# PSEB 9th Class Maths Solutions Chapter 1 Number Systems Ex 1.3

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 1 Number Systems Ex 1.3 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3

Question 1.
Write the following in decimal form and say
what kind of decimal expansion each has:

(i) $$\frac{36}{100}$$
The decimal expansion of $$\frac{36}{100}$$ is terminating.

(ii) $$\frac{1}{11}$$

Thus, $$\frac{1}{11}$$ = 0.0909…. = $$0.\overline{09}$$
The decimal expansion of $$\frac{1}{11}$$ is non- terminating.

(iii) 4$$\frac{1}{8}$$
4$$\frac{1}{8}$$ = $$\frac{33}{8}$$

Thus, 4$$\frac{1}{8}$$ = 4.125
The decimal expansion of 4$$\frac{1}{8}$$ is terminating.

(iv) $$\frac{3}{13}$$

Thus, $$\frac{3}{13}$$ = 0.230769230769…. = $$0.\overline{230769}$$
The decimal expansion of 4$$\frac{2}{11}$$ is non-terminating recurring.

(v) $$\frac{2}{11}$$

Thus, $$\frac{2}{11}$$ = 0.1818….. = $$0 . \overline{18}$$
The decimal expansion of $$\frac{2}{11}$$ is non-terminating recurring.

(vi) $$\frac{329}{400}$$

Thus, $$\frac{329}{400}$$ = 0.8225
The decimal expansion of $$\frac{329}{400}$$ is terminating recurring.

Question 2.
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? [Hint: Study the remainders while finding the value of $$\frac{1}{7}$$ carefully.]
While finding the decimal expansion of $$\frac{1}{7}$$ by long division, we observe that the remainder 3, 2, 6, 4, 5 and 1 repeat in that order and the digits 1, 4, 2, 8, 5 and 7 repeat in that order in the quotient. So, in the decimal expansion of $$\frac{2}{7}$$, the same digits recur beginning with 2. Thus,

Question 3.
Express the following in the form $$\frac{p}{q}$$, where p and q are integers and q ≠ 0:
(i) 0.$$\overline{6}$$
Let x = 0.$$\overline{6}$$
∴ x = 0.6666…
∴ 10x = 6.6666…
∴ 10x = 6 + x
∴ 10x – x = 6
∴ 9x = 6
∴ x = $$\frac{6}{9}$$
∴ x = 
Thus. 0.$$\overline{6}$$ = $$\frac{2}{3}$$

(ii) 0.4$$\overline{7}$$
Let x = 0.4$$\overline{7}$$
∴ x = 0.4777…..
∴ 10x = 4.7777…..
∴ 10x = 4.3 + 0.4777…..
∴ 10x = 4.3 + x
∴ 9x = 4.3
∴ 9x = $$\frac{43}{10}$$
∴ x = $$\frac{43}{90}$$
Thus, 0.4$$\overline{7}$$ = $$\frac{43}{90}$$

(iii) 0.$$\overline{001}$$
Let x = 0.$$0 . \overline{001}$$
∴ x = 0.001001001 ……
∴ 1000 x = 1.001001001 ……
∴ 1000 x = 1 + 0.001001001 ……
∴ 1000 x = 1 + x
∴ 999 x = 1
∴ x = $$\frac{1}{999}$$
Thus, $$0 . \overline{001}$$ = $$\frac{1}{999}$$

Question 4.
Express 0.99999 ….. in the form $$\frac{p}{q}$$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Let x = 0.99999 …..
∴ 10x = 9.99999 ……
∴ 10x = 9 + 0.99999 ……
∴ 10x = 9+x
∴9x = 9
∴ x = 1
Thus, 0.99999 ….. = 1
in other words, 0.$$\overline{9}$$ = 1.
This also follows the rule mentioned in ‘Remember’ section that 0.$$\overline{m}$$ = $$\frac{m}{9}$$. As per rule, 0.$$\overline{9}$$ = $$\frac{9}{9}$$ = 1.

The answer seems to be tricky. Actually, there is no situation that gives the decimal expansion of a rational number $$\left[\frac{p}{q}\right]$$ as 0.$$\overline{9}$$.
This is an imaginary illustration of recurring decimal number.
You may say that 0.$$\overline{3}$$ + 0.$$\overline{6}$$ = 0.$$\overline{9}$$ But in simple fraction form 0.$$\overline{3}$$ and 0.$$\overline{6}$$ add up to give 1. Thus, 0.$$\overline{9}$$ = 1.

Question 5.
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.
The maximum number of digits in the repeating block of digits in the decimal expansion of $$\frac{1}{17}$$ can be 16(17 – 1).

Thus, $$\frac{1}{17}$$ = $$0.\overline{0588235294117647}$$
Here, 17 – 1 16 gives the maximum possible number of digits in the repeating block of digits in the decimal expansion of $$\frac{1}{17}$$. In some other cases, the number of digits in the repeating block of digits in the decimal expansion of $$\frac{1}{m}$$ can be smaller than (m – 1), where m is a natural number.

For example, $$\frac{1}{13}$$ = $$0 . \overline{076923}$$ has only 6 digits (not 12) in the repeating block of digits in the decimal expansion.

Question 6.
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?
$$\frac{1}{2}$$ = 0.5, $$\frac{1}{5}$$ = 0.2, $$\frac{1}{10}$$ = 0.1,
$$\frac{1}{4}$$ = 0.25, $$\frac{1}{8}$$ = 0.125, $$\frac{1}{16}$$ = 0.0625,
$$\frac{1}{25}$$ = 0.04, $$\frac{1}{125}$$ = 0.008, $$\frac{1}{625}$$ = 0.0016,
$$\frac{1}{20}$$ = 0.05, $$\frac{1}{50}$$ = 0.02 etc.
But, $$\frac{1}{3}$$ = 0.$$\overline{3}$$, $$\frac{1}{7}$$ = $$0 . \overline{142857}$$, $$\frac{1}{6}$$ = 0.1$$\overline{6}$$ etc.

This suggest that the decimal expansion of a rational number $$\frac{\boldsymbol{P}}{\boldsymbol{q}}$$ is terminating if and only if q has no prime factors other than 2 and 5. in other words, q = 2m5n, where m and n are whole numbers.

Question 7.
Write three numbers whose decimal expansions are non-terminating non-recurring.
We know that the decimal expansion of an irrational number is non-terminating non recurring. There are infinitely many irrational numbers. We can state few of them as below:
o.o1001000100001 …,
0.02002000200002…., 0.50500500050000… are a few required numbers in the decimal form. The decimal expansions of numbers like √2, √3, √5, $$\sqrt[3]{10}$$, etc. are also non- terminating non-recurring.

Question 8.
Find three different irrational numbers between the rational numbers $$\frac{5}{7}$$ and $$\frac{9}{11}$$
We know that $$\frac{5}{7}$$ = $$0 . \overline{714285}$$ and $$\frac{9}{11}$$ = $$0 . \overline{81}$$
There are actually infinitely many irrational numbers between these recurring decimals. Three irrational numbers between them can be stated as below:
0.720720072000 …, 0.750750075000 …
0.780780078000….
As you see, the above numbers are non-terminating non-recurring. Hence, they are irrational numbers.

Question 9.
Classify the following numbers as rational or irrational:
(i) √23
√23 is an irrational number.

(ii) √225
7.478478…… = $$7 . \overline{478}$$ is a rational number.