Class-IX NCERT Mathematics Solutions Exercise 1.3

Q-1: Write the following in decimal form and say what kind of decimal expansion each has :

(i) $\cfrac{36}{100}$

(ii) $\cfrac{1}{11}$

(iii) $4\cfrac{1}{8}$

(iv) $\cfrac{3}{13}$

(v) $\cfrac{2}{11}$

(vi) $\cfrac{329}{400}$

Solution:

(i) $\cfrac{36}{100}$

$\cfrac{36}{100} = 0.36$

Decimal Expansion is terminating.

(ii) $\cfrac{1}{11}$

Performing long division, we get,

$\cfrac{1}{11} = 0.0909… = 0.\overline{09}$

Decimal expansion is non-terminating and repeating.

(iii) $4\cfrac{1}{8}$

$4\cfrac{1}{8} = \cfrac{33}{8}$

Performing long division, we get,

$4\cfrac{1}{8} = \cfrac{33}{8} = 8.125$

Decimal expansion is terminating.

(iv) $\cfrac{3}{13}$

Performing long division, we get,

$\cfrac{3}{13} = 0.230769… = 0.\overline{230769}$

Decimal expansion is non-terminating and repeating.

(v) $\cfrac{2}{11}$

Performing long division, we get,

Q-4: Express 0.99999…. in the form $\cfrac{p}{q}$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

Solution:

Let $x = 0.9999….$ —————–(1)

Multiply both sides by 10, we get,

$10x = 9.999….$

$\implies 10x = 9 + 0.999…$

$\implies 10x = 9 + x$ (Using 1)

$\implies 10x – x = 9$

$\implies 9x = 9$

$\implies x = \cfrac{9}{9} = 1$

$0.9999999..$ or $0.\overline{9}$ is too close to 1. Therefore, we can consider it to be equal to 1.

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

Solution:

By using long division, we get,

$\therefore \cfrac{1}{17} = 0 . \overline{0588235294117647}$ which means there are 16 digits in the repeating block.

Q-6: Look at several examples of rational numbers in the form $\cfrac{p}{q} (q \ne 0)$, 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?

Solution:

Let’s consider few examples

$\cfrac{5}{2} = 2.5$————–terminating when denominator is 2 (or $2^1$)

$\cfrac{5}{4} = 1.25$————-terminating when denominator is 4 (or $2^2$)

$\cfrac{5}{8} = 0.625$————terminating when denominator is 8 (or $2^3$)

$\cfrac{2}{5} = 0.4$————–terminating when denominator is 5 (or $5^1$)

$\cfrac{2}{25} = 0.08$————terminating when denominator is 25 (or $5^2$)

$\cfrac{2}{200} = 0.01$———–terminating when denominator is 200 (or $2^3\times5^2$)

$\therefore$ We observe that the decimal expansion is terminating when the prime factorization of denominator has power of 2 or 5 or both.

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

Solution:

All irrational numbers are non-terminating non-recurring. three numbers with decimal expansions that are non-terminating and non-recurring are:

1. $\sqrt{3} = 1.732050807568..$
2. $\sqrt{26} = 5.099019513592..$
3. $\sqrt{101} = 10.04987562112..$

Q-8: Find three different irrational numbers between the rational numbers $\cfrac{5}{7}$ and $\cfrac{9}{11}$.

Solution:

$\cfrac{5}{7} = 0.\overline{714285}$

$\cfrac{9}{11} = 0.\overline{81}$

Three different irrational number between $\cfrac{5}{7}$ and $\cfrac{9}{11}$ are

Q-9: Classify the following numbers as rational or irrational according to their type:

(i) $\sqrt{23}$

(ii) $\sqrt{225}$

(iii) $0.3796$

(iv) $7.478478$

(v) $1.101001000100001…$

Solution:

(i) $\sqrt{23}$

$\sqrt{23}$ is an irrational number because it’s decimal expansion is non-terminating and non-recurring. [$\sqrt{23} = 4.79583152331…$]

(ii) $\sqrt{225}$

$\sqrt{225} = 15 = \cfrac{15}{1} $

$\therefore \sqrt{225}$ is a rational number because it can be expressed in form $\cfrac{p}{q}$.

(iii) $0.3796$

The number has a terminating decimal expansion therefore, it is a rational number.

(iv) $7.748748..$

The decimal expansion is non-terminating but recurring therefore, the number is rational.

(v) $1.101001000100001…$

The decimal expansion is non-terminating and non-recurring therefore, the number is irrational.