Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 1 Real Numbers Exercise 1.4

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:—

Solution:

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

Suppose that x = \(\frac{13}{3125}\) ……..(i)

Now comparing (i) with

x = \(\frac{p}{q}\)

Here, p = 13 and q = 3125

Now prime factors of q, i.e.,

3125 = 5 × 5 × 5 × 5 × 5

= 5^{5} × 2^{0}

Which is of the form 5^{m} × 2^{n} Here m = 5 and n = 0 and these are non-negative integers. Hence the decimal expansion of

x = \(\frac{13}{3125}\) is termmatmg.

(ii) \(\frac{7}{8}\)

Suppose that x = \(\frac{7}{8}\) …..(i)

Now comparing (i) with

x = \(\frac{p}{q}\)

Here p = 17 and q = 8

Now prime factors of q, i.e.,

8 = 2 × 2 × 2 = 2^{3}

= 2^{3} × 5^{0}

Which is of the form 2^{n} × 5^{m}. Here n = 3 and m = 0 and these are non-negative integers. Hence the decimal expansion of \(\frac{17}{8}\) is terminating.

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

Suppose that x = \(\frac{64}{455}\) …(i)

Now comparing (i) with x = \(\frac{p}{q}\)

Here p = 64 and q = 455

Now prime factors of q, i.e., 455

= 5 × 7 × 13

Which is not of the form 2^{n} × 5^{m} as in the denominator these are prime factors 7 and 13 also other than 2 or 5.

Hence the decimal expansion of x = \(\frac{64}{455}\) is non-terminating repeating.

(v) \(\frac{15}{1600}\)

Suppose that x = \(\frac{15}{1600}\) …(i)

Now comparing (i) with x = \(\frac{p}{q}\)

Here p = 15 and q = 1600

Now prime factors of q, i.e.

1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

= 2^{6} × 5^{2}

Which is of the form 2^{n} × 5^{m}

Here m = 2 and n = 6 and these are non¬negative integers. Hence the decimal expansion of x = \(\frac{15}{1600}\) is terminating.

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

Suppose that x = \(\frac{29}{343}\) …..(i)

Now comparing (i) with x = \(\frac{p}{q}\)

Here p = 29 and q = 343

Now prime factors of q, i.e., 343

= 7 × 7 × 7

= 7^{3}

Which is not of the form 2^{n} × 5^{m} as in the denominator these are not the prime factors of 2 or 5.

Hence the decimal expansion of x = \(\frac{29}{343}\) is non-terminating repeating.

(vi) \(\frac{23}{2∴{3} 5∴{2}}\)

Suppose that x = \(\frac{23}{2∴{3} 5∴{2}}\)

Now comparing (i) with x = \(\frac{p}{q}\)

Here p = 23 and q = 2^{3}5^{2}

Now prime factors of q, i.e., 2^{3}5^{2} = 2^{3} × 5^{2}

Which is of the form 2^{n} × 5^{m}

Here n = 3 and m = 2 and these are non-negative integers.

Hence the decimal expansion of x = \(\frac{23}{2∴{3} 5∴{2}}\) is terminating.

(vii) \(\frac{129}{2∴{2} 5∴{7} 7∴{5}}\)

Suppose that x = \(\frac{129}{2∴{2} 5∴{7} 7∴{5}}\) …(i)

Now comparing (i) with x = \(\frac{p}{q}\)

Here p = 129 and q = 2^{2}5^{7}7^{5}

2^{2} × 5^{7} × 7^{5}

Which is not of the form 2^{n} × 5^{m} as in the denominator There factor 7 also other than 2 or 5.

Hence the decimal expansion of x = \(\frac{129}{2∴{2} 5∴{7} 7∴{5}}\) is non-terminating repeating.

(viii) x = \(\frac{6}{15}\)

Suppose that x = \(\frac{6}{15}\) = \(\frac{2}{5}\) …(i)

Now comparing (i) with x = \(\frac{p}{q}\)

Here p = 129 and q = 5

Now prime factors of q, i.e., 5

= 2^{0} × 5^{1}

Which is of the form 2^{n} × 5^{m}.

