1}{34}$.

The sum of $\frac{3}{7}$ and $\frac{-5}{14}$

$\frac{3}{7} + \frac{-5}{14}$

LCM of 7 and 14 is 2 x 7 = 14

$$\begin{aligned}= \frac{3 \times 2}{7 \times 2} + \frac{-5 \times 1}{14 \times 1}\\[1em] = \frac{6}{14} + \frac{-5}{14}\\[1em] = \frac{6 + (-5)}{14}\\[1em] = \frac{1}{14}\end{aligned}$$

Dividing the sum of $\frac{3}{7}$ and $\frac{-5}{14}$ by $-\frac{1}{2}$

$$\begin{aligned}\frac{1}{14} ÷ -\frac{1}{2}\\[1em] = \frac{1}{14} \times -\frac{2}{1}\\[1em] = -\frac{1 \times 2}{14 \times 1}\\[1em] = -\frac{2}{14}\\[1em] = -\frac{1}{7}\end{aligned}$$

On dividing the sum of $\frac{3}{7}$ and $\frac{-5}{14}$ by $-\frac{1}{2}$ we get $-\frac{1}{7}$.

$$\begin{aligned}(m + n) ÷ (m - n)\\[1em] = \Big(\frac{2}{3} + \frac{3}{2}\Big) ÷ \Big(\frac{2}{3} - \frac{3}{2}\Big)\end{aligned}$$

LCM of 3 and 2 is 2 x 3 = 6

$$\begin{aligned}= \Big(\frac{2 \times 2}{3 \times 2} + \frac{3 \times 3}{2 \times 3}\Big) ÷ \Big(\frac{2 \times 2}{3 \times 2} - \frac{3 \times 3}{2 \times 3}\Big)\\[1em] = \Big(\frac{4}{6} + \frac{9}{6}\Big) ÷ \Big(\frac{4}{6} - \frac{9}{6}\Big)\\[1em] = \Big(\frac{4 + 9}{6}\Big) ÷ \Big(\frac{4 - 9}{6}\Big)\\[1em] = \Big(\frac{13}{6}\Big) ÷ \Big(\frac{-5}{6}\Big)\\[1em] = \frac{13}{6} \times \frac{6}{-5}\\[1em] = \frac{13 \times 6}{6 \times -5}\\[1em] = -\frac{78}{30}\\[1em] = -\frac{13}{5}\end{aligned}$$

If $m$ = $\frac{2}{3}$ and $n$ = $\frac{3}{2}$ then $(m + n) ÷ (m - n) = -\frac{13}{5}$.

$$\begin{aligned}(m + n) ÷ (m - n)\\[1em] = \Big(\frac{3}{4} + \frac{4}{3}\Big) ÷ \Big(\frac{3}{4} - \frac{4}{3}\Big)\end{aligned}$$

LCM of 4 and 3 is 2 x 2 x 3 = 12

$$\begin{aligned}= \Big(\frac{3 \times 3}{4 \times 3} + \frac{4 \times 4}{3 \times 4}\Big) ÷ \Big(\frac{3 \times 3}{4 \times 3} - \frac{4 \times 4}{3 \times 4}\Big)\\[1em] = \Big(\frac{9}{12} + \frac{16}{12}\Big) ÷ \Big(\frac{9}{12} - \frac{16}{12}\Big)\\[1em] = \Big(\frac{9 + 16}{12}\Big) ÷ \Big(\frac{9 - 16}{12}\Big)\\[1em] = \Big(\frac{25}{12}\Big) ÷ \Big(\frac{-7}{12}\Big)\\[1em] = \frac{25}{12} \times \frac{12}{-7}\\[1em] = \frac{25 \times 12}{12 \times -7}\\[1em] = -\frac{300}{84}\\[1em] = -\frac{25}{7}\end{aligned}$$

If $m$ = $\frac{3}{4}$ and $n$ = $\frac{4}{3}$ then $(m + n) ÷ (m - n) = -\frac{25}{7}$.

$$\begin{aligned}(m + n) ÷ (m - n)\\[1em] =\Big[\frac{4}{5} + \Big(-\frac{3}{10}\Big)\Big] ÷ \Big[\frac{4}{5} - \Big(-\frac{3}{10}\Big)\Big]\end{aligned}$$

LCM of 5 and 10 is 2 x 5 = 10

$$\begin{aligned}= \Big[\frac{4 \times 2}{5 \times 2} + \Big(-\frac{3 \times 1}{10 \times 1}\Big)\Big] ÷ \Big[\frac{4 \times 2}{5 \times 2} - \Big(-\frac{3 \times 1}{10 \times 1}\Big)\Big]\\[1em] = \Big[\frac{8}{10} + \Big(-\frac{3}{10}\Big)\Big] ÷ \Big[\frac{8}{10} - \Big(-\frac{3}{10}\Big)]\\[1em] = \Big[\frac{8 + (-3)}{10}\Big] ÷ \Big[\frac{8 - (-3)}{10}\Big]\\[1em] = \Big[\frac{5}{10}\Big] ÷ \Big[\frac{11}{10}\Big]\\[1em] = \frac{5}{10} \times \frac{10}{11}\\[1em] = \frac{5 \times 10}{10 \times 11}\\[1em] = \frac{50}{110}\\[1em] = \frac{5}{11}\end{aligned}$$

If $m$ = $\frac{4}{5}$ and $n$ = -$\frac{3}{10}$ then $(m + n) ÷ (m - n) = \frac{5}{11}$.

