Express the following in the form "p/q", where p and q are integers and q != 0. (i) 0.6666... (ii) 0.4777... (iii) 0.001...

Express the following in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

(i) $0 . \overline{6}$

(ii) $0.4 \overline{7}$

(iii) $0 . \overline{001}$

To express these repeating decimals in the form $\frac{p}{q}$, we use the following technique for each:

1. For $0.\overline{6}$:

Let $x = 0.\overline{6}$.

This implies $10x = 6.\overline{6}$.

Subtracting these two equations gives: $9x = 6$.

Thus, $x = \frac{6}{9}$ which simplifies to $\frac{2}{3}$.

2. For $0.4\overline{7}$:

Let $y = 0.4\overline{7}$.

This implies $100y = 47.\overline{7}$.

Subtracting the first from the second (after multiplying the first by 10) gives: $90y = 47.3$.

Thus, $y = \frac{47.3}{90}$.

But we'll need to work this out precisely.

3. For $0.\overline{001}$:

Let $z = 0.\overline{001}$.

This means $1000z = 1.\overline{001}$.

Subtracting the first equation from the second gives: $999z = 1$.

Thus, $z = \frac{1}{999}$.

Let's now compute the precise form of $0.4\overline{7}$ using algebraic manipulation. For the given repeating decimals, they can be expressed in the form $\frac{p}{q}$ as follows:

(i) $0.\overline{6} = \frac{2}{3}$

(ii) $0.4\overline{7} = \frac{493}{900}$

(iii) $0.\overline{001} = \frac{1}{999}$