Can a repeating decimal be equal to an integer?

Jun 11, 2015

No, it will allways turn out to be a fraction.

Explanation:

I will not delve into how you turn a repeating decimal into a fraction, but just one example:

$0.333 \ldots . = \frac{1}{3}$

There is one exeption though (see example above):

$0.999 \ldots . = 3 \cdot 0.333 \ldots . = 3 \cdot \frac{1}{3} = 1$

Dec 18, 2016

Yes

Explanation:

The general term of a geometric series can be written:

${a}_{n} = a \cdot {r}^{n - 1}$

where $a$ is the initial term and $r$ the common ratio.

When $\left\mid r \right\mid < 1$ then its sum to infinity converges and is given by the formula:

${\sum}_{n = 1}^{\infty} a {r}^{n - 1} = \frac{a}{1 - r}$

So for example:

$0.999 \ldots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \ldots$

is given by $a = 9$ and $r = \frac{1}{10}$

which has sum:

${\sum}_{n = 1}^{\infty} \frac{9}{10} \cdot {\left(\frac{1}{10}\right)}^{n - 1} = \frac{\frac{9}{10}}{1 - \frac{1}{10}} = \frac{\frac{9}{10}}{\frac{9}{10}} = 1$

So $0. \overline{9} = 0.999 \ldots = 1$

In fact, any integer can be expressed as a repeating decimal using $9$'s.

For example:

$12345 = 12344.999 \ldots = 12344. \overline{9}$

$- 5 = - 4.999 \ldots = - 4. \overline{9}$