Let the number be $x$.

$$\begin{aligned}\frac{-7}{15} \times x = -5\\[1em] \Rightarrow x = -5 ÷ \frac{-7}{15}\\[1em] \Rightarrow x = -\frac{5}{1} \times \frac{-15}{7}\\[1em] \Rightarrow x = \frac{5 \times 15}{1 \times 7}\\[1em] \Rightarrow x = \frac{75}{7}\\[1em] \Rightarrow x = 10\frac{5}{7}\end{aligned}$$

The other number is $10\frac{5}{7}$.

The sum of $\frac{5}{8}$ and $\frac{-11}{12}$

$\frac{5}{8} + \frac{-11}{12}$

LCM of 8 and 12 is 2 x 2 x 2 x 3 = 24

$$\begin{aligned}= \frac{5 \times 3}{8 \times 3} + \frac{-11 \times 2}{12 \times 2}\\[1em] = \frac{15}{24} + \frac{-22}{24}\\[1em] = \frac{15 + (-22)}{24}\\[1em] = \frac{-7}{24}\end{aligned}$$

The difference of $\frac{3}{7}$ and $\frac{5}{14}$

$\frac{3}{7} - \frac{5}{14}$

LCM of 7 and 14 is 2 x 7 = 14

$$\begin{aligned}= \frac{3 \times 2}{7 \times 2} - \frac{5 \times 1}{14 \times 1}\\[1em] = \frac{6}{14} - \frac{5}{14}\\[1em] = \frac{6 - 5}{14}\\[1em] = \frac{1}{14}\end{aligned}$$

Dividing the sum of $\frac{5}{8}$ and $\frac{-11}{12}$ by the difference of $\frac{3}{7}$ and $\frac{5}{14}$,

$$\begin{aligned}\frac{-7}{24} ÷ \frac{1}{14}\\[1em] = \frac{-7}{24} \times \frac{14}{1}\\[1em] = -\frac{7 \times 14}{24 \times 1}\\[1em] = -\frac{98}{24}\\[1em] = -\frac{49}{12}\\[1em] = -4\frac{1}{12}\end{aligned}$$

$\Big(\frac{5}{8} + \frac{-11}{12}\Big) ÷ \Big(\frac{3}{7} - \frac{5}{14}\Big) = -4\frac{1}{12}$.

Area of a rectangular plate = $5\frac{5}{7}$ m² = $\frac{40}{7}$ m²

Length of a rectangular plate = $3\frac{3}{4}$ m = $\frac{15}{4}$ m

Let the breadth of the rectangular plate be b.

Area = length x breadth

$$\begin{aligned}\frac{40}{7} = \frac{15}{4} \times b\\[1em] \Rightarrow b = \frac{40}{7} ÷ \frac{15}{4}\\[1em] \Rightarrow b = \frac{40}{7} \times \frac{4}{15}\\[1em] \Rightarrow b = \frac{40 \times 4}{7 \times 15}\\[1em] \Rightarrow b = \frac{160}{105}\\[1em] \Rightarrow b = \frac{32}{21}\\[1em] \Rightarrow b = 1\frac{11}{21}\end{aligned}$$

Perimeter = 2(length + breadth)

$= 2 \times \Big(\frac{15}{4} + \frac{32}{21}\Big)$

LCM of 4 and 21 is 2 x 2 x 3 x 7 = 84

$$\begin{aligned}= 2 \times \Big(\frac{15 \times 21}{4 \times 21} + \frac{32 \times 4}{21 \times 4}\Big)\\[1em] = 2 \times \Big(\frac{315}{84} + \frac{128}{84}\Big)\\[1em] = 2 \times \Big(\frac{315 + 128}{84}\Big)\\[1em] = 2 \times \Big(\frac{443}{84}\Big)\\[1em] = \Big(\frac{443 \times 2}{84}\Big)\\[1em] = \Big(\frac{886}{84}\Big)\\[1em] = \Big(\frac{443}{42}\Big)\\[1em] = 10\frac{23}{42}\end{aligned}$$

Hence, breadth = $1\frac{11}{21}$ and perimeter = $10\frac{23}{42}$

Area of a rectangular paper = $7\frac{3}{26}$ cm² = $\frac{185}{26}$ cm²

Breadth of a rectangular paper = $2\frac{9}{13}$ m = $\frac{35}{13}$ cm

Let the length of the piece of paper be l.

