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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 2n5m, 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 2n5m, 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 2n5m 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 2n5m 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 2n5m 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 2n5m 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 2n5m only.

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 2n5m 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 2n5m 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 2n5m 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 2n5m 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 2n5m 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 2n5m, 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 2n5m, 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 2n5m only, where n, m are non-negative integers

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