Rational Numbers — Exercise 1.1

Add the following:

$\dfrac{4}{7} \text{ and }\dfrac{-5}{7}$

Answer

Both rational numbers have the same (positive) denominator 7.

$\Rightarrow \dfrac{4}{7} + \dfrac{-5}{7} \\[1em] \Rightarrow \dfrac{4 + (-5)}{7} \\[1em] \Rightarrow \dfrac{-1}{7}$

Hence, $\mathbf{\dfrac{4}{7} + \dfrac{-5}{7} = \dfrac{-1}{7}}$.

Add the following:

$\dfrac{7}{-13} \text{ and }\dfrac{4}{-13}$

Answer

First express each rational number with a positive denominator.

$\Rightarrow \dfrac{7}{-13} = \dfrac{7 \times (-1)}{-13 \times (-1)} = \dfrac{-7}{13} \\[1em] \Rightarrow \dfrac{4}{-13} = \dfrac{4 \times (-1)}{-13 \times (-1)} = \dfrac{-4}{13}$

Both rational numbers now have the same (positive) denominator 13.

$\Rightarrow \dfrac{-7}{13} + \dfrac{-4}{13} \\[1em] \Rightarrow \dfrac{(-7) + (-4)}{13} \\[1em] \Rightarrow \dfrac{-11}{13}$

Hence, $\mathbf{\dfrac{7}{-13} + \dfrac{4}{-13} = \dfrac{-11}{13}}$.

Write the following rational number in Standard Form:

$\dfrac{5}{-3}$

Answer

A rational number $\dfrac{p}{q}$ is in standard form if $q \gt 0 \text{ and } p, q$ have no common factor other than 1.

First express rational number with a positive denominator.

$\dfrac{5}{-3} = \dfrac{5 \times (-1)}{-3 \times (-1)} = \dfrac{-5}{3}$

HCF of 5 and 3 = 1.

Hence, the standard form of$\mathbf{\dfrac{5}{-3} \text{ is } \dfrac{-5}{3}}$

Write the following rational number in Standard Form:

$\dfrac{-12}{36}$

Answer

A rational number $\dfrac{p}{q}$ is in standard form if $q \gt 0 \text{ and } p, q$ have no common factor other than 1.

Here the denominator 36 is already positive.

HCF of 12 and 36 = 12.

Dividing both numerator and denominator by their HCF,

$\Rightarrow \dfrac{-12}{36} \\[1em] \Rightarrow \dfrac{-12 \div 12}{36 \div 12} \\[1em] \Rightarrow \dfrac{-1}{3}$

Hence, the standard form of $\mathbf{\dfrac{-12}{36} \text{ is } \dfrac{-1}{3}}$

Write the following rational number in Standard Form:

$\dfrac{2}{7}$

Answer

A rational number $\dfrac{p}{q}$ is in standard form if $q \gt 0 \text{ and } p, q$ have no common factor other than 1.

In $\dfrac{2}{7}$, the denominator 7 is already positive.

HCF of 2 and 7 = 1.

So, the numerator and denominator have no common factor other than 1 and $\dfrac{2}{7}$ is already in standard form.

Hence, the standard form of $\mathbf{\dfrac{2}{7} \text{ is } \dfrac{2}{7}}$.

Simplify:

$\dfrac{-4}{9} + 2\dfrac{12}{13}$

Answer

Simplifying,

$\Rightarrow \dfrac{-4}{9} + 2\dfrac{12}{13} \\[1em] \Rightarrow \dfrac{-4}{9} + \dfrac{38}{13}\\[1em] [\text{ LCM of } 9 \text{ and }13 = 117] \\[1em] \Rightarrow \dfrac{(-4) \times 13 + 38 \times 9}{117} \\[1em] \Rightarrow \dfrac{-52 + 342}{117} \\[1em] \Rightarrow \dfrac{290}{117}\\[1em] \Rightarrow 2\dfrac{56}{117}$

Hence, $\mathbf{\dfrac{-4}{9} + 2\dfrac{12}{13} = 2\dfrac{56}{117}}$

Simplify:

$\dfrac{11}{-7} + 8\dfrac{2}{3}$

Answer

First express the rational number with a positive denominator and convert the mixed fraction into an improper fraction.