Area = length x breadth

$$\begin{aligned}\frac{185}{26} = l \times \frac{35}{13}\\[1em] \Rightarrow l = \frac{185}{26} ÷ \frac{35}{13}\\[1em] \Rightarrow l = \frac{185}{26} \times \frac{13}{35}\\[1em] \Rightarrow l = \frac{185 \times 13}{26 \times 35}\\[1em] \Rightarrow l = \frac{2405}{910}\\[1em] \Rightarrow l = \frac{37}{14}\\[1em] \Rightarrow l = 2\frac{9}{14}\end{aligned}$$

Perimeter = 2(length + breadth)

$= 2 \times \Big(\frac{37}{14} + \frac{35}{13}\Big)$

LCM of 14 and 13 is 2 x 7 x 13 = 182

$$\begin{aligned}= 2 \times \Big(\frac{37 \times 13}{14 \times 13} + \frac{35 \times 14}{13 \times 14}\Big)\\[1em] = 2 \times \Big(\frac{481}{182} + \frac{490}{182}\Big)\\[1em] = 2 \times \Big(\frac{481 + 490}{182}\Big)\\[1em] = 2 \times \Big(\frac{971}{182}\Big)\\[1em] = \Big(\frac{971 \times 2}{182}\Big)\\[1em] = \Big(\frac{1942}{182}\Big)\\[1em] = \Big(\frac{971}{91}\Big)\\[1em] = 10\frac{61}{91}\end{aligned}$$

Length = $2\frac{9}{14}$ and perimeter = $10\frac{61}{91}$

In this number line, there are 5 small lines between every 2 consecutive integers, which means moving one step left from 0 gives $-\frac{1}{5}$.

Point A is 4 step left from -1. So, A = $-1\frac{4}{5}$

Point B is 3 step right from 2. So, B = $2\frac{3}{5}$

Hence, Option 3 is the correct option.

As we know, A = $-1\frac{4}{5}$ and B = $2\frac{3}{5}$.

Length = $\frac{9}{5} + \frac{13}{5}$ $$\begin{aligned}= \frac{9 + 13}{5}\\[1em] = \frac{22}{5}\\[1em] = 4\frac{2}{5}\end{aligned}$$

Hence, Option 3 is the correct option.

For any two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, $\Big(\frac{a + c}{b + d}\Big)$ is also a rational number with its value lying between $\frac{a}{b}$ and $\frac{c}{d}$.

Hence, Option 3 is the correct option.

As we know that, for any two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, $\Big(\frac{a + c}{b + d}\Big)$ is also a rational number with its value lying between $\frac{a}{b}$ and $\frac{c}{d}$.

The rational number between $\frac{1}{3}$ and $\frac{1}{2}$ is $\Big(\frac{1 + 1}{3 + 2}\Big)$

= $\Big(\frac{2}{5}\Big)$

The rational number between $\frac{1}{3}$ and $\frac{2}{5}$ is $\Big(\frac{1 + 2}{3 + 5}\Big)$

= $\Big(\frac{3}{8}\Big)$

The rational number between $\frac{1}{2}$ and $\frac{2}{5}$ is $\Big(\frac{1 + 2}{2 + 5}\Big)$

= $\Big(\frac{3}{7}\Big)$

$\frac{3}{7}$ and $\frac{3}{8}$ are two rational number between $\frac{1}{3}$ and $\frac{1}{2}$

Hence, option 1 is the correct option.

In this number line, there are 4 small lines between every 2 consecutive integers, which means moving one step towards left from 0 gives $-\frac{1}{4}$.

Point C is 3 step left from -1. So, C = $-1\frac{3}{4} = -\frac{7}{4}$

Point E is 3 step right from 0. So, E = $\frac{3}{4}$

∴ $-\frac{7}{4}$ and $\frac{3}{4}$ are represented by C and E, respectively.

Hence, option 3 is the correct option.

Draw a number line as shown below:

In this number line

OA = AB = ...............= OA' = A'B' = 1 unit

Since the denominator of each given rational number is 4, divide each OA, AB, BC, OA', A'B',etc into four equal parts.

To represent $\frac{1}{4}$, move one step towards the right side of 0 to reach P as shown.

As, OA = 1 unit, therefore OP = $\frac{1}{4}$ unit.

Hence, to represent $\frac{3}{4}$, move 3 steps towards the right side of 0 to reach point Q. So, Q represent $\frac{3}{4}$.

In the same way to represent $\frac{-3}{4}$, move 3 steps towards the left side of 0 to reach point R. So, R represent $\frac{-3}{4}$.

So, S represent $\frac{7}{4}$ and T represent $\frac{-7}{4}$.

Draw a number line as shown below:

In this number line

OA = AB = ...............= OA' = A'B' = 1 unit

Since the denominator of each given rational number is 3, divide each OA, AB, BC, OA', A'B', etc into three equal parts.

To represent $\frac{1}{3}$, move one step towards the right side of 0 to reach P as shown.

As, OA = 1 unit, therefore OP = $\frac{1}{3}$ unit.