Here n = 0 and m = 1 and These are non-negative integers.

Hence the decimal expansion of x = \(\frac{6}{15}\) in terminating.

(ix) x = \(\frac{35}{50}\)

Suppose that x = \(\frac{35}{50}\) = \(\frac{7}{10}\) …. (i)

Now comparing (i) with x = \(\frac{p}{q}\)

Here p = 7 and q = 10

Now prime factors of q, i.e.,

10 = 2 × 5 = 2^{×} × 5^{×}

Which is of the form 2^{n} × 5^{m}

Here n = 1 and m = 1 and These are non-negative integers.

Hence the decimal expansion of x = \(\frac{7}{10}\) or \(\frac{35}{50}\) is terminating.

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

Suppose that x = \(\frac{77}{210}\) = \(\frac{11}{30}\) ….(i)

Now comparing (i) with x = \(\frac{p}{q}\)

Here p = 11 and q = 30

Now prime factors of q, i.e., 30 = 2 × 3 × 5

Which is not of the form 2^{n} × 5^{m} as in the denominator there is factor 3 also other than 2 or 5.

Hence the decimal expansion of x = \(\frac{77}{210}\) is non-terminating repeating.

Question 2.

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

Solution:

(To make denominator a power of 10 multiplying and dividing by 2^{5})

Hence the decimal expansion of is \(\frac{13}{3125}\) is 0.00416

(To make denominator a power of 10 multiplying and dividing by 5^{3})

Hence the decimal expansion of \(\frac{17}{8}\) is 2.125.

(iv) Suppose that x = \(\frac{15}{1600}\)

(To make denominator a power of 10 multiplying and dividing by 5^{4})

Hence the decimal expansion of \(\frac{15}{1600}\) is 0.009375.

(To make denominator a power of 10 multiplying and dividing by 5)

Hence the decimal expansion of \(\frac{23}{2∴{3} 5∴{2}}\) is 0.115.

(viii) Suppose that x = \(\frac{6}{15}=\frac{2}{5}\)

= \(\frac{2 \times 2^{1}}{2^{1} \times 5^{1}}\)

(To make denominator a power of 10 multiplying and dividing by 2^{1})

x = \(\frac{4}{(2 \times 5)^{1}}\) = \(\frac{4}{10}\)

∴ x = 0.4

Hence the decimal expansion of \(\frac{6}{15}\) is 0.4.

(ix) Suppose that x = \(\frac{35}{50}\) = \(\frac{7}{10}\)

∴ x = \(\frac{7}{2^{1} \times 5^{1}}\) = \(\frac{7}{(2 \times 5)^{1}}\) = \(\frac{7}{(10)^{1}}\)

∴ x = 0.7

Hence the decimal expansion of \(\frac{35}{50}\) is 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) \(43 . \overline{123456789}\)

Solution:

(i) Suppose that x = 43.123456789

Which, .being of the form \(\frac{p}{q}\), is a rational number. Hence the prime factors of q will be 2 or 5 or both.

(ii) Suppose that

x = 0.120120012000120000…

From, the given number it is clear that it is an irrational number because the decimal expansion of this number is non-terminating and non-recurring.

∴ It cannot be written in the form of \(\frac{p}{q}\)

∴ This number is not rational.

(iii) Suppose that x = \(43 . \overline{123456789}\)

x = 43.123456789123456789… ……(i)

Multiplying both sides by 109

10^{9} x = 43123456789.123456789 ……..(ii)

Subtracting (i) from (ii)

(10^{–} – 1)x = 43123456746

or 999999999 x = 43123456746

∴ x = \(\frac{43123456746}{999999999}\)

Which, being of the form \(\frac{p}{q}\), is a rational number.

or x = \(\frac{4791495194}{111111111}\)

x = \(\frac{4791495194}{9(12345679)}\)

Hence the prime factors of q will be other than 2 or 5. Hence the given number is rational and the prime factors of q are other than 2 or 5.

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