Simplifying,

$\Rightarrow \dfrac{11}{-7} + 8\dfrac{2}{3} \\[1em] \Rightarrow \dfrac{-11}{7} + \dfrac{26}{3} \\[1em] [\text{ LCM of } 7 \text{ and } 3 = 21] \\[1em] \Rightarrow \dfrac{(-11) \times 3 + 26 \times 7}{21} \\[1em] \Rightarrow \dfrac{-33 + 182}{21} \\[1em] \Rightarrow \dfrac{149}{21} \\[1em] \Rightarrow 7\dfrac{2}{21}$

Hence, $\mathbf{\dfrac{11}{-7} + 8\dfrac{2}{3} = 7\dfrac{2}{21}}$

Verify commutative property of addition for the following pair of rational numbers:

$\dfrac{-4}{3} \text{ and } \dfrac{3}{7}$

Answer

To verify the commutative property of addition, we have to show that $\dfrac{-4}{3} + \dfrac{3}{7} = \dfrac{3}{7} + \dfrac{-4}{3}$

LCM of 3 and 7 = 21

Solving L.H.S.,

$\Rightarrow \dfrac{-4}{3} + \dfrac{3}{7} \\[1em] \Rightarrow \dfrac{(-4) \times 7 + 3 \times 3}{21} \\[1em] \Rightarrow \dfrac{-28 + 9}{21} \\[1em] \Rightarrow \dfrac{-19}{21}$

Solving R.H.S.,

$\Rightarrow \dfrac{3}{7} + \dfrac{-4}{3} \\[1em] \Rightarrow \dfrac{3 \times 3 + (-4) \times 7}{21} \\[1em] \Rightarrow \dfrac{9 - 28}{21} \\[1em] \Rightarrow \dfrac{-19}{21}$

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

$\Rightarrow \dfrac{-4}{3} + \dfrac{3}{7} = \dfrac{3}{7} + \dfrac{-4}{3}$

Hence, the commutative property of addition is verified.

Verify commutative property of addition for the following pair of rational numbers:

$\dfrac{-2}{-5} \text{ and } \dfrac{1}{3}$

Answer

First express $\dfrac{-2}{-5}$ with a positive denominator.

$\dfrac{-2}{-5} = \dfrac{-2 \times (-1)}{-5 \times (-1)} = \dfrac{2}{5}$

To verify the commutative property of addition, we have to show that $\dfrac{2}{5} + \dfrac{1}{3} = \dfrac{1}{3} + \dfrac{2}{5}$

LCM of 5 and 3 = 15

Solving L.H.S.,

$\Rightarrow \dfrac{2}{5} + \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{2 \times 3 + 1 \times 5}{15} \\[1em] \Rightarrow \dfrac{6 + 5}{15} \\[1em] \Rightarrow \dfrac{11}{15}$

Solving R.H.S.,

$\Rightarrow \dfrac{1}{3} + \dfrac{2}{5} \\[1em] \Rightarrow \dfrac{1 \times 5 + 2 \times 3}{15} \\[1em] \Rightarrow \dfrac{5 + 6}{15} \\[1em] \Rightarrow \dfrac{11}{15}$

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

$\Rightarrow \dfrac{-2}{-5} + \dfrac{1}{3} = \dfrac{1}{3} + \dfrac{-2}{-5}$

Hence, the commutative property of addition is verified.

Verify commutative property of addition for the following pair of rational numbers:

$\dfrac{9}{11} \text{ and } \dfrac{2}{13}$

Answer

To verify the commutative property of addition, we have to show that $\dfrac{9}{11} + \dfrac{2}{13} = \dfrac{2}{13} + \dfrac{9}{11}$

LCM of 11 and 13 = 143

Solving L.H.S.,

$\Rightarrow \dfrac{9}{11} + \dfrac{2}{13} \\[1em] \Rightarrow \dfrac{9 \times 13 + 2 \times 11}{143} \\[1em] \Rightarrow \dfrac{117 + 22}{143} \\[1em] \Rightarrow \dfrac{139}{143}$

Solving R.H.S.,

$\Rightarrow \dfrac{2}{13} + \dfrac{9}{11} \\[1em] \Rightarrow \dfrac{2 \times 11 + 9 \times 13}{143} \\[1em] \Rightarrow \dfrac{22 + 117}{143} \\[1em] \Rightarrow \dfrac{139}{143}$

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

$\Rightarrow \dfrac{9}{11} + \dfrac{2}{13} = \dfrac{2}{13} + \dfrac{9}{11}$

Hence, the commutative property of addition is verified.

Find the additive inverse of the following rational number:

$\dfrac{2}{-3}$

Answer

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

First express the rational number with a positive denominator.

$\Rightarrow \dfrac{2}{-3} \\[1em] \Rightarrow \dfrac{2 \times (-1)}{-3 \times (-1)} \\[1em] \Rightarrow \dfrac{-2}{3}$

Changing the sign of $\dfrac{-2}{3}$, we get

$\Rightarrow -\left(\dfrac{-2}{3}\right) \\[1em] \Rightarrow \dfrac{2}{3}$

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

Find the additive inverse of the following rational number:

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

Answer

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

First express the rational number with a positive denominator.