Hence, to represent $\frac{2}{3}$, move 2 steps towards the right side of 0 to reach point Q. So, Q represents $\frac{2}{3}$.

In the same way to represent $\frac{-2}{3}$, move 2 steps towards the left side of 0 to reach point R. So, R represents $\frac{-2}{3}$.

Similarly, S represents $\frac{-8}{3}$, T represents $\frac{7}{3}$ and B' represents -2.

As we know that, for any two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, $\Big(\frac{a + c}{b + d}\Big)$ is also a rational number with its value lying between $\frac{a}{b}$ and $\frac{c}{d}$.

The rational number between $\frac{3}{5}$ and $\frac{5}{8}$ is $\Big(\frac{3 + 5}{5 + 8}\Big)$

= $\Big(\frac{8}{13}\Big)$

Hence, one rational number between $\frac{3}{5}$ and $\frac{5}{8}$ is $\frac{8}{13}$.

As we know that, for any two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, $\Big(\frac{a + c}{b + d}\Big)$ is also a rational number with its value lying between $\frac{a}{b}$ and $\frac{c}{d}$.

The rational number between $\frac{1}{2}$ and $\frac{2}{1}$ is $\Big(\frac{1 + 2}{2 + 1}\Big)$

= $\Big(\frac{3}{3}\Big)$

=1

Hence, one rational number between $\frac{1}{2}$ and 2 is 1.

As we know that, for any two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, $\Big(\frac{a + c}{b + d}\Big)$ is also a rational number with its value lying between $\frac{a}{b}$ and $\frac{c}{d}$.

Given numbers = $\frac{5}{7}$ and $\frac{3}{8}$

$$\begin{aligned}= \frac{5}{7}, \frac{5 + 3}{7 + 8} ,\frac{3}{8} \\[1em] = \frac{5}{7}, \frac{8}{15}, \frac{3}{8} \\[1em] = \frac{5}{7}, \frac{5 + 8}{7 + 15}, \frac{8}{15}, \frac{3}{8} \\[1em] = \frac{5}{7}, \frac{13}{22}, \frac{8}{15}, \frac{3}{8}\end{aligned}$$

Hence, required rational numbers between $\frac{5}{7}$ and $\frac{3}{8}$ are : $\frac{13}{22}$ and $\frac{8}{15}$

As we know that, for any two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, $\Big(\frac{a + c}{b + d}\Big)$ is also a rational number with its value lying between $\frac{a}{b}$ and $\frac{c}{d}$.

Given numbers = $\frac{8}{11}$ and $\frac{4}{9}$

$$\begin{aligned}= \frac{8}{11}, \frac{8 + 4}{11 + 9} ,\frac{4}{9} \\[1em] = \frac{8}{11}, \frac{12}{20}, \frac{4}{9} \\[1em] = \frac{8}{11}, \frac{3}{5}, \frac{4}{9} \\[1em] = \frac{8}{11}, \frac{8 + 3}{11 + 5}, \frac{3}{5}, \frac{4}{9} \\[1em] = \frac{8}{11}, \frac{11}{16}, \frac{3}{5}, \frac{4}{9}\\[1em] = \frac{8}{11}, \frac{11}{16}, \frac{3}{5},\frac{3 + 4}{5 + 9}, \frac{4}{9}\\[1em] = \frac{8}{11}, \frac{11}{16}, \frac{3}{5},\frac{7}{14}, \frac{4}{9}\\[1em] = \frac{8}{11}, \frac{11}{16}, \frac{3}{5},\frac{1}{2}, \frac{4}{9}\end{aligned}$$

Hence, required rational numbers between $\frac{8}{11}$ and $\frac{4}{9}$ are : $\frac{11}{16},\frac{3}{5}$ and $\frac{1}{2}$

LCM of 5 and 3 is 3 x 5 = 15

Make denominator of each given rational number equal to 15 (the LCM).

$\frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15}$

and

$\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}$

Since five rational numbers are required between $\frac{3}{5}$ and $\frac{2}{3}$; multiply the numerator and the denominator of each rational number by 5 + 1 = 6.

$\therefore \frac{9}{15} = \frac{9 \times 6}{15 \times 6} = \frac{54}{90}$

and

$\frac{10}{15} = \frac{10 \times 6}{15 \times 6} = \frac{60}{90}$

⇒ Required rational numbers between $\frac{3}{5}$ and $\frac{2}{3}$ are : $\frac{54}{90} , \frac{55}{90} , \frac{56}{90} , \frac{57}{90} , \frac{58}{90} , \frac{59}{90} , \frac{60}{90}$

= $\frac{3}{5} , \frac{11}{18} , \frac{28}{45} , \frac{19}{30} , \frac{29}{45} , \frac{59}{90} , \frac{2}{3}$

Hence, $\frac{11}{18} , \frac{28}{45} , \frac{19}{30} , \frac{29}{45}$ and $\frac{59}{90}$ lie between $\frac{3}{5}$ and $\frac{2}{3}$.

