Get step-by-step Rbse Solutions for Class 9 Maths Chapter 1, Exercise 1.3. Master terminating and recurring decimal expansions and learn to express $0.\overline{6}$ and other repeating decimals in the $\frac{p}{q}$ form. Essential for RBSE/CBSE students.
1. Write the following in decimal form and say what kind of decimal expansion each has:
(i) $\frac{36}{100}$
$$\frac{36}{100} = 0.36$$
Decimal Expansion: Terminating. (The division ends after a finite number of steps.)
(ii) $\frac{1}{11}$
$$\frac{1}{11} = 0.\overline{09}$$
Decimal Expansion: Non-terminating recurring (or repeating). (The block ’09’ repeats indefinitely.)
(iii) $4\frac{1}{8}$
$$4\frac{1}{8} = \frac{(4 \times 8) + 1}{8} = \frac{33}{8}$$
$$\frac{33}{8} = 4.125$$
Decimal Expansion: Terminating.
(iv) $\frac{3}{13}$
$$\frac{3}{13} = 0.\overline{230769}$$
Decimal Expansion: Non-terminating recurring. (The block ‘230769’ repeats indefinitely.)
(v) $\frac{2}{11}$
$$\frac{2}{11} = 0.\overline{18}$$
Decimal Expansion: Non-terminating recurring. (The block ’18’ repeats indefinitely.)
(vi) $\frac{329}{400}$
$$\frac{329}{400} = 0.8225$$
Decimal Expansion: Terminating.
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?
Answer:
Yes, we can predict the decimal expansions without performing long division by observing the repeating block of digits in $\frac{1}{7}$.
Since $\frac{1}{7} = 0.\overline{142857}$, the decimal expansions of $\frac{2}{7}, \frac{3}{7}, \dots$ will use the same sequence of digits, but they will start from a different digit in that sequence. This phenomenon is caused by the sequence of remainders in the long division of $1 \div 7$.
| Fraction | Calculation | Prediction | Starting Digit |
| $\frac{2}{7}$ | $2 \times \frac{1}{7}$ | $0.\overline{285714}$ | Starts with 2 (the second digit) |
| $\frac{3}{7}$ | $3 \times \frac{1}{7}$ | $0.\overline{428571}$ | Starts with 4 |
| $\frac{4}{7}$ | $4 \times \frac{1}{7}$ | $0.\overline{571428}$ | Starts with 5 |
| $\frac{5}{7}$ | $5 \times \frac{1}{7}$ | $0.\overline{714285}$ | Starts with 7 |
| $\frac{6}{7}$ | $6 \times \frac{1}{7}$ | $0.\overline{857142}$ | Starts with 8 |
3. Express the following in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
(i) $0.\overline{6}$
Let $x = 0.\overline{6}$
$$x = 0.6666\dots \quad (1)$$
Since one digit is repeating, multiply by $10$:
$$10x = 6.6666\dots \quad (2)$$
Subtract equation (1) from equation (2):
$$10x – x = (6.666\dots) – (0.666\dots)$$
$$9x = 6$$
$$x = \frac{6}{9} = \frac{2}{3}$$
(ii) $0.4\overline{7}$
Let $x = 0.4\overline{7}$
$$x = 0.4777\dots \quad (1)$$
Multiply by $10$ to move the non-repeating digit to the left of the decimal:
$$10x = 4.777\dots \quad (2)$$
Multiply by $100$ (since 1 digit is repeating in total after the decimal):
$$100x = 47.777\dots \quad (3)$$
Subtract equation (2) from equation (3):
$$100x – 10x = (47.777\dots) – (4.777\dots)$$
$$90x = 43$$
$$x = \frac{43}{90}$$
(iii) $0.\overline{001}$
Let $x = 0.\overline{001}$
$$x = 0.001001001\dots \quad (1)$$
Since three digits are repeating, multiply by $1000$:
$$1000x = 1.001001001\dots \quad (2)$$
Subtract equation (1) from equation (2):
$$1000x – x = (1.001001\dots) – (0.001001\dots)$$
$$999x = 1$$
$$x = \frac{1}{999}$$
4. Express $0.\overline{9}$ 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 $x = 0.\overline{9}$
$$x = 0.9999\dots \quad (1)$$
Since one digit is repeating, multiply by $10$:
$$10x = 9.9999\dots \quad (2)$$
Subtract equation (1) from equation (2):
$$10x – x = (9.999\dots) – (0.999\dots)$$
$$9x = 9$$
$$x = \frac{9}{9} = 1$$
Surprise and Justification:
The answer $1$ is surprising because $0.\overline{9}$ appears to be slightly less than 1. However, the difference between $1$ and $0.\overline{9}$ is $1 – 0.\overline{9} = 0.\overline{0}$. Since $0.\overline{0}$ is mathematically equal to 0, there is no difference between the two numbers.
This makes sense because $0.\overline{9}$ is not merely close to 1, it is exactly equal to 1.
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 a rational number $\frac{p}{q}$ is always less than $q$.
