Question

​What can you say about the prime factorisations of the denominators of the following rationals:
$\left(i\right)43.123456789$
$\left(ii\right)43.\overline{123456789}$
$\left(iii\right)27.\overline{142857}$
$\left(iv\right)\mathrm{0.120120012000120000....}$

Answer 1

(i) Since, 43.123456789 has terminating decimal expansion.

$=\frac{43.123456789}{{10}^9}$

$=\frac{43.123456789}{2^9\times 5^9}$

as $43.123456789$ is in form of $\frac{p}{q}$ and $q$ is of form of $2^{m}and{5}^n$.

Hence, prime factors of $q$ will be $2and5$ .

(ii) Since, $43.\overline{123456789}$ is not terminating, but repeating.

$=\frac{43.\overline{123456789}}{{10}^9}$

$=\frac{43.\overline{123456789}}{2^9\times 5^9}$

as $43.\overline{123456789}$ is in the form of $\frac{p}{q}$ and $q$ is of form of $2^{n}and{5}^m$.

Hence, prime factors of $q$ will be $2and5$ .
(iii) Since, $27.\overline{142857}$ is not terminating, but repeating.

$=\frac{27.\overline{142857}}{{10}^6}$

$=\frac{27.\overline{142857}}{2^6\times 5^6}$

as, $27.\overline{142857}$ is in form of $\frac{p}{q}$ , and $q$ is of form $2^{m}and{5}^n$

Hence, prime factors of $q$ will be $2and5$.
(iv) Since, $\mathrm{0.120120012000120000....}$ as this is not terminating and non repeating.

So, it is not a rational number.