NCERT Solutions for Class 9 Maths Chapter 3 — Exercise Set 3.5#

Each fraction is already in its lowest terms. We factorise each denominator:

$$20 = 2^2 \times 5$$

$$15 = 3 \times 5$$

$$250 = 2 \times 5^3$$

• $\dfrac{7}{20}$: the denominator has only 2s and 5s, so the decimal terminates.
• $\dfrac{4}{15}$: the denominator has the factor 3, so the decimal repeats.
• $\dfrac{13}{250}$: the denominator has only 2s and 5s, so the decimal terminates.

$\dfrac{7}{20}$: we bring down zeros and divide by 20.

Bring down	÷ 20	Remainder
70	3	10
100	5	0

The remainder becomes 0, so $\dfrac{7}{20} = 0.35$ — terminating ✓.

$\dfrac{4}{15}$: we divide by 15.

Bring down	÷ 15	Remainder
40	2	10
100	6	10
100	6	10

The remainder 10 keeps coming back, so the digit 6 repeats: $\dfrac{4}{15} = 0.2666\ldots = 0.2\overline{6}$ — repeating ✓.

$\dfrac{13}{250}$: we divide by 250.

Bring down	÷ 250	Remainder
130	0	130
1300	5	50
500	2	0

The remainder becomes 0, so $\dfrac{13}{250} = 0.052$ — terminating ✓.

Bring down	÷ 13	Remainder
10	0	10
100	7	9
90	6	12
120	9	3
30	2	4
40	3	1

The remainder 1 has come back, so the division starts again from the beginning:

$$\dfrac{1}{13} = 0.076923\,076923\ldots = 0.\overline{076923}$$

The repeating block is 076923 (6 digits).

For $\dfrac{2}{13}$:

$$\dfrac{2}{13} = 0.\overline{153846}$$

The block 153846 is not a rotation of 076923 (the rotations of 076923 are 769230, 692307, 923076, 230769 and 307692). So, unlike $\dfrac{1}{7}$, the block of $\dfrac{1}{13}$ is not fully cyclic.

Now we compute more multiples:

$$\dfrac{3}{13} = 0.\overline{230769}$$

$$\dfrac{4}{13} = 0.\overline{307692}$$

Both of these are rotations of 076923.

What we notice: the twelve fractions $\dfrac{1}{13}, \dfrac{2}{13}, \ldots, \dfrac{12}{13}$ split into two families of six:

Rotations of 076923	Rotations of 153846
$\dfrac{1}{13}, \dfrac{3}{13}, \dfrac{4}{13}, \dfrac{9}{13}, \dfrac{10}{13}, \dfrac{12}{13}$	$\dfrac{2}{13}, \dfrac{5}{13}, \dfrac{6}{13}, \dfrac{7}{13}, \dfrac{8}{13}, \dfrac{11}{13}$

The repeating block of $\dfrac{1}{13}$ has only 6 digits, but there are 12 possible remainders (1 to 12). One block can visit only 6 of them, so the other 6 remainders form a second cycle. For $\dfrac{1}{7}$, the block has 6 digits and there are 6 possible remainders, so a single block covers all of them — that is why 142857 is cyclic.

$$\sqrt{81} = 9$$

9 is an integer, so $\sqrt{81}$ is rational:

$$\sqrt{81} = \dfrac{9}{1}$$

$$\sqrt{12} = \sqrt{4 \times 3}$$

$$\sqrt{12} = 2\sqrt{3}$$

$\sqrt{3}$ is irrational (3 is not a perfect square), and 2 times an irrational number is irrational. So $\sqrt{12}$ is irrational.

The digit 3 repeats, so the number is rational.

Let $x = 0.333\ldots$. One digit repeats, so we multiply by 10:

$$x = 0.333\ldots \quad \cdots (1)$$

$$10x = 3.333\ldots \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} 10x &=& 3.333\ldots \\ x &=& 0.333\ldots \\ \hline 9x &=& 3 \end{array}$$

$$\Rightarrow x = \dfrac{3}{9}$$

$$\Rightarrow x = \dfrac{1}{3}$$

∴ $0.33333\ldots = \dfrac{1}{3}$.

The block 12345 repeats, so the number is rational: $0.\overline{12345}$.

Let $x = 0.\overline{12345}$. Five digits repeat, so we multiply by $10^5 = 100000$:

$$x = 0.12345\,12345\ldots \quad \cdots (1)$$

$$100000x = 12345.12345\ldots \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} 100000x &=& 12345.12345\ldots \\ x &=& 0.12345\ldots \\ \hline 99999x &=& 12345 \end{array}$$

$$\Rightarrow x = \dfrac{12345}{99999}$$

Both 12345 and 99999 are divisible by 3:

$$x = \dfrac{4115}{33333}$$

∴ $0.123451234512345\ldots = \dfrac{4115}{33333}$.

After each 1, the number of 0s increases by one: 1, 2, 3, 4, … zeros. So no single block of digits repeats forever, and the decimal never ends.

∴ $1.01001000100001\ldots$ is irrational.

This decimal terminates (it has 21 digits after the decimal point and then stops), so it is rational.

We write it over $10^{21}$:

$$23.560185612239874790120 = \dfrac{23560185612239874790120}{10^{21}}$$

The numerator ends in 120, so it is divisible by 40, and $10^{21} \div 40 = 25 \times 10^{18}$. Dividing both by 40:

$$23.560185612239874790120 = \dfrac{589004640305996869753}{25000000000000000000}$$

The new numerator ends in 3, so it has no factor 2 or 5 left in common with the denominator, and the fraction is in lowest terms.

Let $x = 0.\overline{9}$. One digit repeats, so we multiply by 10:

$$x = 0.999\ldots \quad \cdots (1)$$

$$10x = 9.999\ldots \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} 10x &=& 9.999\ldots \\ x &=& 0.999\ldots \\ \hline 9x &=& 9 \end{array}$$

$$\Rightarrow x = 1$$

The digits after the decimal point are the same in both numbers and cancel completely, leaving exactly $9x = 9$.

∴ $0.\overline{9} = 1$ exactly. It is not “slightly less than 1”: there is no number between $0.999\ldots$ and 1, so they are the same number written in two ways.

The block of $\dfrac{1}{n}$ is cyclic when it is as long as possible, that is, when it has $n - 1$ digits and so passes through every remainder $1, 2, \ldots, n - 1$. This happens for these primes below 100:

$$n = 7, 17, 19, 23, 29, 47, 59, 61, 97$$

Two examples:

$$\dfrac{1}{17} = 0.\overline{0588235294117647} \quad (16 \text{ digits})$$

$$\dfrac{1}{19} = 0.\overline{052631578947368421} \quad (18 \text{ digits})$$

For example, $\dfrac{2}{17} = 0.\overline{1176470588235294}$, which is the block of $\dfrac{1}{17}$ starting from a different place, just as with $\dfrac{1}{7}$.

$13$ does not work (Question 2): its block has only 6 digits instead of 12.