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) $\frac{77}{210}$

Answer 1

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 $2^n{5}^m$ so $\frac{6}{15}$ is not terminating.