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

Last Updated : 03 Apr, 2024
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This exercise has been deleted as per the new NCERT syllabus

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, and 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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