LCM of 6 and 9 is 2 x 3 x 3 = 18

Make denominator of each given rational number equal to 18 (the LCM).

$\frac{5}{6} = \frac{5 \times 3}{6 \times 3} = \frac{15}{18}$

and

$\frac{8}{9} = \frac{8 \times 2}{9 \times 2} = \frac{16}{18}$

Since six rational numbers are required between $\frac{5}{6}$ and $\frac{8}{9}$; multiply the numerator and the denominator of each rational number by 6 + 1 = 7.

$\frac{15}{18} = \frac{15 \times 7}{18 \times 7} = \frac{105}{126}$

and

$\frac{16}{18} = \frac{16 \times 7}{18 \times 7} = \frac{112}{126}$

Required rational numbers between $\frac{5}{6}$ and $\frac{8}{9}$ are : $\frac{105}{126} , \frac{106}{126} , \frac{107}{126} , \frac{108}{126} , \frac{109}{126} , \frac{110}{126} , \frac{111}{126} , \frac{112}{126}$

= $\frac{5}{6} , \frac{53}{63} , \frac{107}{126} , \frac{6}{7} , \frac{109}{126} , \frac{55}{63} , \frac{37}{42} , \frac{8}{9}$

Hence, $\frac{5}{6} , \frac{53}{63} , \frac{107}{126} , \frac{6}{7} , \frac{109}{126}$ and $\frac{37}{42}$ lie between $\frac{5}{6}$ and $\frac{8}{9}$.

2 = $\frac{2}{1}$

3 = $\frac{3}{1}$

Since seven rational numbers are required between $\frac{2}{1}$ and $\frac{3}{1}$; multiply the numerator and the denominator of each rational number by 7 + 1 = 8.

$\frac{2}{1} = \frac{2 \times 8}{1 \times 8} = \frac{16}{8}$

and

$\frac{3}{1} = \frac{3 \times 8}{1 \times 8} = \frac{24}{8}$

Required rational numbers between $\frac{2}{1}$ and $\frac{3}{1}$ are :

$\frac{16}{8} , \frac{17}{8} , \frac{18}{8} , \frac{19}{8} , \frac{20}{8} , \frac{21}{8} , \frac{22}{8} , \frac{23}{8} , \frac{24}{8}$

= $\frac{2}{1} , \frac{17}{8} , \frac{9}{4} , \frac{19}{8} , \frac{5}{2} , \frac{21}{8} , \frac{11}{4} , \frac{23}{8} , \frac{3}{1}$

= $2 , 2\frac{1}{8} , 2\frac{1}{4} , 2\frac{3}{8} , 2\frac{1}{2} , 2\frac{5}{8} , 2\frac{3}{4} , 2\frac{7}{8} , 3$

Hence, $2\frac{1}{8} , 2\frac{1}{4} , 2\frac{3}{8} , 2\frac{1}{2} , 2\frac{5}{8} , 2\frac{3}{4}$ and $2\frac{7}{8}$ lie between 2 and 3.

$$\begin{aligned}\frac{0}{5} ÷ \frac{0}{8}\\[1em] = \frac{0}{5} \times \frac{8}{0}\\[1em] = \frac{0 \times 8}{5 \times 0}\\[1em] = \frac{0}{0}\\[1em]\end{aligned}$$ $\frac{0}{0}$ is undefined.

Hence, option 4 is the correct option.

a + b = 0

a = 0 - b

a = -b

a and b are numerically equal but opposite in sign.

Hence, option 3 is the correct option

Additive inverse of $\frac{3}{8}$ = $-\frac{3}{8}$

$$\begin{aligned}\frac{3}{8} \times -\frac{3}{8}\\[1em] = -\frac{3 \times 3}{8 \times 8}\\[1em] = -\frac{9}{64}\end{aligned}$$

Hence, option 4 is the correct option.

Reciprocal of $\frac{2}{3}$ = $\frac{3}{2}$

We need to find the sum of $\frac{2}{3}$ and $\frac{3}{2}$

$\frac{2}{3} + \frac{3}{2}$

LCM of 3 and 2 is 2 x 3 = 6

$$\begin{aligned}\frac{2 \times 2}{3 \times 2} + \frac{3 \times 3}{2 \times 3}\\[1em] = \frac{4}{6} + \frac{9}{6}\\[1em] = \frac{4 + 9}{6}\\[1em] = \frac{13}{6}\\[1em] = 2\frac{1}{6}\end{aligned}$$

Hence, option 2 is the correct option.

Let the number be $x$.

$$\begin{aligned}\frac{2}{5} \times x = -1\\[1em] ⇒ \frac{2}{5} \times x = -\frac{1}{1}\\[1em] ⇒ x = -\frac{1}{1} ÷ \frac{2}{5}\\[1em] ⇒ x = -\frac{1}{1} \times \frac{5}{2}\\[1em] ⇒ x = -\frac{1 \times 5}{1 \times 2}\\[1em] ⇒ x = -\frac{5}{2}\end{aligned}$$

Hence, option 3 is the correct option.

