Can a repeating decimal be equal to an integer?

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Answer 1 · 2015-06-11T14:06:50

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

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

There is one exeption though (see example above):

Answer 2 · 2016-12-18T09:20:16

Yes

The general term of a geometric series can be written:

$a_{n} = a \cdot r^{n - 1}$ $a_n = a*r^(n-1)$

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

When $| r | < 1$ $abs(r) < 1$ then its sum to infinity converges and is given by the formula:

$\infty \sum n = 1 a r^{n - 1} = \frac{a}{1 - r}$ $sum_(n=1)^oo ar^(n-1) = a/(1-r)$

So for example:

$0.999... = 9/10+9/100+9/1000+...$

is given by $a=9$ and $r=1/10$

which has sum:

$sum_(n=1)^oo 9/10*(1/10)^(n-1) = (9/10)/(1-1/10) = (9/10)/(9/10) = 1$

So $0.bar(9) = 0.999... = 1$

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

For example:

$12345 = 12344.999... = 12344.bar(9)$

$-5 = -4.999... = -4.bar(9)$