Rational Numbers — Exercise 1.2

Subtract:

$2\dfrac{3}{5} \text{ from } \dfrac{-3}{7}$

Answer

Subtracting one rational number from another is the same as adding the additive inverse of the rational number that is being subtracted.

$\Rightarrow \dfrac{-3}{7} - 2\dfrac{3}{5} \\[1em] \Rightarrow \dfrac{-3}{7} - \dfrac{13}{5} \\[1em] \Rightarrow \dfrac{-3}{7} + \left(\dfrac{-13}{5}\right) \\[1em] [\text{ LCM of } 7 \text{ and } 5 = 35] \\[1em] \Rightarrow \dfrac{(-3) \times 5 + (-13) \times 7}{35} \\[1em] \Rightarrow \dfrac{-15 - 91}{35} \\[1em] \Rightarrow \dfrac{-106}{35} \\[1em] \Rightarrow -3\dfrac{1}{35}$

Hence, $\mathbf{\dfrac{-3}{7} - 2\dfrac{3}{5} = -3\dfrac{1}{35}}$

Subtract:

$\dfrac{-4}{9} \text{ from } 3\dfrac{5}{8}$

Answer

Subtracting one rational number from another is the same as adding the additive inverse of the rational number that is being subtracted.

$\Rightarrow 3\dfrac{5}{8} - \left(\dfrac{-4}{9}\right) \\[1em] \Rightarrow \dfrac{29}{8} + \dfrac{4}{9} \\[1em] [\text{ LCM of } 8 \text{ and } 9 = 72] \\[1em] \Rightarrow \dfrac{29 \times 9 + 4 \times 8}{72} \\[1em] \Rightarrow \dfrac{261 + 32}{72} \\[1em] \Rightarrow \dfrac{293}{72} \\[1em] \Rightarrow 4\dfrac{5}{72}$

Hence, $\mathbf{3\dfrac{5}{8} - \left(\dfrac{-4}{9}\right) = 4\dfrac{5}{72}}$

Subtract:

$-3\dfrac{1}{5} \text{ from } -4\dfrac{7}{9}$

Answer

Subtracting one rational number from another is the same as adding the additive inverse of the rational number that is being subtracted.

$\Rightarrow -4\dfrac{7}{9} - \left(-3\dfrac{1}{5}\right) \\[1em] \Rightarrow \dfrac{-43}{9} - \left(\dfrac{-16}{5}\right) \\[1em] \Rightarrow \dfrac{-43}{9} + \dfrac{16}{5} \\[1em] [\text{ LCM of } 9 \text{ and } 5 = 45] \\[1em] \Rightarrow \dfrac{(-43) \times 5 + 16 \times 9}{45} \\[1em] \Rightarrow \dfrac{-215 + 144}{45} \\[1em] \Rightarrow \dfrac{-71}{45} \\[1em] \Rightarrow -1\dfrac{26}{45}$

Hence, $\mathbf{-4\dfrac{7}{9} - \left(-3\dfrac{1}{5}\right) = -1\dfrac{26}{45}}$

Sum of two rational numbers is $\dfrac{3}{5}$. If one of them is $\dfrac{-2}{7}$, find the other.

Answer

Let the other rational number be $x$.

As the sum of the two rational numbers is $\dfrac{3}{5}$,

$\Rightarrow \dfrac{-2}{7} + x = \dfrac{3}{5} \\[1em] \Rightarrow x = \dfrac{3}{5} - \left(\dfrac{-2}{7}\right) \\[1em] \Rightarrow x = \dfrac{3}{5} + \dfrac{2}{7} \\[1em] [\text{ LCM of } 5 \text{ and } 7 = 35] \\[1em] \Rightarrow x = \dfrac{3 \times 7 + 2 \times 5}{35} \\[1em] \Rightarrow x = \dfrac{21 + 10}{35} \\[1em] \Rightarrow x = \dfrac{31}{35}$

Hence, the other rational number is $\mathbf{\dfrac{31}{35}}$

What rational number should be added to $\dfrac{-5}{11}$ to get $\dfrac{-7}{8}$?

Answer

Let the required rational number be $x$.

According to the given condition,

$\Rightarrow \dfrac{-5}{11} + x = \dfrac{-7}{8} \\[1em] \Rightarrow x = \dfrac{-7}{8} - \left(\dfrac{-5}{11}\right) \\[1em] \Rightarrow x = \dfrac{-7}{8} + \dfrac{5}{11} \\[1em] [\text{ LCM of } 8 \text{ and } 11 = 88] \\[1em] \Rightarrow x = \dfrac{(-7) \times 11 + 5 \times 8}{88} \\[1em] \Rightarrow x = \dfrac{-77 + 40}{88} \\[1em] \Rightarrow x = \dfrac{-37}{88}$

Hence, the required rational number is $\mathbf{\dfrac{-37}{88}}$

What rational number should be subtracted from $-4\dfrac{3}{5}$ to get $-3\dfrac{1}{2}$?

Answer

Let the required rational number be $x$.