$\Rightarrow \dfrac{-7}{-12} \\[1em] \Rightarrow \dfrac{-7 \times (-1)}{-12 \times (-1)} \\[1em] \Rightarrow \dfrac{7}{12}$

Changing the sign of $\dfrac{7}{12}$, we get

$\Rightarrow -\left(\dfrac{7}{12}\right) \\[1em] \Rightarrow \dfrac{-7}{12}$

Hence, the additive inverse of $\mathbf{\dfrac{-7}{-12} \text{ is } \dfrac{-7}{12}}$

Using appropriate properties of addition, find the following:

$\dfrac{4}{5} + \dfrac{11}{7} + \dfrac{-7}{5} + \dfrac{-2}{7}$

Answer

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

$\Rightarrow \dfrac{4}{5} + \dfrac{11}{7} + \dfrac{-7}{5} + \dfrac{-2}{7} \\[1em] \Rightarrow \left(\dfrac{4}{5} + \dfrac{-7}{5}\right) + \left(\dfrac{11}{7} + \dfrac{-2}{7}\right) \\[1em] \Rightarrow \dfrac{4 + (-7)}{5} + \dfrac{11 + (-2)}{7} \\[1em] \Rightarrow \dfrac{-3}{5} + \dfrac{9}{7} \\[1em] [\text{ LCM of } 5 \text{ and } 7 = 35] \\[1em] \Rightarrow \dfrac{(-3) \times 7 + 9 \times 5}{35} \\[1em] \Rightarrow \dfrac{-21 + 45}{35} \\[1em] \Rightarrow \dfrac{24}{35}$

Hence, $\mathbf{\dfrac{4}{5} + \dfrac{11}{7} + \dfrac{-7}{5} + \dfrac{-2}{7} = \dfrac{24}{35}}$

Using appropriate properties of addition, find the following:

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

Answer

Using commutativity and associativity of addition, group together the rational numbers whose denominators have a small LCM.

$\Rightarrow \dfrac{3}{7} + \dfrac{4}{9} + \dfrac{-5}{21} + \dfrac{2}{3} \\[1em] \Rightarrow \left(\dfrac{3}{7} + \dfrac{-5}{21}\right) + \left(\dfrac{4}{9} + \dfrac{2}{3}\right) \\[1em] \Rightarrow \dfrac{3 \times 3 + (-5)}{21} + \dfrac{4 + 2 \times 3}{9} \\[1em] \Rightarrow \dfrac{9 - 5}{21} + \dfrac{4 + 6}{9} \\[1em] \Rightarrow \dfrac{4}{21} + \dfrac{10}{9} \\[1em] [\text{ LCM of } 21 \text{ and } 9 = 63] \\[1em] \Rightarrow \dfrac{4 \times 3 + 10 \times 7}{63} \\[1em] \Rightarrow \dfrac{12 + 70}{63} \\[1em] \Rightarrow \dfrac{82}{63} \\[1em] \Rightarrow 1\dfrac{19}{63}$

Hence, $\mathbf{\dfrac{3}{7} + \dfrac{4}{9} + \dfrac{-5}{21} + \dfrac{2}{3} = 1\dfrac{19}{63}}$

If $a = \dfrac{-11}{27}, b = \dfrac{4}{9} \text{ and } c = \dfrac{-5}{18}$, then verify that a + (b + c) = (a + b) + c

Answer

To verify the associative property of addition, we have to show that $a + (b + c) = (a + b) + c$

Solving L.H.S.,

$\Rightarrow b + c = \dfrac{4}{9} + \dfrac{-5}{18} \\[1em] \Rightarrow \dfrac{4 \times 2 + (-5)}{18} \\[1em] \Rightarrow \dfrac{8 - 5}{18} = \dfrac{3}{18} = \dfrac{1}{6} \\[1em] \therefore a + (b + c) = \dfrac{-11}{27} + \dfrac{1}{6} \\[1em] [\text{ LCM of } 27 \text{ and } 6 = 54] \\[1em] \Rightarrow \dfrac{(-11) \times 2 + 1 \times 9}{54} \\[1em] \Rightarrow \dfrac{-22 + 9}{54} \\[1em] \Rightarrow \dfrac{-13}{54}$

Solving R.H.S.,

$\Rightarrow a + b = \dfrac{-11}{27} + \dfrac{4}{9} \\[1em] \Rightarrow \dfrac{-11 + 4 \times 3}{27} \\[1em] \Rightarrow \dfrac{-11 + 12}{27} = \dfrac{1}{27} \\[1em] \therefore (a + b) + c = \dfrac{1}{27} + \dfrac{-5}{18} \\[1em] [\text{ LCM of } 27 \text{ and } 18 = 54] \\[1em] \Rightarrow \dfrac{1 \times 2 + (-5) \times 3}{54} \\[1em] \Rightarrow \dfrac{2 - 15}{54} \\[1em] \Rightarrow \dfrac{-13}{54}$

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

$\Rightarrow a + (b + c) = (a + b) + c$

Hence, proved that $\bm{a + (b + c) = (a + b) + c}$.