NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.3 - Number Systems

Q1 (i) Write the following in decimal form and say what kind of decimal expansion each has : (i) $\frac{36}{100}$

Answer:

$\frac{36}{100} $ = 0.36

Since the decimal expansion ends after a finite number of steps. Hence, it is terminating

Q1 (ii) Write the following in decimal form and say what kind of decimal expansion each has : (ii) $\frac{1}{11}$

Answer:

$\frac{1}{11} = 0.\overline{09}$

Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion.

Q1 (iii) Write the following in decimal form and say what kind of decimal expansion each has : (iii) $4\frac{1}{8}$

Answer:

$\Rightarrow 4\frac{1}{8} = \frac{33}{8}= 4.125$

Since the decimal expansion ends after a finite number. Therefore, it is terminating

Q1 (iv) Write the following in decimal form and say what kind of decimal expansion each has : (iv) $\frac{3}{13}$

Answer:

$\Rightarrow \frac{3}{13} = 0.230769230769 = 0.\overline{230769}$

Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion.

Q1 (v) Write the following in decimal form and say what kind of decimal expansion each has: (v) $\frac{2}{11}$

Answer:

$\Rightarrow \frac{2}{11} = 0.181818......= 0.\overline{18}$

Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion.

Q1 (vi) Write the following in decimal form and say what kind of decimal expansion each has : (vi) $\frac{329}{400}$

Answer:

$\Rightarrow \frac{329}{400}= 0.8225$

Since decimal expansion ends after finite number of figures. Hence, it is terminating.

Q2 You know that $\frac{1}{7}=0.\overline{142857}$ Can you predict what the decimal expansions of $\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7}$ are, without actually doing the long division? If so, how?

Answer:

Yes, the decimal expansion of 1/7 is:

$\frac{1}{7} = 0.\overline{142857}$

So for other multiples:

$\frac{2}{7} = 0.\overline{285714}$

$\frac{3}{7} = 0.\overline{428571}$

$\frac{4}{7} = 0.\overline{571428}$

$\frac{5}{7} = 0.\overline{714285}$

$\frac{6}{7} = 0.\overline{857142}$

All are cyclic permutations of the repeating part 142857.

Q3 (i) Express the following in the form $\frac{p}{q}$ , where p and q are integers and q ≠ 0. (i) $0.\bar{6}$

Answer:

Let x = $0.\overline{6}$

Then 10x = $6.\overline{6}$

Subtract:

10x - x = $6.\overline{6}$ - $0.\overline{6}$ $\Rightarrow 9x = 6 \Rightarrow x = \frac{2}{3}$

Hemce, $\frac{2}{3}$

Q3 (ii) Express the following in the form $\frac{p}{q}$ , where p and q are integers and q ≠ 0. (ii) $0.4\bar{7}$

Answer:

Let x = $0.4\overline{7}$
Then 100x = $47.\overline{7}$
Subtract:

Let x = 0.477777...
Or x = 0.47 + 0.00777...

Now:

• 0.47 = 47/100
  
  
• 0.00777... = y

Let y = 0.00777... = $0.00\overline{7}$

Now write y as:

Y = $\frac{7}{900}$

So:

$x = \frac{47}{100} + \frac{7}{900} = \frac{(423 + 7)}{900} = \frac{430}{900} = \boxed{\frac{43}{90}}$

Hence, $\frac{43}{90}$

Q3 (iii) Express the following in the form $\frac{p}{q}$ , where p and q are integers and q ≠ 0. (iii) $0.\overline{001}$

Answer:

Let $x = 0.\overline{001}= 0.001001....$

Then $1000x = 1.001001...$

We can write as:

$\Rightarrow 1000x = 1 + x $

After Subtraction:

$\Rightarrow 999x = 1$

$\Rightarrow x = \frac{1}{999}$

Therefore, $\frac{p}{q}$ form of $0.\overline{001}$ is $\frac{1}{999}$

Q4 Express 0.99999 .... in the form $\frac{p}{q}$ . Are you surprised by your answer?

Answer:

Let x = $0.\overline{9}$
⇒ 10x = $9.\overline{9}$
⇒ 10x - x = 9
⇒ 9x = 9 ⇒ x = 1

So, 0.9999… = 1

Yes, it’s surprising but mathematically correct. The difference between 1 and 0.999… is infinitely small — effectively zero.

Q5 What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$ ? Perform the division to check your answer.

Answer:

Performing the division:

1 ÷ 17 = 0.0588235294117647… → Repeats after 16 digits

So, maximum repeating block length = 16

Q6 Look at several examples of rational numbers in the form $\frac{p}{q}$ (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

Answer:

We can observe that when q is 2, 4, 5, 8, 10… then the decimal expansion is terminating. For example:

$\frac{3}{2}$ = 1.5, denominator q = 2¹

$\frac{8}{5}$ = 1.6, denominator q = 5¹

$\frac{15}{10}$ = 1.5, denominator q = 10 = 2 х 5 = 2¹, 5¹

Therefore,

It is evident that when the denominator of the given fractions is prime factorized to a power of either 2 or 5, or both, the terminating decimal can be attained.

Q7 Write three numbers whose decimal expansions are non-terminating non-recurring.

Answer:

Three numbers whose decimal expansions are non-terminating non-recurring are
1) 0.02002000200002......
2) 0.15115111511115.......
3) 0.27227222722227.......

Q8 Find three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$ .

Answer:

$\Rightarrow \frac{5}{7} = 0.714285714285.... = 0.\overline{714285}$

And $\frac{9}{11}$:

$\Rightarrow \frac{9}{11} = 0.818181.... = 0.\overline{81}$

Therefore, three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$ are

1) 0.72737475....
2) 0.750760770780...
3) 0.790780770760....

Q9 (i) Classify the following numbers as rational or irrational : $\sqrt{23}$

Answer:

Writing $\sqrt{23}$ in decimal form, we get:

$\Rightarrow \sqrt{23} = 4.7958152....$

Since, the decimal expansion of the number obtained is non-terminating and non-recurring. It is an irrational number.

Q9 (ii) Classify the following numbers as rational or irrational : $\sqrt{225}$

Answer:

Writing $\sqrt{225}$ in decimal form, we get:

$\Rightarrow \sqrt{225} = 15$

It is clear that it is a rational number because we can represent it in $\frac{p}{q}$ form.

Q9 (iii) Classify the following numbers as rational or irrational : 0.3796

Answer:

Writing 0.3796 in fraction form, we get:

$\Rightarrow 0.3796 = \frac{3796}{10000}$

It is clear that it is a rational number as the decimal expansion of this number is terminating and we can also write it in $\frac{p}{q}$ form.

Q9 (iv) Classify the following numbers as rational or irrational : 7.478478....

Answer:

We can rewrite 7.478478.... as

$\Rightarrow 7.478478.... = 7.\overline{478}$

Now, as the decimal expansion of this number is non-terminating recurring. Therefore, it is a rational number.

Q9 (v) Classify the following numbers as rational or irrational : 1.101001000100001...

Answer:

In the case of number 1.101001000100001...
It can be observed that the decimal expansion of this number is non-terminating and non-repeating. Therefore, it is an irrational number.