RBSE Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 1 Number Systems Exercise 1.3

Question 1.
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, terminating.

(ii) \(\frac{1}{11}\)
Answer:
By long division method, we have:

∴ \(\frac{1}{11}\) = 0.090909… = \(0 . \overline{09}\), non terminating and repeating.

(iii) 4\(\frac{1}{8}\)
Answer:
4\(\frac{1}{8}\) = \(\frac{4 \times 8+1}{8}\) = \(\frac{33}{8}\)
By long division method, we have:

∴ \(\frac{33}{8}\) = 4.125, terminating decimal.

(iv) \(\frac{3}{13}\)
Answer:
By long division method, we have:

∴ \(\frac{3}{13}\) = 0.23076923…… = \(0 . \overline{230769}\),
non-terminating and repeating decimal.

(v) \(\frac{2}{11}\)
Answer:
By long division method, we have:

∴ \(\frac{2}{11}\) = 0.181818 = \(0 . \overline{18}\),
non-terminating and repeating decimal.

(vi) \(\frac{329}{400}\)
Answer:
By long division method, we have:

∴ \(\frac{329}{400}\) = 0.8225, terminating decimal.

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 y carefully.)
Answer:
Yes, the decimal expansion of \(\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}\) and \(\frac{6}{7}\) can be predicted as all of the above will have repeating decimals which are permutations of 1, 4, 2, 8, 5, 7. For example, here

To find \(\frac{2}{7}\), locate when the remainder becomes 2 and respective quotient (here it is 2), then write the new quotient beginning from there (the arrows drawn in the above division) using the repeating digits 1, 4, 2, 8, 5, 7.

Question 3.
Express the following in the form of \(\frac{6}{7}\), where p and q are integers and q ≠ 0.
(i) \(0 . \overline{6}\)
(ii) \(0.4 \overline{7}\)
(iii) \(0 . \overline{001}\)
Answer:
(i) Let x = \(0 . \overline{6}\)
Then, x = 0.666 ……… (1)
Here, we have only one repeating digit. So, we multiply both sides of (1) by 10 to get
10x = 6.66 ……….(2)
Subtracting (1) from (2), we get
10x – x = (6.66…) – (0.66…)
⇒ 9x = 6 ⇒ x = \(\frac{2}{3}\)
Hence, \(0 . \overline{6}\) = \(\frac{2}{3}\)

(ii) Let x = \(0.4 \overline{7}\) …(1)
Clearly, there is just one digit on the right side of the decimal point which is without bar. So, we multiply both sides by 10 so that only the repeating decimal is left on the right side of the decimal point.
10x = 4.\(\overline{7}\) …..(2)
Multiplying (i) by 100 both side
we get: 100x = 47.\(\overline{7}\) …… (3)
Subtracting equation (2) from (3), we get:
100x – 10x = 47.\(\overline{7}\) – 4.\(\overline{7}\)
Hence, 0.4\(\overline{7}\) = \(\frac{43}{90}\)

(iii) Let x = \(0 . \overline{001}\)
∴ x = 0.001001001 ………. (1)
Here, we have a block of three repeating digits after the decimal point. So, we multiply (1) by 10³ = 1000, we get:
1000x = 1.001001… (2)
Subtracting (1) from (2), we get:
1000x – x = (1.001001…) – (0.001001…)
⇒ 999x = 1
⇒ x = \(\frac{1}{999}\)
Hence, \(0 . \overline{001}\) = \(\frac{1}{999}\)

Question 4.
Express 0.99999…. in the form of \(\frac{p}{q}\). Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Answer:
Let x = 0.9999 …………. (1)
Here, we have only one repeating digit. So, we multiply both sides of (1) by 10 to get:
10x = 9.999… (2)
Subtracting (1) from (2), we get:
10x – x = (9.999…) – (0.999…)
⇒ 9x = 9
⇒ x = 1
Hence, 0.9999… = 1
Since 0.9999… goes on infinitely. So, 0.9999 is approaching towards 1, i.e. it is equal to 1 in the limiting form and hence it makes sense.

Question 5.
What can the maximum number of digits be in the repeating block of digits in the decimal expansion \(\frac{1}{17}\)? Perform the division to check your answer.
Answer:
The maximum number of digits in the quotient while computing \(\frac{1}{17}\) is 17 – 1 = 16.

Thus, \(\frac{1}{17}\) = \(0 . \overline{0588235294117647 \ldots . .}\) In it, a block of 16 digits is repeated.

Question 6.
Look at several examples of rational numbers in the form of \(\frac{p}{q}\) (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:
Consider several rational numbers in the form of \(\frac{p}{q}\) (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations.
Let the various such rational numbers be \(\frac{1}{2}, \frac{1}{4}, \frac{7}{8}, \frac{37}{25}, \frac{8}{125}, \frac{17}{20}, \frac{31}{16}\), etc.

In all cases, we think of the natural numbers which when multiplied by their respective denominators gives 10 or a power of 10.

We have seen that those rational numbers whose denominators when multiplied by a suitable integer produce a power of 10 are expressible in the terminating decimal form. But this can always be done only when the denominator of the given rational number has either 2 or 5 or both of them as the only prime factors. Thus, we obtain the following property possessed by q.

If the denominator of a rational number in standard form has no prime factors other . than 2 or 5, then and only then it can be represented as a terminating decimal.

Question 7.
Write three numbers whose decimal expansions are non-terminating and non-recurring.
Answer:
Three numbers whose decimal representations are non-terminating and non-recurring are:
0.1001000100001…., 0.202002000200002… and 0.003000300003…

Question 8.
Find three different irrational numbers between the rational numbers \(\frac{5}{7}\) and \(\frac{9}{11}\).
Answer:
We have:

∴ \(\frac{5}{7}\) = \(0 . \overline{714285}\) and \(\frac{9}{11}\) = \(0 . \overline{81}\)
Thus, three different irrational numbers between \(\frac{5}{7}\) and \(\frac{9}{11}\) are:
0.75675667566675666675…, 0.767876788767888… and 0.80800800080000…

Question 9.
Classify the following numbers as rational or irrational:
(i) √23
Answer:
√23 is an irrational number as 23 is not a perfect square.

(ii) √225
Answer:
√225 = \(\sqrt{3 \times 3 \times 5 \times 5}\)
= 3 × 5
= 15
Thus, √225 is a rational number.

(iii) 0.3796
Answer:
0.3796 is a rational number as it is terminating decimal.

(iv) 7.478478…
Answer:
7.478478… is non-terminating but repeating. So, it is a rational number.

(v) 1.101001000100001…
Answer:
1.101001000100001… is non-terminating and non-repeating. So, it is an irrational number.