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 = 2m5n, 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.