Rational & Irrational Numbers — Exercise 1(A)

Let zero = $\dfrac{p}{q}$, where p and q are integers. What additional condition will make 0 = $\dfrac{p}{q}$ a rational number :

1. q = 0

2. p ≠ 0

3. q ≠ 0

4. p ≠ 0 and q ≠ 0.

Answer

$\dfrac{p}{q} = 0$ is a rational number. If,

p and q are integers and q ≠ 0.

Hence, Option 3 is the correct option.

Every non-terminating decimal number is a :

1. recurring decimal

2. real number

3. non-recurring decimal

4. circulating decimal

Answer

Every non-terminating decimal number is a real number.

Hence, Option 2 is the correct option.

7.478478.... is a :

1. terminating rational

2. recurring

3. neither rational non-terminating

4. not real

Answer

7.478478 is a recurring decimal.

Hence, Option 2 is the correct option.

$\dfrac{71}{75}$ is :

1. terminating

2. non-terminating

3. periodic decimal

4. not a rational number

Answer

On solving,

$\dfrac{71}{75}$ = 0.94666.......

∴ $\dfrac{71}{75}$ is a recurring or periodic decimal.

Hence, Option 3 is the correct option.

Which of the following is terminating:

$\dfrac{13}{85}, \dfrac{51}{405} \text{ and } \dfrac{9}{524}$

1. $\dfrac{13}{85}$

2. $\dfrac{9}{524}$

3. $\dfrac{51}{405}$

4. None of these

Answer

On solving,

$\dfrac{13}{85} = 0.152941$........

$\dfrac{51}{405} = 0.1259259$......

$\dfrac{9}{524}$ = 0.0171755......

None of the these fraction is terminating.

Hence, Option 4 is the correct option.

Are the following statements true or false ? Give reasons for your answers ?

(i) Every whole number is a natural number.

(ii) Every whole number is a rational number.

(iii) Every integer is a rational number.

(iv) Every rational number is a whole number.

Answer

(i) False, as zero is a whole number but not a natural number.

(ii) True

(iii) True

(iv) False, as $\dfrac{2}{5}$ is a rational number but not a whole number.

Arrange $-\dfrac{5}{9}, \dfrac{7}{12}, -\dfrac{2}{3} \text{ and } \dfrac{11}{18}$ in the ascending order of their magnitudes.

Also, find the difference between the largest and the smallest of these rational numbers. Express this difference as a decimal fraction correct to one decimal place.

Answer

L.C.M. of 9, 12, 3 and 18 is 36.

So, converting denominator of each fraction $-\dfrac{5}{9}, \dfrac{7}{12}, -\dfrac{2}{3} \text{ and } \dfrac{11}{18}$ into 36.

$\Rightarrow -\dfrac{5}{9} \times \dfrac{4}{4} = -\dfrac{20}{36} \\[1em] \Rightarrow \dfrac{7}{12} \times \dfrac{3}{3} = \dfrac{21}{36} \\[1em] \Rightarrow -\dfrac{2}{3} \times \dfrac{12}{12} = -\dfrac{24}{36} \\[1em] \Rightarrow \dfrac{11}{18} \times \dfrac{2}{2} = \dfrac{22}{36}.$

Since, -24 < -20 < 21 < 22

∴ -$\dfrac{24}{36} \lt -\dfrac{20}{36} \lt \dfrac{21}{36} \lt \dfrac{22}{36}$

$\Rightarrow -\dfrac{2}{3} \lt -\dfrac{5}{9} \lt \dfrac{7}{12} \lt \dfrac{11}{18}$

Difference between largest and smallest fraction :

$\Rightarrow \dfrac{11}{18} - \Big(-\dfrac{2}{3}\Big) \\[1em] \Rightarrow \dfrac{11}{18} + \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{11}{18} + \dfrac{12}{18} \\[1em] \Rightarrow \dfrac{23}{18} \\[1em] \Rightarrow 1.3$

Hence, fractions in ascending order are $-\dfrac{2}{3} \lt -\dfrac{5}{9} \lt \dfrac{7}{12} \lt \dfrac{11}{18}$ and required difference = 1.3

Arrange $\dfrac{5}{8}, -\dfrac{3}{16}, -\dfrac{1}{4} \text{ and } \dfrac{17}{32}$ in the descending order of their magnitudes.

