Question

Without actual division, show that each of the following rational numbers is a nonterminating repeating decimal.

$\left(i\right)\frac{11}{(2^3\times 3)}\left(ii\right)\frac{73}{(2^2\times 3^3\times 5)}\left(iii\right)\frac{129}{(2^2\times 5^3\times 7^2)}\left(iv\right)\frac{9}{35}\left(v\right)\frac{77}{210}\left(vi\right)\frac{32}{147}\left(vii\right)\frac{29}{343}\left(viii\right)\frac{64}{455}$

Answer 1

(i) $\frac{11}{2^3\times 3}$

We know either 2 or 3 is not a factor of 11, so it is in its simplest form.

Moreover, ($2^3\times 3)\ne (2^m\times 5^n$)

Hence, the given rational is non-terminating repeating decimal.

(ii) We know 2, 3 or 5 is not a factor of 73, so it is in its simplest form.

Moreover, ($2^2\times 3^3\times 5)\ne (2^m\times 5^n$)

Hence, the given rational is non-terminating repeating decimal.

(iii) We know 2, 5 or 7 is not a factor of 129, so it is in its simplest form.

Moreover, ($2^3\times 5^7\times 7^5)\ne (2^m\times 5^n$)

Hence, the given rational is non-terminating repeating decimal.

(iv) $\frac{9}{35}$ = $\frac{9}{7\times 5}$

We know either 5 or 7 is not a factor of 9, so it is in its simplest form. Moreover, ($5\times 7)\ne (2^m\times 5^n$)

Hence, the given rational is non-terminating repeating decimal.

(v) $\frac{77}{210}$ = $\frac{77}{7\times 30}$ = $\frac{11}{30}$

$\frac{11}{2\times 3\times 5}$

We know 2, 3 or 5 is not a factor of 11, so $\frac{11}{30}$ is in its simplest form.

Moreover, ($2\times 3\times 5)\ne (2^m\times 5^n$)

Hence, the given rational is non-terminating repeating decimal.

(vi) We know either 3 or 7 is not a factor of 32, so it is in its simplest form.

Moreover, $(3\times 7^2)\ne (2^m\times 5^n)$

Hence, the given rational is non-terminating repeating decimal.

(vii) We know 7 is not a factor of 29, so it is in its simplest form.

Moreover, $7^3\ne (2^m\times 5^n)$

Hence, the given rational is non-terminating repeating decimal.

(viii)$\frac{64}{455}=\frac{64}{5\times 7\times 13}$

We know 5, 7 or 13 is not a factor of 64, so it is in its simplest form.

Moreover, ($5\times 7\times 13)\ne (2^m\times 5^n)$

Hence, the given rational is non-terminating repeating decimal.