Class 10 NCERT Solutions- Chapter 1 Real Numbers – Exercise 1.4

Difficulty Level : Medium
Last Updated : 04 Mar, 2021

NCERT Theorem 1.5 : Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form, $\frac{p}{q}$ where p and q are coprime, and the prime factorization of q is of the form 2ⁿ5^m, where n, m are non-negative integers.
NCERT Theorem 1.6 : Let x = $\frac{p}{q}$ be a rational number, such that the prime factorization of q is of the form 2ⁿ5^m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

Question 1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i) $\frac{13}{3125}$

(ii) $\frac{17}{8}$

(iii) $\frac{64}{455}$

(iv) $\frac{15}{1600}$

(v) $\frac{29}{343}$

(vi) $\frac{23}{2^35^2}$

(vii) $\frac{129}{2^25^77^5}$

(viii) $\frac{6}{15}$

(ix) $\frac{35}{50}$

(x) $\frac{77}{210}$

Solution:

(i) $\frac{13}{3125}$
By doing prime factorization of denominator, we get
3125 = 5×5×5 = 5³
As denominator is in the form 2ⁿ5^m only where n=0 and m=3.
According to Theorem 1.6,
$\frac{13}{3125}$ will have a terminating decimal expansion.
(ii) $\frac{17}{8}$
By doing prime factorization of denominator, we get
8 = 2×2×2 = 2³
As denominator is in the form 2ⁿ5^m only where n=3 and m=0.
According to Theorem 1.6,
$\frac{17}{8}$ will have a terminating decimal expansion.
(iii) $\frac{64}{455}$
By doing prime factorization of denominator, we get
455 = 5×7×13
As denominator is not in the form 2ⁿ5^m only.
According to contradiction of Theorem 1.6,
$\frac{64}{455}$ will have a non-terminating decimal expansion.
(iv) $\frac{15}{1600}$
By doing prime factorization of denominator, we get
1600 = 2×2×2×2×2×2×5×5 = 2⁶5²
As denominator is in the form 2ⁿ5^m only where n=6 and m=2.
According to Theorem 1.6, 1
$\frac{15}{1600}$ will have a terminating decimal expansion.
(v) $\frac{29}{343}$
By doing prime factorization of denominator, we get
343 = 7×7×7 = 7³
As denominator is not in the form 2ⁿ5^monly.
According to contradiction of Theorem 1.6,
$\frac{29}{343}$ will have a non-terminating decimal expansion.
(vi) $\frac{23}{2^35^2}$
Prime factorization of denominator, we have
= 2³5²
As denominator is in the form 2ⁿ5^m only where n=3 and m=2.
According to Theorem 1.6,
$\frac{23}{2^35^2}$ will have a terminating decimal expansion.
(vii) $\frac{129}{2^25^77^5}$
Prime factorization of denominator, we have
= 2²5⁷7⁵
As denominator is not in the form 2ⁿ5^m only.
According to contradiction of Theorem 1.6,
$\frac{129}{2^25^77^5}$ will have a non-terminating decimal expansion.
(viii) $\frac{6}{15}$
$\frac{6}{15} = \frac{3}{5}$
by doing prime factorization of denominator, we get
5 = 5¹
As denominator is in the form 2ⁿ5^m only where n=0 and m=1.
According to Theorem 1.6,
$\frac{6}{15}$ will have a terminating decimal expansion.
(ix) $\frac{35}{50}$
by doing prime factorization of denominator, we get
50= 2×5×5 = 2¹5²
As denominator is in the form 2ⁿ5^m where n=1 and m=2.
According to Theorem 1.6,
$\frac{35}{50}$ will have a terminating decimal expansion.
(x) $\frac{77}{210}$
by doing prime factorization of denominator, we get
210 = 2×3×5×7
As denominator is not in the form 2ⁿ5^m only.
According to contradiction of Theorem 1.6,
$\frac{77}{210}$ will have a non-terminating decimal expansion.

Question 2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

(i) $\frac{13}{3125}$

(ii) $\frac{17}{8}$

(iv) $\frac{15}{1600}$

(vi) $\frac{23}{2^35^2}$

(viii) $\frac{6}{15}$

(ix) $\frac{35}{50}$

Solution:

(i) $\frac{13}{3125}$
= 0.00416
(ii) $\frac{17}{8}$
= 2.125
(iv) $\frac{15}{1600}$
= 0.009375
(vi) $\frac{23}{2^35^2}$
= 0.0115
(viii) $\frac{6}{15}$
= 0.4
(ix) $\frac{35}{50}$
= 0.7

Question 3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form, $\frac{p}{q}$ what can you say about the prime factors of q?

(i) 43.123456789

(ii) 0.120120012000120000. . .

(iii) 43.123456789

Solution:

(i) 43.123456789
As this is a rational number whose decimal expansion terminates. Then it can be expressed in the form, $\frac{p}{q}$ where p and q are coprime, and the prime factorization of q is of the form 2ⁿ5^m, where n, m are non-negative integers.
= 43123456789 / 10⁹
= 43123456789 / 2⁹ × 5⁹
(ii) 0.120120012000120000…………..
As given decimal number expansion is non-terminating and non-repeating, then it is not a rational number. Then it can’t be expressed in the form, $\frac{p}{q}$ where p and q are coprime, and the prime factorization of q is of the form 2ⁿ5^m, where n, m are non-negative integers.
(iii) 43. $\overline{123456789}$ [Tex] [/Tex]
As given decimal number expansion is non-terminating and repeating, then it is a rational number. Then it can be expressed in the form, $\frac{p}{q}$ where p and q are coprime, but the prime factorization of q is not in the form of 2ⁿ5^monly, where n, m are non-negative integers