Exercise-1.3, Class 9th Maths, Chapter 1, NCERT

1. Write the following in decimal form and say what kind of decimal expansion each has :
(i) ${\displaystyle \frac{36}{100}}$

(ii) ${\displaystyle \frac{1}{11}}$

(iii) ${\displaystyle \frac{1}{48}}$

(iv) ${\displaystyle \frac{3}{13}}$

(v) ${\displaystyle \frac{2}{11}}$

(vi) ${\displaystyle \frac{329}{400}}$

Solution.

(i) ${\displaystyle \frac{36}{100}}=0.36$ — terminating decimal.

(ii) ${\displaystyle \frac{1}{11}}=0.090909\dots =0.\overline{09}$ — non-terminating recurring (repeating).

(iii) ${\displaystyle \frac{1}{48}}=0.02083333\dots =0.02083\bar{3}$ — non-terminating recurring.

(Quick check: $48=16\times 3$, so decimal does not terminate; remainders repeat → repeating block.)

(iv) ${\displaystyle \frac{3}{13}}=0.230769230769\dots =0.\overline{230769}$ — non-terminating recurring.

(v) ${\displaystyle \frac{2}{11}}=0.181818\dots =0.\overline{18}$ — non-terminating recurring.

(vi) ${\displaystyle \frac{329}{400}}=0.8225$ — terminating decimal.

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2. You know that ${\displaystyle \frac{1}{7}}=0.\overline{142857}$. Can you predict the decimal expansions of ${\displaystyle \frac{2}{7}},{\displaystyle \frac{3}{7}},{\displaystyle \frac{4}{7}},{\displaystyle \frac{5}{7}},{\displaystyle \frac{6}{7}}$without doing long division? If so, how?

Solution.

The repeating block for ${\displaystyle \frac{1}{7}}$ is $142857$. Multiplying that repeating block by $2,3,\dots ,6$ (mod 7 arithmetic / cyclic rotation) produces the other fractions — the digits cycle (remainders permute). So:

• ${\displaystyle \frac{2}{7}}=0.\overline{285714}$

• ${\displaystyle \frac{3}{7}}=0.\overline{428571}$

• ${\displaystyle \frac{4}{7}}=0.\overline{571428}$

• ${\displaystyle \frac{5}{7}}=0.\overline{714285}$

• ${\displaystyle \frac{6}{7}}=0.\overline{857142}$

(Reason: long-division remainders cycle; multiplying numerator rotates the recurring block.)

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3. Express the following in the form ${\displaystyle \frac{p}{q}}$, where $p$ and $q$ are integers and $q\mathrm{\ne }0$.

(i) $0.\bar{6}$
(ii) $0.\overline{47}$
(iii) $0.\overline{001}$

Solution.

(i) Let $x=0.\bar{6}$. Then $10x=6.\bar{6}$. Subtract: $10x-x=6$ ⇒ $9x=6$ ⇒ $x={\displaystyle \frac{6}{9}}={\displaystyle \frac{2}{3}}$

So $0.\bar{6}={\displaystyle \frac{2}{3}}$

(ii) Let $x=0.\overline{47}$. The repeating block has 2 digits, so multiply by 100: $100x=47.\overline{47}$. Subtract: $100x-x=47$ ⇒ $99x=47$ ⇒ $x={\displaystyle \frac{47}{99}}$

So $0.\overline{47}={\displaystyle \frac{47}{99}}$.

(iii) Let $x=0.\overline{001}$ (repeating 001, a block of 3 digits). Multiply by $1000$: $1000x=1.\overline{001}$. Subtract: $1000x-x=1$ ⇒ $999x=1$ ⇒ $x={\displaystyle \frac{1}{999}}$

So $0.\overline{001}={\displaystyle \frac{1}{999}}$

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4. Express $0.99999\dots$ in the form ${\displaystyle \frac{p}{q}}$. Are you surprised by your answer?

Solution.