Also, find the sum of the lowest and the largest of these rational numbers. Express the result obtained as a decimal fraction correct to two decimal places.

Answer

L.C.M. of 4, 8, 16 and 32 is 32.

So, converting denominator of each fraction $\dfrac{5}{8}, -\dfrac{3}{16}, -\dfrac{1}{4} \text{ and } \dfrac{17}{32}$ into 32.

$\Rightarrow \dfrac{5}{8} \times \dfrac{4}{4} = \dfrac{20}{32} \\[1em] \Rightarrow -\dfrac{3}{16} \times \dfrac{2}{2} = -\dfrac{6}{32} \\[1em] \Rightarrow -\dfrac{1}{4} \times \dfrac{8}{8} = -\dfrac{8}{32} \\[1em] \Rightarrow \dfrac{17}{32} \times \dfrac{1}{1} = \dfrac{17}{32}.$

Since, -8 < -6 < 17 < 20.

$\therefore -\dfrac{8}{32} \lt -\dfrac{6}{32} \lt \dfrac{17}{32} \lt \dfrac{20}{32}$

$\therefore -\dfrac{1}{4} \lt -\dfrac{3}{16} \lt \dfrac{17}{32} \lt \dfrac{5}{8}$

So, in descending order.

$\Rightarrow \dfrac{5}{8} \gt \dfrac{17}{32} \gt -\dfrac{3}{16} \gt -\dfrac{1}{4}$

Sum of largest and lowest :

$= \dfrac{5}{8} + \Big(-\dfrac{1}{4}\Big) \\[1em] = \dfrac{5}{8} - \dfrac{1}{4} \\[1em] = \dfrac{5 \times 4}{32} - \dfrac{1 \times 8}{32} \\[1em] = \dfrac{20 - 8}{32} \\[1em] = \dfrac{12}{32} \\[1em] = \dfrac{3}{8} \\[1em] = 0.38$

Hence, fractions in descending order are : $\dfrac{5}{8} \gt \dfrac{17}{32} \gt -\dfrac{3}{16} \gt -\dfrac{1}{4}$ and required sum = 0.38

Without doing any actual division, find which of the following rational numbers have terminating decimal representation :

(i) $\dfrac{7}{16}$

(ii) $\dfrac{23}{125}$

(iii) $\dfrac{9}{14}$

(iv) $\dfrac{32}{45}$

(v) $\dfrac{43}{50}$

Answer

In rational numbers, if the denominator of the fraction can be expressed in the form of 2^m × 5ⁿ, then it is a terminating decimal.

(i) $\dfrac{7}{16} = \dfrac{7}{2^4}$

So, 16 can be expressed as 2⁴ × 5⁰.

Hence, it is a terminating decimal number.

(ii) $\dfrac{23}{125} = \dfrac{23}{5^3}$

So, 125 can be expressed as 2⁰ × 5³.

Hence, it is a terminating decimal number.

(iii) $\dfrac{9}{14} = \dfrac{9}{2 \times 7}$

So, 14 cannot be expressed in form of 2^m × 5ⁿ.

Hence, it is not a terminating decimal number.

(iv) $\dfrac{32}{45} = \dfrac{32}{3^2 \times 5}$

So, 45 cannot be expressed in form of 2^m × 5ⁿ.

Hence, it is not a terminating decimal number.

(v) $\dfrac{43}{50} = \dfrac{43}{2 \times 5^2}$

So, 50 can be expressed in form of 2¹ × 5².

Hence, it is a terminating decimal number.