For a rational number $\frac{7}{9}, \frac{7}{9} - 0 = \frac{7}{9} \text{ and } 0 - \frac{7}{9} = - \frac{7}{9}$

We know that,

For any number a,

a - 0 = a and 0 - a ≠ 0

Thus, subtraction only has right identity.

So, statement 1 is true.

Subtraction has 0 as right identity element.

So, statement 2 is false.

Hence, option 3 is the correct option.

The additive inverse of a number a is a number -a such that :

⇒ a + (-a) = 0.

So, reason (R) is true.

According to Assertion: Additive inverse of $\frac{2}{5}$ is $-\frac{5}{2}$.

$$\begin{aligned}\Rightarrow \frac{2}{5} + (-\frac{5}{2})\\[1em] \Rightarrow \frac{2}{5} - \frac{5}{2}\\[1em] \Rightarrow \frac{4}{10} - \frac{25}{10}\\[1em] \Rightarrow \frac{4 - 25}{10}\\[1em] \Rightarrow \frac{-21}{10}\\[1em] \ne 0\end{aligned}$$

So, assertion (A) is false.

Hence, option 4 is the correct option.

We know that,

The multiplicative inverse of any number a is it's reciprocal i.e. $\frac{1}{a}$.

So, reason (R) is true.

The reciprocal of $-\frac{7}{5} = -\frac{1}{\frac{7}{5}} = -\frac{5}{7}$.

So, assertion (A) is true.

∴ Both A and R are true and R is correct reason for A.

Hence, option 1 is the correct option.

According to Assertion:

$$\begin{aligned}\Rightarrow \frac{1}{2} + 2 \\[1em] \Rightarrow \frac{1}{2} + \frac{4}{2} \\[1em] \Rightarrow \frac{1 + 4}{2} \\[1em] \Rightarrow \frac{5}{2}\end{aligned}$$

A number is rational if it can be written in the form $\frac{p}{q}$, where p and q are integers.

Since, $\frac{5}{2}$ is in the form of $\frac{p}{q}$ as well as 5 and 2 are integers.

So, assertion (A) is true.

According to commutative property of addition: When two numbers are added together, then a change in their positions does not change the result.

When $\frac{p}{q} \text{ and } \frac{r}{s}$ are any two rational numbers then $\frac{p}{q} + \frac{r}{s} = \frac{r}{s} + \frac{p}{q}$, as addition of rational numbers is a commutative property.

So, reason (R) is true but it does not explain assertion.

Hence, option 2 is the correct option.

When 0 is divided by any non-zero number, the result is 0.

⇒ 0 ÷ $\frac{11}{12} = 0$

Since, 0 = $\frac{0}{1}$ is in the form of $\frac{p}{q}$.

So, assertion (A) is true.

The division of a rational number $\frac{a}{b}$ by another non-zero rational number $\frac{c}{d}$ is:

$$\begin{aligned}\Rightarrow \frac{a}{b} ÷ \frac{c}{d}\\[1em] \Rightarrow \frac{a}{b} \times \frac{d}{c}\\[1em] \Rightarrow \frac{ad}{bc}\end{aligned}$$

Since, $\frac{ad}{bc}$ is in the form of $\frac{p}{q}$.

So, reason is true. But it does not explains about assertion.

Hence, option 2 is the correct option.

The rational number is $\frac{0}{1}$

The reciprocal of $\frac{0}{1} = \frac{1}{0}$ (not defined).

Hence, 0 is the rational number that does not have a reciprocal.

Reciprocal of $\frac{1}{1} = \frac{1}{1} = 1$.

Reciprocal of $\frac{-1}{1} = \frac{1}{-1} = -1$.

The numbers 1 and -1 are their own reciprocal.

$$\begin{aligned}-\frac{8}{17} + \frac{-8}{17}\\[1em] = \frac{-8 + (-8)}{17}\\[1em] = \frac{-16}{17}\end{aligned}$$

The reciprocal of $\frac{-16}{17} = -\frac{17}{16}$.

LCM of 2 and 3 is 2 x 3 = 6

Make denominator of each given rational number equal to 6 (the LCM).

$-\frac{3}{2} = -\frac{3 \times 3}{2 \times 3} = -\frac{9}{6}$

and

$\frac{5}{3} = \frac{5 \times 2}{3 \times 2} = \frac{10}{6}$

The rational number between $-\frac{9}{6}$ and $\frac{10}{6}$ are $-\frac{8}{6}, -\frac{7}{6}, -\frac{6}{6}, -\frac{5}{6}, -\frac{4}{6}, -\frac{3}{6}, -\frac{2}{6}, -\frac{1}{6}, \frac{0}{6}, \frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}, \frac{6}{6}, \frac{7}{6}, \frac{8}{6}, \frac{9}{6}$

From these rational numbers we can take any five rational number.

