Class 10 NCERT Solutions Chapter 1 Real Numbers Exercise 1.4 (original) (raw)
****(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.
****(ii) 0.120120012000120000. . .**
****(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}
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
Exercise 1.4 of NCERT Class 10 Chapter 1 - Real Numbers focuses on revisiting rational numbers and their decimal expansions, as well as introducing irrational numbers. It reinforces the concept that rational numbers have either terminating or non-terminating repeating decimal expansions, while irrational numbers have non-terminating, non-repeating decimal expansions. The exercise also covers operations with irrational numbers, proving irrationality, and understanding the properties of rational and irrational numbers on the number line.