According to the given condition,

$\Rightarrow -4\dfrac{3}{5} - x = -3\dfrac{1}{2} \\[1em] \Rightarrow \dfrac{-23}{5} - x = \dfrac{-7}{2} \\[1em] \Rightarrow x = \dfrac{-23}{5} - \left(\dfrac{-7}{2}\right) \\[1em] \Rightarrow x = \dfrac{-23}{5} + \dfrac{7}{2} \\[1em] [\text{ LCM of } 5 \text{ and } 2 = 10] \\[1em] \Rightarrow x = \dfrac{(-23) \times 2 + 7 \times 5}{10} \\[1em] \Rightarrow x = \dfrac{-46 + 35}{10} \\[1em] \Rightarrow x = \dfrac{-11}{10} = -1\dfrac{1}{10}$

Hence, the required rational number is $\mathbf{\dfrac{-11}{10}}$

Subtract the sum of $\dfrac{-5}{7} \text{ and } \dfrac{-8}{3} \text{ from the sum of } \dfrac{5}{2} \text{ and } \dfrac{-11}{12}$

Answer

First find the sum of $\dfrac{-5}{7} \text{ and } \dfrac{-8}{3}$.

$\Rightarrow \dfrac{-5}{7} + \dfrac{-8}{3} \\[1em] [\text{ LCM of } 7 \text{ and } 3 = 21] \\[1em] \Rightarrow \dfrac{(-5) \times 3 + (-8) \times 7}{21} \\[1em] \Rightarrow \dfrac{-15 - 56}{21} \\[1em] \Rightarrow \dfrac{-71}{21}$

Now find the sum of $\dfrac{5}{2} \text{ and } \dfrac{-11}{12}$.

$\Rightarrow \dfrac{5}{2} + \dfrac{-11}{12} \\[1em] [\text{ LCM of } 2 \text{ and } 12 = 12] \\[1em] \Rightarrow \dfrac{5 \times 6 + (-11)}{12} \\[1em] \Rightarrow \dfrac{30 - 11}{12} \\[1em] \Rightarrow \dfrac{19}{12}$

Subtracting the first sum from the second sum,

$\Rightarrow \dfrac{19}{12} - \left(\dfrac{-71}{21}\right) \\[1em] \Rightarrow \dfrac{19}{12} + \dfrac{71}{21} \\[1em] [\text{ LCM of } 12 \text{ and } 21 = 84] \\[1em] \Rightarrow \dfrac{19 \times 7 + 71 \times 4}{84} \\[1em] \Rightarrow \dfrac{133 + 284}{84} \\[1em] \Rightarrow \dfrac{417}{84} = \dfrac{139}{28} \\[1em] \Rightarrow 4\dfrac{27}{28}$

Hence, the required difference is $\mathbf{4\dfrac{27}{28}}$

If $x = \dfrac{-4}{7} \text{ and } y = \dfrac{2}{5}$, then verify that:

(i) $x - y ≠ y - x$

(ii) $-(x + y) = (-x) + (-y)$

Answer

(i) To verify that subtraction of rational numbers is not commutative, we have to show that $x - y \neq y - x$

LCM of 7 and 5 = 35

Solving $x - y$,

$\Rightarrow \dfrac{-4}{7} - \dfrac{2}{5} \\[1em] \Rightarrow \dfrac{(-4) \times 5 - 2 \times 7}{35} \\[1em] \Rightarrow \dfrac{-20 - 14}{35} \\[1em] \Rightarrow \dfrac{-34}{35}$

Solving $y - x$,

$\Rightarrow \dfrac{2}{5} - \left(\dfrac{-4}{7}\right) \\[1em] \Rightarrow \dfrac{2}{5} + \dfrac{4}{7} \\[1em] \Rightarrow \dfrac{2 \times 7 + 4 \times 5}{35} \\[1em] \Rightarrow \dfrac{14 + 20}{35} \\[1em] \Rightarrow \dfrac{34}{35}$

As $\dfrac{-34}{35} \neq \dfrac{34}{35}$

$\Rightarrow x - y \neq y - x$

Hence, proved that $\bm{x - y \neq y - x}$.

(ii) To verify that the additive inverse of a sum equals the sum of the additive inverses, we have to show that $-(x + y) = (-x) + (-y)$

LCM of 7 and 5 = 35

Solving L.H.S.,

$\Rightarrow x + y = \dfrac{-4}{7} + \dfrac{2}{5} \\[1em] \Rightarrow \dfrac{(-4) \times 5 + 2 \times 7}{35} \\[1em] \Rightarrow \dfrac{-20 + 14}{35} = \dfrac{-6}{35} \\[1em] \therefore -(x + y) = -\left(\dfrac{-6}{35}\right) \\[1em] \Rightarrow \dfrac{6}{35}$

Solving R.H.S.,

$\Rightarrow (-x) + (-y) = \dfrac{4}{7} + \left(\dfrac{-2}{5}\right) \\[1em] \Rightarrow \dfrac{4 \times 5 + (-2) \times 7}{35} \\[1em] \Rightarrow \dfrac{20 - 14}{35} \\[1em] \Rightarrow \dfrac{6}{35}$

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

$\Rightarrow -(x + y) = (-x) + (-y)$

Hence, proved that $\bm{-(x + y) = (-x) + (-y)}$.

