Number Systems

NCERT Solutions • Class 9 Maths • Chapter 1
Exercise 1.3
1. Write the following in decimal form and say what kind of decimal expansion each has:
Solution:
  1. (i) $\frac{36}{100}$ $= 0.36$.
    Type: Terminating decimal.
  2. (ii) $\frac{1}{11}$ When we divide 1 by 11, we get $0.090909…$
    $= 0.\overline{09}$.
    Type: Non-terminating Repeating decimal.
  3. (iii) $4\frac{1}{8}$ $= \frac{33}{8} = 4.125$.
    Type: Terminating decimal.
  4. (iv) $\frac{3}{13}$ $= 0.230769230769…$
    $= 0.\overline{230769}$.
    Type: Non-terminating Repeating decimal.
  5. (v) $\frac{2}{11}$ $= 0.181818…$
    $= 0.\overline{18}$.
    Type: Non-terminating Repeating decimal.
  6. (vi) $\frac{329}{400}$ $= \frac{329}{4 \times 100} = \frac{82.25}{100} = 0.8225$.
    Type: Terminating decimal.
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?
Solution: Yes, we can predict them by multiplying the decimal expansion of $\frac{1}{7}$.

Given, $\frac{1}{7} = 0.\overline{142857}$
  • $\frac{2}{7} = 2 \times \frac{1}{7} = 2 \times 0.\overline{142857} = \mathbf{0.\overline{285714}}$
  • $\frac{3}{7} = 3 \times \frac{1}{7} = 3 \times 0.\overline{142857} = \mathbf{0.\overline{428571}}$
  • $\frac{4}{7} = 4 \times \frac{1}{7} = 4 \times 0.\overline{142857} = \mathbf{0.\overline{571428}}$
  • $\frac{5}{7} = 5 \times \frac{1}{7} = 5 \times 0.\overline{142857} = \mathbf{0.\overline{714285}}$
  • $\frac{6}{7} = 6 \times \frac{1}{7} = 6 \times 0.\overline{142857} = \mathbf{0.\overline{857142}}$
Note: The digits cycle in the same order (1-4-2-8-5-7), just starting from a different digit each time.
3. Express the following in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
Solution:
  1. (i) $0.\overline{6}$ Let $x = 0.6666…$ — (1)
    Multiply by 10:
    $10x = 6.6666…$ — (2)
    Subtract (1) from (2):
    $9x = 6 \Rightarrow x = \frac{6}{9} = \mathbf{\frac{2}{3}}$
  2. (ii) $0.4\overline{7}$ Let $x = 0.4777…$
    Multiply by 10 (to move non-repeating part):
    $10x = 4.777…$ — (1)
    Multiply by 10 again (to move one repeating digit):
    $100x = 47.777…$ — (2)
    Subtract (1) from (2):
    $90x = 43 \Rightarrow x = \mathbf{\frac{43}{90}}$
  3. (iii) $0.\overline{001}$ Let $x = 0.001001…$ — (1)
    Multiply by 1000 (since 3 digits repeat):
    $1000x = 1.001001…$ — (2)
    Subtract (1) from (2):
    $999x = 1 \Rightarrow x = \mathbf{\frac{1}{999}}$
4. Express $0.99999…$ in the form $\frac{p}{q}$. Are you surprised by your answer? Discuss why the answer makes sense.
Solution: Let $x = 0.99999…$ — (1)
Multiply by 10:
$10x = 9.9999…$ — (2)
Subtract (1) from (2):
$9x = 9$
$x = \frac{9}{9} = \mathbf{1}$
Discussion: Yes, it might be surprising that $0.999…$ equals exactly $1$. However, this makes sense because the difference between 1 and $0.999…$ is infinitely small ($0.000…1$), which mathematically tends to zero. Thus, $0.\overline{9}$ and $1$ represent the same number.
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: The maximum number of digits in the repeating block is always divisor minus 1.
Here divisor is 17. So, maximum digits = $17 – 1 = \mathbf{16}$.

Performing the division:
After performing the long division, we get: $$ \frac{1}{17} = 0.\overline{0588235294117647} $$ There are exactly 16 digits in the repeating block.
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. Can you guess what property $q$ must satisfy?
Solution: For a rational number $\frac{p}{q}$ to have a terminating decimal representation, the prime factorization of the denominator $q$ must contain only powers of 2, powers of 5, or both.

Property: $q$ must be of the form $2^m \times 5^n$, where $m$ and $n$ are non-negative integers.
7. Write three numbers whose decimal expansions are non-terminating non-recurring.
Solution: Numbers that are non-terminating and non-recurring are Irrational Numbers.

Examples:
  1. $0.101001000100001…$
  2. $0.202002000200002…$
  3. $0.73073007300073…$ (Any pattern that doesn’t repeat cyclically)
  4. $\pi = 3.1415926535…$
8. Find three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$.
Solution: First, convert the fractions to decimals:
$\frac{5}{7} = 0.\overline{714285} \approx 0.714$
$\frac{9}{11} = 0.\overline{81} \approx 0.818$

We need irrational numbers between $0.71…$ and $0.81…$.
Three examples are:
  1. $0.720720072000…$
  2. $0.750750075000…$
  3. $0.80800800080000…$
9. Classify the following numbers as rational or irrational:
Solution:
  1. (i) $\sqrt{23}$ Irrational. (Because 23 is not a perfect square).
  2. (ii) $\sqrt{225}$ $= 15$. Rational. (Because it is a perfect square).
  3. (iii) $0.3796$ Rational. (Because the decimal is terminating).
  4. (iv) $7.478478…$ $= 7.\overline{478}$. Rational. (Non-terminating but recurring).
  5. (v) $1.101001000100001…$ Irrational. (Non-terminating and non-recurring).
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