Hence, required rational numbers between $-\frac{3}{2}$ and $\frac{5}{3}$ are :

$-\frac{8}{6} , -\frac{6}{6} , -\frac{4}{6} , -\frac{2}{6} , \frac{2}{6}$

= $-\frac{4}{3} , -\frac{1}{1} , -\frac{2}{3} , -\frac{1}{3} , \frac{1}{3}$

Hence, $-1\frac{1}{3} , -1 , -\frac{2}{3} , -\frac{1}{3}$ and $\frac{1}{3}$ lies between $-\frac{3}{2}$ and $\frac{5}{3}$.

There are infinite many rational number between -4 and ∞.

Hence, -3, -2, -1, 1, 2 are five rational number greater than -4.

Let $x$ be added to $-2\frac{1}{2}$.

$$\begin{aligned}-2\frac{1}{2} + x = -3\frac{1}{3}\\[1em] \Rightarrow -\frac{5}{2} + x = -\frac{10}{3}\\[1em] \Rightarrow x = -\frac{10}{3} - \frac{-5}{2}\end{aligned}$$

LCM of 3 and 2 is 2 x 3 = 6

$$\begin{aligned}\Rightarrow x = \frac{-10 \times 2}{3 \times 2} - \frac{-5 \times 3}{2 \times 3}\\[1em] \Rightarrow x = \frac{-20}{6} - \frac{-15}{6}\\[1em] \Rightarrow x = \frac{-20 - (-15)}{6}\\[1em] \Rightarrow x = \frac{-20 + 15}{6}\\[1em] \Rightarrow x = \frac{-5}{6}\end{aligned}$$

The number added to $-2\frac{1}{2}$ to get $-3\frac{1}{3}$ is $\frac{-5}{6}$.

Let $x$ be subtracted from $2\frac{1}{2}$.

$$\begin{aligned}2\frac{1}{2} - x = -3\frac{1}{3}\\[1em] \Rightarrow \frac{5}{2} - x = -\frac{10}{3}\\[1em] \Rightarrow x = \frac{5}{2} - \frac{-10}{3}\end{aligned}$$

LCM of 2 and 3 is 2 x 3 = 6

$$\begin{aligned}\Rightarrow x = \frac{5 \times 3}{2 \times 3} - \frac{-10 \times 2}{3 \times 2}\\[1em] \Rightarrow x = \frac{15}{6} - \frac{-20}{6}\\[1em] \Rightarrow x = \frac{15 - (-20)}{6}\\[1em] \Rightarrow x = \frac{15 + 20}{6}\\[1em] \Rightarrow x = \frac{35}{6}\\[1em] \Rightarrow x = 5\frac{5}{6}\end{aligned}$$

The number subtracted from $2\frac{1}{2}$ to get $-3\frac{1}{3}$ is $5\frac{5}{6}$.

To prove:

m - n ≠ n - m

LHS:

$$\begin{aligned}m - n\\[1em] -\frac{7}{9} - \frac{5}{6}\end{aligned}$$

LCM of 9 and 6 is 2 x 3 x 3 = 18

$$\begin{aligned}-\frac{7 \times 2}{9 \times 2} - \frac{5 \times 3}{6 \times 3}\\[1em] = -\frac{14}{18} - \frac{15}{18}\\[1em] = \frac{-14 - 15}{18}\\[1em] = \frac{-29}{18}\\[1em] = -1\frac{11}{18}\end{aligned}$$

RHS:

$$\begin{aligned}n - m\\[1em] \frac{5}{6} - \Big(-\frac{7}{9}\Big)\\[1em] = \frac{5}{6} + \frac{7}{9}\end{aligned}$$

LCM of 6 and 9 is 2 x 3 x 3 = 18

$$\begin{aligned}= \frac{5 \times 3}{6 \times 3} + \frac{7 \times 2}{9 \times2}\\[1em] = \frac{15}{18} + \frac{14}{18}\\[1em] = \frac{15 + 14}{18}\\[1em] = \frac{29}{18}\\[1em] = 1\frac{11}{18}\end{aligned}$$

Hence, LHS ≠ RHS

m - n ≠ n - m

To prove:

-(m + n) = (-m) + (-n)

LHS:

$$\begin{aligned}-(m + n)\\[1em] -\Big(-\frac{7}{9} + \frac{5}{6}\Big)\end{aligned}$$

LCM of 9 and 6 is 2 x 3 x 3 = 18

$$\begin{aligned}= -\Big(-\frac{7 \times 2}{9 \times 2} + \frac{5 \times 3}{6 \times 3}\Big)\\[1em] = -\Big(-\frac{14}{18} + \frac{15}{18}\Big)\\[1em] = -\Big(\frac{-14 + 15}{18}\Big)\\[1em] = -\Big(\frac{1}{18}\Big)\end{aligned}$$

RHS:

$$\begin{aligned}(-m) + (-n)\\[1em] = -\Big(-\frac{7}{9}\Big) + \Big(-\frac{5}{6}\Big)\\[1em] = \Big(\frac{7}{9}\Big) + \Big(-\frac{5}{6}\Big)\end{aligned}$$

LCM of 9 and 6 is 2 x 3 x 3 = 18

$$\begin{aligned}= \Big(\frac{7 \times 2}{9 \times 2}\Big) + \Big(-\frac{5 \times 3}{6 \times 3}\Big)\\[1em] = \Big(\frac{14}{18}\Big) + \Big(-\frac{15}{18}\Big)\\[1em] = \Big(\frac{14 + (-15)}{18}\Big)\\[1em] = \Big(\frac{-1}{18}\Big)\end{aligned}$$

Hence, LHS = RHS

$\therefore -(m + n) = (-m) + (-n)$

The rational numbers are $-\frac{7}{3}$ and $\frac{7}{4}$.

