Vachmi

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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) $\dfrac{13}{3125}$ (ii) $\dfrac{17}{8}$ (iii) $\dfrac{64}{455}$ (iv) $\dfrac{15}{1600}$ (v) $\dfrac{29}{343}$ (vi) $\dfrac{23}{2^{3}5^{2}}$ (vii) $\dfrac{129}{2^{2}5^{7}7^{5}}$ (viii) $\dfrac{6}{15}$ (ix) $\dfrac{35}{50}$ (x) $\dfrac{77}{210}$

Solution

(i) $\dfrac{13}{3125}$ = $\dfrac{13}{5^{5}}$
The denominator is of form $5^{m}$.
Hence the decimal expansion of $\dfrac{13}{3125}$ is terminating.

(ii) $\dfrac{17}{8}$ = $\dfrac{17}{2^{3}}$
The denominator is of the form $2^{n}$.
Hence the decimal expansion of $\dfrac{17}{8}$ is terminating.

(iii) $\dfrac{64}{455}$ = $\dfrac{64}{5 × 7 × 13}$
Since the denominator is not of the form $2^m × 5^n$, the decimal expansion of $\dfrac{64}{455}$ will be non-terminating repeating.

(iv) $\dfrac{15}{1600}$ = $\dfrac{15}{2^{6} × 5^{2}}$ = $\dfrac{3\times 5}{2^{6}\times 5^{2}}$
The denominator is of the form $2^m × 5^n$, the decimal expansion of $\dfrac{64}{455}$ is terminating.

(v) $\dfrac{29}{343}$ = $\dfrac{29}{7^{3}}$
Since the denominator is not of the form $2^m × 5^n$, the decimal expansion of $\dfrac{29}{343}$ will be non-terminating repeating.

(vi) $\dfrac{23}{2^{3} × 5^{2}}$
The denominator is of the form $2^m × 5^n$, the decimal expansion of $\dfrac{23}{2^{3} × 5^{2}}$ is terminating.

(vii) $\dfrac{129}{2^{2} × 5^{7} × 7^{5}}$
Since the denominator is not of the form $2^m × 5^n$, the decimal expansion of $\dfrac{129}{2^{2} × 5^{7} × 7^{5}}$ will be non-terminating repeating.

(viii) $\dfrac{6}{15}$ = $\dfrac{2 × 3}{3 × 5}$ = $\dfrac{2}{5}$
The denominator is of the form $5^{m}$.
Hence the decimal expansion of $\dfrac{6}{15}$ is terminating.

(ix) $\dfrac{35}{50}$ = $\dfrac{5 × 7}{2 × 5 × 5}$ = $\dfrac{7}{2 × 5}$
The denominator is of the form $2^m × 5^n$, the decimal expansion of $\dfrac{35}{50}$ is terminating.

(x) $\dfrac{77}{210}$ = $\dfrac{7 × 11}{2 × 3 × 5 × 7}$
Since the denominator is not of the form $2^m × 5^n$, the decimal expansion of $\dfrac{77}{210}$ will be non-terminating repeating.

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2

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

Solution

(i) $\dfrac{13}{3125}$ = $0.00416$

(ii) $\dfrac{17}{8}$ = $2.125$

(iv) $\dfrac{15}{1600}$ = $0.009375$

(vi) $\dfrac{23}{2^{3} × 5^{2}}$ = $\dfrac{23}{200}$ = $0.115$

(viii) $\dfrac{6}{15}$ = $\dfrac{2 × 3}{3 × 5}$ = $\dfrac{2}{5}$ = $0.4$

(ix) $\dfrac{35}{50}$ = $\dfrac{5 × 7}{2 × 5 × 5}$ = $\dfrac{7}{2 × 5}$ = $0.7$

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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 $\dfrac{p}{q}$, what can you say about the prime factor of q?
(i) $43.123456789$ (ii) $0.120120012000120000...$ (iii) $43.\overline{123456789}$

Solution

(i) $43.123456789$
Since this number has a terminating decimal expansion, it is a rational number of the form $\dfrac{p}{q}$ and q is of the form $2^m5^n$
So the prime factors of q will be either 2 or 5 or both.

(ii) $0.120120012000120000...$
The decimal expansion is neither terminating or recurring. So the given number is an irrational number.

(iii) $43.\overline{123456789}$
Since this number is non-terminating recurring, the given number is a rational number of the form $\dfrac{p}{q}$ and q is not of the form $2^m5^n$
So the prime factors of q will also have factor or factors other than 2 or 5.