Question #de166

Sep 2, 2016

$0. \overline{3} = \frac{1}{3}$

Explanation:

Using the notation that a bar over a digit or string of digits denotes them repeating infinitely, let $x = 0. \overline{3}$

$\implies 10 x = 3. \overline{3}$

$\implies 10 x - x = 3. \overline{3} - 0. \overline{3}$

$\implies 9 x = 3$

$\therefore x = \frac{3}{9} = \frac{1}{3}$

This technique will actually work for converting any repeating decimal into a fraction. Suppose we have a repeating decimal $x = 0. \overline{{a}_{1} {a}_{2.} . . {a}_{n}}$ (that is, with $n$ repeating digits). We multiply $x$ by ${10}^{n}$, subtract $x$, and then divide by ${10}^{n} - 1$ to obtain the fractional form:

${10}^{n} x = {a}_{1} {a}_{2.} . . {a}_{n} . \overline{{a}_{1} {a}_{2.} . . {a}_{n}}$

$\implies {10}^{n} x - x = {a}_{1} {a}_{2.} . . {a}_{n} . \overline{{a}_{2} {a}_{2.} . . {a}_{n}} - 0. \overline{{a}_{1} {a}_{2.} . . {a}_{n}}$

$\implies x \left({10}^{n} - 1\right) = {a}_{1} {a}_{2.} . . {a}_{n}$

$\therefore x = \frac{{a}_{1} {a}_{2.} . . {a}_{n}}{{10}^{n} - 1}$

Note that once we know the technique, we can skip to the end as a shortcut: $0. \overline{{a}_{1} {a}_{2.} . . {a}_{n}} = \frac{{a}_{1} {a}_{2.} . . {a}_{n}}{{10}^{n} - 1}$. In the given problem, only a single digit, $3$, is repeating, so the shortcut gives us: $0. \overline{3} = \frac{3}{{10}^{1} - 1} = \frac{3}{9} = \frac{1}{3}$.

We can also modify the technique to handle cases where the repeating portion does not begin until after a finite number of non-repeating digits, such as numbers like $0.1232323 \ldots = 0.1 \overline{23}$. The idea is still to multiply by an appropriate power of $10$ and then subtracting away the repeating portion. As an exercise, a student can try finding the fractional form of such a number.