Well-designed AP 9th Class Maths Textbook Solutions Chapter 1 Number System Exercise 1.3 offers step-by-step explanations to help students understand problem-solving strategies.
Number System Class 9 Exercise 1.3 Solutions – 9th Class Maths 1.3 Exercise Solutions
Question 1.
Write the following In decimal form and say what kind of decimal expansion each has:
i) \(\frac{36}{100}\)
ii) \(\frac{1}{11}\)
iii) \(4 \frac{1}{8}\)
iv) \(\frac{3}{13}\)
v) \(\frac{2}{11}\)
vi) \(\frac{329}{400}\)
Solution:
i) \(\frac{36}{100}\) = 0.36 terminating decimal hence it Is a rational number.
ii) \(\frac{1}{11}\) = 0.090909… = \(0 . \overline{09}\) is a non-terminating recurring decimal. Hence,
it is a rational number.
iii) \(4 \frac{1}{8}\) = 4 + \(\frac{1}{8}\)
\(\frac{1}{8}\) = 0.1254 = 4 + 0.125 = 4.125 is a terminating decimal. Hence, a rational number.
iv) \(\frac{3}{13}\) = 0.230769230769.. = \(0 . \overline{230769}\) is a non-terminating recurring decimal.
Hence, a rational number.
v) \(\frac{2}{11}\) = 0.181818… = \(0 . \overline{18}\) is a non-terminat ¡ng recurring decimal. Hence, a rational number.
vi) \(\frac{329}{400}\) = 0.8225 is a terminating decimal.
Question 2.
You know that \(\frac{1}{7}\) = 0.142857. 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 f inding the value of \(\frac{1}{7}\) carefully.]
Solution:
Given that \(\frac{1}{7}\) = 0.142857 this is non-terminating, recurring decimal expansion.
Then \(\frac{2}{7}\) = 2 × \(\frac{1}{7}\) = (0.142857142857… × 2)
= 0.285714285714 ……..
= \(0 . \overline{285714}\)
Now, \(\frac{3}{7}\) = 3 × \(\frac{1}{7}\) = (0.142857142857… × 3)
= 0.428571428571…
= \(0 . \overline{428571}\)
\(\frac{4}{7}\) = 4 ×\(\frac{1}{7}\) = (0.142857142857… × 4)
= 0.571428571428…
= \(0 . \overline{571428}\)
\(\frac{5}{7}\) = 5 × \(\frac{1}{7}\) = (0.142857142857… × 5)
= 0.714285714285…
.= \(0 . \overline{714285}\)
and \(\frac{6}{7}\) = 6 × \(\frac{1}{7}\) = (0.142857142857… × 6)
= 0.85714857142
= \(0 . \overline{857142}\)
We observe in the above, same dig-its in same order with different starting digit follows.
Question 3.
Express the following in the form \(\frac{p}{q}\) where p and q are integers and q ≠ 0.
i) \(0 . \overline{6}\)
ii) \(0 . \overline{47}\)
iii) \(0 . \overline{001}\)
Solution:
Expressing the following in the form \(\frac{p}{q}\), where ‘p’ and ‘q’ are integers and q * 0.
i) Let x = \(0 . \overline{6}\) = 0.6666……….
Here we observe only one digit (6) is repeating. Hence multiply ‘x’ by 10 to get
∴ x = \(0 . \overline{6}\) = \(\frac{2}{3}\) which is in required \(\frac{p}{q}\) form.
ii) Given \(0 . \overline{47}\)
∴ x = 0.47777… then, multiply by 100
⇒ x = \(\frac{43}{90}\) which is in required \(\frac{p}{q}\) form.
iii) x = \(0 . \overline{001}\)
.’. x = 0.001001001 ….. here 3 digits (001) are repeating.
So, multiply both sides by 1000, to get
∴ x = \(\frac{1}{999}\) which is in required \(\frac{p}{q}\) form.
∴ \(0 . \overline{001}\) = \(\frac{1}{999}\)
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.
Solution:
Let given 0.99999 is ‘x’„
∴ x = 0.99999 here only one digit (9) is recurring.
Hence multiply both sides by 10, to get
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.
Solution:
Given x = \(\frac{1}{17}\)
Let us divide ‘1’ by 17 to get its decimal expansion.
Here onwards same division repeats.
Hence \(\frac{1}{17}\) = \(0 . \overline{0588235294117647}\)
So, there are 16 digits in repeating block.
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 termi-nating decimal representations (expansions). Can you guess what prop-erty q must satisfy ?
Solution:
We can see the following are terminating decimal expansions.
Ex : \(\frac{1}{5}\) = 0.2; \(\frac{1}{8}\) = 0.125, \(\frac{2}{25}\) = 0.08, \(\frac{1}{100}\) = 0.01, etc.
After observing number of examples, we can conclude that in the form of (p/q), where ’q’ can be written in the form of
(q = 2m.5n), where m, n are.whole numbers.
Question 7.
Write three numbers whose decimal expansions are non-terminating nonrecurring.
Solution:
Examples for non-terminating, non-recurring decimal expansions.
3.458458045800458000…..
2.03003000300003….
1.10203040………
Question 8.
Find three different irrational numbers between the rational numbers \(\frac{5}{7}\) and \(\frac{9}{11}\).
Solution:
So, recurring starts from here,
∴ \(\frac{5}{7}\) = \(0 . \overline{714285}\) and
So, recurring starts from here,
∴ \(\frac{9}{11}\) = \(0 . \overline{81}\) = 0.8181818181……..
So, we can write many irrationals in between \(0 . \overline{714285}\) and \(0 . \overline{81}\) like
i) 0.720720072000….
ii) 0.770770077000….
iii) 0.790790079000…..
iv) 0.801801080100…….
Thus we can write as many as required irrationals in between given two rationals.
Question 9.
Classify the following numbers as rational or irrational :
i) \(\sqrt{23}\)
ii) \(\sqrt{225}\)
iii) 0.3796
iv) 7.478478…
v) l.101001000100001…
Solution:
I) \(\sqrt{23}\) here 23 is a prime hence square root of prime number is always irrational.
ii) \(\sqrt{225}\) = 15 which can be written terminating decimal (15.0). Hence, rational.
iii)0.3796 it Is a terminating decimal. Hence it is rational.
iv)7.478478… it is non-terminating, recurring decimal expansion. Hence, it is an irrational.
v) 1.10100100010001… It Is non-terminating non-recurring decimal expansion. Hence, it is an irrational.