LCM of 3 and 4 is 2 x 2 x 3 = 12

Hence, the rational number will be-

$$\begin{aligned}-\frac{7 \times 4}{3 \times 4} , \frac{7 \times 3}{4 \times 3}\\[1em] = -\frac{28}{12} , \frac{21}{12}\\[1em] = -2\frac{4}{12} , 1\frac{9}{12}\end{aligned}$$

Draw a number line as shown below:

In this number line OA = AB = ........... = OA' = A'B' = 1 unit

Since, the denominator of each given rational numbers is 12, divide each of OA, AB,...,OA', A'B', etc. into twelve equal parts.

To represent $\frac{1}{12}$, moves one step towards the right side of O to reach point P as shown.

Hence, OA = 1 unit , therefore OP = $\frac{1}{12}$ unit and so P represents $\frac{1}{12}$.

In the same way, to represent $-2\frac{4}{12}$ , move 4 steps toward the left side of B' to reach point Q. Clearly, Q represents $-\frac{7}{3}$.

Similarly, to represent $1\frac{9}{12}$ , move 9 steps toward the right side of A to reach point R. Clearly, R represents $\frac{7}{4}$.

LCM of 6 and 8 is 2 x 2 x 2 x 3 = 24

$$\begin{aligned}-\frac{5 \times 4}{6 \times 4} + \frac{3 \times 3}{8 \times 3} \\[1em] =-\frac{20}{24} + \frac{9}{24} \\[1em] =\frac{-20 + 9}{24} \\[1em] =\frac{-11}{24}\end{aligned}$$

∴ $-\frac{5}{6} + \frac{3}{8} = -\frac{11}{24}$

LCM of 8 and 6 is 2 x 2 x 2 x 3 = 24

$$\begin{aligned}-\frac{5 \times 4}{6 \times 4} - \frac{3 \times 3}{8 \times 3} \\[1em] =-\frac{20}{24} - \frac{9}{24} \\[1em] =\frac{-20 - 9}{24} \\[1em] =\frac{-29}{24}\\[1em] =-1\frac{5}{24}\end{aligned}$$

∴ $-\frac{5}{6} - \frac{3}{8} = -1\frac{5}{24}$

$$\begin{aligned}-\frac{5}{6} \times \frac{3}{8}\\[1em] = \frac{-5 \times 3}{6 \times 8}\\[1em] = \frac{-15}{48}\\[1em] = -\frac{5}{16}\end{aligned}$$

$-\frac{5}{6} \times \frac{3}{8} = -\frac{5}{16}$

$$\begin{aligned}\frac{3}{8} ÷ -\frac{5}{6}\\[1em] = \frac{3}{8} \times -\frac{6}{5}\\[1em] = \frac{3 \times (-6)}{8 \times 5}\\[1em] = \frac{-18}{40}\\[1em] = -\frac{9}{20}\end{aligned}$$

$\frac{3}{8} ÷ -\frac{5}{6} = -\frac{9}{20}$

Let the number be $x$

$$\begin{aligned}\frac{12}{17} \times x = -\frac{4}{7}\\[1em] ⇒ x = -\frac{4}{7} ÷ \frac{12}{17}\\[1em] ⇒ x = -\frac{4}{7} \times \frac{17}{12}\\[1em] ⇒ x = -\frac{4 \times 17}{7 \times 12}\\[1em] ⇒ x = -\frac{68}{84}\\[1em] ⇒ x = -\frac{17}{21}\end{aligned}$$

$-\frac{17}{21}$ must be multiplied by $\frac{12}{17}$ so that the product is $-\frac{4}{7}$.

Let the number be $x$

$$\begin{aligned}\frac{12}{17} ÷ x = -\frac{4}{7}\\[1em] \Rightarrow \frac{12}{17} \times \frac{1}{x} = -\frac{4}{7} \\[1em] \Rightarrow x = \frac{12}{17} \times -\frac{7}{4}\\[1em] \Rightarrow x = -\frac{12 \times 7}{17 \times 4}\\[1em] \Rightarrow x = -\frac{84}{68}\\[1em] \Rightarrow x = -\frac{21}{17}\\[1em] \Rightarrow x = -1\frac{4}{17}\end{aligned}$$

$-1\frac{4}{17}$ must be divided by $\frac{12}{17}$ so that the product is $-\frac{4}{7}$.