If $x = \dfrac{4}{9}, y = \dfrac{-7}{12} \text{ and } z = \dfrac{-2}{3}$, then verify that

x - (y - z) ≠ (x - y) - z

Answer

To verify that subtraction of rational numbers is not associative, we have to show that $x - (y - z) \neq (x - y) - z$

Solving L.H.S.,

$\Rightarrow y - z = \dfrac{-7}{12} - \left(\dfrac{-2}{3}\right) \\[1em] \Rightarrow \dfrac{-7}{12} + \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{-7 + 2 \times 4}{12} = \dfrac{-7 + 8}{12} = \dfrac{1}{12} \\[1em] \therefore x - (y - z) = \dfrac{4}{9} - \dfrac{1}{12} \\[1em] [\text{ LCM of } 9 \text{ and } 12 = 36] \\[1em] \Rightarrow \dfrac{4 \times 4 - 1 \times 3}{36} \\[1em] \Rightarrow \dfrac{16 - 3}{36} \\[1em] \Rightarrow \dfrac{13}{36}$

Solving R.H.S.,

$\Rightarrow x - y = \dfrac{4}{9} - \left(\dfrac{-7}{12}\right) \\[1em] \Rightarrow \dfrac{4}{9} + \dfrac{7}{12} \\[1em] \Rightarrow \dfrac{4 \times 4 + 7 \times 3}{36} = \dfrac{16 + 21}{36} = \dfrac{37}{36} \\[1em] \therefore (x - y) - z = \dfrac{37}{36} - \left(\dfrac{-2}{3}\right) \\[1em] \Rightarrow \dfrac{37}{36} + \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{37 + 2 \times 12}{36} \\[1em] \Rightarrow \dfrac{37 + 24}{36} \\[1em] \Rightarrow \dfrac{61}{36}$

As $\dfrac{13}{36} \neq \dfrac{61}{36}$

$\Rightarrow x - (y - z) \neq (x - y) - z$

Hence, proved that $\bm{x - (y - z) \neq (x - y) - z}$.

Which of the following statement is true/false?

(i) $\dfrac{2}{3} - \dfrac{4}{5}$ is not a rational number.

(ii) $\dfrac{-5}{7} \text{ is the additive inverse of } \dfrac{5}{7}$

(iii) 0 is the additive inverse of its own.

(iv) Commutative property holds for subtraction of rational numbers.

(v) Associative property does not hold for subtraction of rational numbers.

(vi) 0 is the identity element for subtraction of rational numbers.

Answer

(i) Rational numbers are closed under subtraction, so the difference of two rational numbers is always a rational number.

$\Rightarrow \dfrac{2}{3} - \dfrac{4}{5} \\[1em] [\text{ LCM of } 3 \text{ and } 5 = 15] \\[1em] \Rightarrow \dfrac{2 \times 5 - 4 \times 3}{15} \\[1em] \Rightarrow \dfrac{10 - 12}{15} \\[1em] \Rightarrow \dfrac{-2}{15}$

As $\dfrac{-2}{15}$ is a rational number, the given statement is false.

Hence, the given statement is false.

(ii) To find the additive inverse of any rational number, change its sign. The sum of a rational number and its additive inverse is 0.

$\Rightarrow \dfrac{5}{7} + \left(\dfrac{-5}{7}\right) \\[1em] \Rightarrow \dfrac{5 + (-5)}{7} \\[1em] \Rightarrow \dfrac{0}{7} = 0$

Hence, the given statement is true.

(iii) The additive inverse of 0 is obtained by changing its sign, and $-0 = 0$.

As $0 + 0 = 0$, the additive inverse of 0 is 0 itself.

Hence, the given statement is true.

(iv) For any two rational numbers $\dfrac{p}{q} \text{ and } \dfrac{r}{s}$, we have $\dfrac{p}{q} - \dfrac{r}{s} \neq \dfrac{r}{s} - \dfrac{p}{q}$

So, subtraction of rational numbers is not commutative.

Hence, the given statement is false.

(v) For any three rational numbers

$\dfrac{p}{q}, \dfrac{r}{s} \text{ and } \dfrac{t}{u}$, we have $\left(\dfrac{p}{q} - \dfrac{r}{s}\right) - \dfrac{t}{u} \neq \dfrac{p}{q} - \left(\dfrac{r}{s} - \dfrac{t}{u}\right)$

So, subtraction of rational numbers is not associative.

Hence, the given statement is true.

(vi) Although $\dfrac{p}{q} - 0 = \dfrac{p}{q}$, we have $0 - \dfrac{p}{q} = \dfrac{-p}{q} \neq \dfrac{p}{q}$

So, subtraction of rational numbers has no identity element.

Hence, the given statement is false.