Let $x=0.99999\dots$ Then $10x=9.99999\dots$. Subtract: $10x-x=9$ ⇒ $9x=9$⇒ $x=1$. Thus $0.99999\dots =1={\displaystyle \frac{1}{1}}$

Not surprising once you note that the infinite repeating 9s is the limit of the sequence $0.9,0.99,0.999,\dots$which tends to $1$. (Both notations represent the same real number.)

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5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of ${\displaystyle \frac{1}{17}}$? Perform the division to check your answer.

Solution.

For ${\displaystyle \frac{1}{q}}$ (with $q$ integer and $\gcd (q,10)=1$), the maximum possible length of the repeating block (period) is at most $q-1$. For $q=17$ the maximum possible period length is $16$. Actually ${\displaystyle \frac{1}{17}}$ has period $16$.

Indeed:

$$\frac{1}{17}=0.\overline{0588235294117647}$$

the repeating block 0588235294117647 has 16 digits.

So the maximum is 16 and ${\displaystyle \frac{1}{17}}$ attains this maximum.

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6. Look at several examples of rationals ${\displaystyle \frac{p}{q}}$ (in lowest terms) that have terminating decimal expansions. Can you guess what property $q$ must satisfy?

Solution.

A rational number ${\displaystyle \frac{p}{q}}$ (in lowest terms) has a terminating decimal expansion iff the denominator $q$ has no prime factors other than $2$ and $5$. Equivalently, $q$ divides some power of 10: $q\mid {10}^n$ for some $n$.
Examples: ${\displaystyle \frac{3}{8}}$ (denominator $8=2^3$) terminates; ${\displaystyle \frac{7}{25}}$ (denominator $25=5^2$) terminates; ${\displaystyle \frac{1}{6}}$ (denominator $6=2\cdot 3$) does not terminate because of the factor 3.

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7. Write three numbers whose decimal expansions are non-terminating non-recurring.

Solution. (Any three irrationals; their decimal expansions are non-terminating and non-recurring.) Examples:

• $\sqrt{2}=1.4142135623\dots$

• $\pi =3.1415926535\dots$
  

• $1.101001000100001\dots$ (the explicitly constructed non-recurring decimal given in the book)

All three are non-terminating non-recurring (irrational).

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8. Find three different irrational numbers between the rational numbers ${\displaystyle \frac{5}{7}}$ and ${\displaystyle \frac{9}{11}}$.

Solution.

First get decimal approximations:

$$\frac{5}{7}\approx 0.714285\dots ,\frac{9}{11}\approx 0.818181\dots$$

Any irrational numbers whose decimal value lies strictly between $0.714285\dots$ and $0.818181\dots$ will do. Examples (all clearly irrational):

1. ${\displaystyle \frac{\sqrt{5}}{3}\approx \frac{2.2360679}{3}\approx 0.745356}$ (between the two).

2. ${\displaystyle \frac{\pi }{4}\approx 0.785398}$ (between the two).

3. ${\displaystyle \sqrt{2}-0.6\approx 1.41421356-0.6\approx 0.81421356}$ (between the two).

Each is irrational (square root or π divided by integer or irrational minus rational) and lies between ${\displaystyle \frac{5}{7}}$ and ${\displaystyle \frac{9}{11}}$.

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9. Classify the following numbers as rational or irrational :

(i) $\sqrt{23}$
(ii) $\sqrt{225}$
(iii) $0.3796$
(iv) $7.478478\dots$
(v) $1.101001000100001\dots$

Solution.

(i) $\sqrt{23}$ is not a perfect square ⇒ $\sqrt{23}$ is irrational.

(ii) $\sqrt{225}=\sqrt{{15}^2}=15$ ⇒ rational (an integer).

(iii) $0.3796$ — terminating decimal ⇒ can be written as ${\displaystyle \frac{3796}{10000}}$ ⇒ rational.

(iv) $7.478478\dots$ — decimal has repeating block 478 ⇒ $7.\overline{478}$ ⇒ rational (non-terminating recurring).

(v) $1.101001000100001\dots$ — decimal constructed to have longer and longer blocks of zeros, not repeating ⇒ irrational (non-terminating non-recurring).