Here, $q=17$, so the maximum number of digits in the repeating block can be $17 – 1 = 16$.
Division Check:
Performing the long division of $1 \div 17$:
$$\frac{1}{17} = 0.\overline{0588235294117647}$$
The repeating block is ‘0588235294117647’, which contains 16 digits. This confirms the theoretical maximum.
6. Look at several examples of rational numbers in the form $\frac{p}{q}$ ($q \neq 0$), 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:
The property that the denominator $q$ must satisfy for a rational number $\frac{p}{q}$ (in simplest form) to have a terminating decimal expansion is:
The prime factorization of the denominator ($q$) must contain only powers of 2, only powers of 5, or powers of both 2 and 5.
Mathematically, $q$ must be of the form:
$$q = 2^m \times 5^n$$
where $m$ and $n$ are non-negative integers.
Examples:
- $\frac{3}{8}$ (Terminating): $8 = 2^3$.
- $\frac{7}{20}$ (Terminating): $20 = 2^2 \times 5^1$.
- $\frac{5}{6}$ (Non-terminating): $6 = 2^1 \times 3^1$. (Contains a factor of 3).
7. Write three numbers whose decimal expansions are non-terminating non-recurring.
Answer:
Non-terminating non-recurring decimal expansions define Irrational Numbers. Any number that cannot be written in the form $\frac{p}{q}$ and whose decimal representation never ends and never repeats is an irrational number.
Three examples are:
- $\sqrt{2} \approx 1.41421356\dots$
- $\pi \approx 3.14159265\dots$
- A constructed irrational number: $0.101001000100001\dots$ (The pattern of increasing zeros ensures it never repeats.)
8. Find three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$.
Answer:
First, find the decimal expansions of the two given rational numbers:
- $\frac{5}{7} = 0.\overline{714285}$
- $\frac{9}{11} = 0.\overline{81}$
We need to find three numbers that are non-terminating non-recurring (irrational) and lie between $0.714285\dots$ and $0.818181\dots$.
Three such irrational numbers are:
- $0.75075007500075\dots$ (Starts with 0.75, which is between the two limits, and is non-recurring.)
- $0.781122334455\dots$ (Starts with 0.78, which is between the two limits, and is non-recurring.)
- $0.80800800080000\dots$ (Starts with 0.80, which is between the two limits, and is non-recurring.)
9. Classify the following numbers as rational or irrational:
(i) $\sqrt{23}$
Classification: Irrational.
Reason: 23 is not a perfect square, so $\sqrt{23}$ is non-terminating and non-recurring.
(ii) $\sqrt{225}$
Classification: Rational.
Reason: $225$ is a perfect square; $\sqrt{225} = 15$. Since $15 = \frac{15}{1}$, it is rational.
(iii) $0.3796$
Classification: Rational.
Reason: The decimal expansion is terminating (it ends). Any terminating decimal can be expressed as a fraction (e.g., $\frac{3796}{10000}$).
(iv) $7.478478\dots$
Classification: Rational.
Reason: This can be written as $7.\overline{478}$. The decimal expansion is non-terminating but recurring (the block ‘478’ repeats). All non-terminating recurring decimals are rational.
(v) $1.101001000100001\dots$
Classification: Irrational.
Reason: The decimal expansion is non-terminating and non-recurring. The pattern of increasing zeros (1 zero, 2 zeros, 3 zeros, etc.) ensures the block of digits never truly repeats.
❓ Frequently Asked Questions (FAQs) on Decimal Expansions
Q: What is the difference between a terminating and a non-terminating decimal?
A: A terminating decimal is one whose division ends after a finite number of steps (e.g., $\frac{3}{8} = 0.375$). A non-terminating decimal is one whose division never ends (e.g., $\frac{1}{3} = 0.333\dots$). Non-terminating decimals can be either recurring (rational) or non-recurring (irrational).
Q: How can I tell if a fraction $\frac{p}{q}$ will have a terminating decimal expansion?
A: A rational number $\frac{p}{q}$ (in its simplest form, where $p$ and $q$ have no common factors) will have a terminating decimal expansion if and only if the prime factors of the denominator ($q$) are only 2, only 5, or both 2 and 5. If the denominator contains any other prime factor (like 3, 7, 13), the decimal will be non-terminating recurring.
Q: Is $0.9999\dots$ (or $0.\overline{9}$) equal to 1?
A: Yes, mathematically $0.\overline{9}$ is exactly equal to 1. This can be proven by expressing $0.\overline{9}$ in the form $\frac{p}{q}$, which results in $\frac{9}{9}$, or 1. The concept makes sense because the difference between 1 and $0.\overline{9}$ is $0.\overline{0}$, which is zero.
Q: What defines an irrational number in terms of its decimal expansion?
A: An irrational number is defined by its decimal expansion being non-terminating and non-recurring (or non-repeating). This means the digits after the decimal point go on forever without ever settling into a fixed, repeating block (e.g., $\sqrt{2}$ or $\pi$).