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^{3} 5^{2}} (vii) \frac{129}{2^{2} 5^{7} 7^{5}} (viii) \frac{6}{15} (ix) \frac{35}{50} (x) \frac{77}{210}

Foundation

Mathematics Foundation

Properties of Real Number

Recurring Decimal Number

Rational Numbers

Question

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^{3} 5^{2}}

(vii)

\frac{129}{2^{2} 5^{7} 7^{5}}

(viii)

\frac{6}{15}

(ix)

\frac{35}{50}

(x)

\frac{77}{210}

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Theorem: Let

x = \frac{p}{q}

be a rational number, such that the prime factorisation of q is of the form

2^{n} 5^{m}

, where n, m are non-negative integers. Then,

x

has a decimal expansion which terminates.

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

Factorise the denominator, we get

3125 = 5 \times 5, \times 5, \times 5, \times 5 = 5^{5}

So, denominator is in form of

5^{m}

so,

\frac{13}{3125}

is terminating.

(ii) $\frac{17}{8}$

Factorise the denominator, we get

8 = 2, \times 2, \times 2 = 2^{3}

So, denominator is in form of

2^{n}

so,

\frac{17}{8}

is terminating.

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

Factorise the denominator, we get

455 = 5, \times 7, \times 13

So, denominator is not in form of

2^{n} 5^{m}

so,

\frac{64}{455}

is not terminating.

(iv) $\frac{15}{1600}$

Factorise the denominator, we get

1600 = 2, \times 2, \times 2, \times 2, \times 2, \times 2, \times 5, \times 5 = 2^{6} 5^{2}

So, denominator is in form of

2^{n} 5^{m}

so,

\frac{15}{1600}

is terminating.

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

Factorise the denominator, we get

343 = 7, \times 7, \times 7 = 7^{3}

So, denominator is not in form of

2^{n} 5^{m}

so,

\frac{29}{343}

is not terminating.

(vi) $\frac{23}{2^{3} 5^{2}}$

Here, the denominator is in form of

2^{n} 5^{m}

so,

\frac{23}{2^{3} 5^{2}}

is terminating.

(vii) $\frac{129}{2^{2} 5^{7} 7^{5}}$

Here, the denominator is not in form of

2^{n} 5^{m}

so,

\frac{129}{2^{2} 5^{7} 7^{5}}

is not terminating.

(viii) $\frac{6}{15}$

Divide nominator and denominator both by 3 we get

\frac{3}{15}

So, denominator is in form of

5^{m}

so,

\frac{6}{15}

is terminating.

(ix) $\frac{35}{50}$

Divide nominator and denominator both by 5 we get

\frac{7}{10}

Factorise the denominator, we get

10 = 2 \times 5

So, denominator is in form of

2^{n} 5^{m}

so,

\frac{35}{50}

is terminating.

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

Divide nominator and denominator both by 7 we get

\frac{11}{30}

Factorise the denominator, we get

30 = 2 \times 3 \times 5

So, denominator is not in the form of

2^{n} 5^{m}

\frac{6}{15}

is not terminating.