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) \(\frac{36}{100}\)

Answer:

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

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

Answer:

Thus, \(\frac{1}{11}\) = 0.0909…. = \(0.\overline{09}\)

The decimal expansion of \(\frac{1}{11}\) is non- terminating.

(iii) 4\(\frac{1}{8}\)

Answer:

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

Answer:

Thus, \(\frac{3}{13}\) = 0.230769230769…. = \(0.\overline{230769}\)

The decimal expansion of 4\(\frac{2}{11}\) is non-terminating recurring.

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

Answer:

Thus, \(\frac{2}{11}\) = 0.1818….. = \(0 . \overline{18}\)

The decimal expansion of \(\frac{2}{11}\) is non-terminating recurring.

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

Answer:

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.]

Answer:

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.\(\overline{6}\)

Answer:

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\(\overline{7}\)

Answer:

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.\(\overline{001}\)

Answer:

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.

Answer:

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.

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?

Answer:

\(\frac{1}{2}\) = 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 = 2^{m}5^{n}, where m and n are whole numbers.

Question 7.

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

Answer:

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 \(\frac{9}{11}\)

Answer:

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

Answer:

√23 is an irrational number.

(ii) √225

Answer:

√225 = 15 is a rational number

(iii) 0.3796

Answer:

0.3796 is a rational number.

(iv) 7.478478……..

Answer:

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

(v) 1.101001000100001…….

Answer:

1.101001000100001……. is an irrational number.