Question

How do you find the repeating decimal 0.82 with 82 repeated as a fraction?

Answer 1

`0.bar(82) = 82/99`

Explanation:

In case you have not encountered it, let me introduce you to some notation:

A repeating decimal can be represented using a bar placed over the repeating pattern of decimals.

So instead of writing `0.828282...`, you can write `0.bar(82)`

If we multiply `0.bar(82)` by `(100-1)` we get an integer:

`(100-1) 0.bar(82) = 100 * 0.bar(82) - 1 * 0.bar(82)`

`color(white)((100-1) 0.bar(82)) = 82.bar(82) - 0.bar(82)`

`color(white)((100-1) 0.bar(82)) = 82`

Notice that the `100` shifts the number two places to the left - the length of the repeating pattern. Then the `-1` cancels out the repeating tail.

Next divide both ends by `(100-1)` and simplify:

`0.bar(82) = 82/(100-1) = 82/99`

`82` and `99` have no common factor, so this is in simplest terms.

`color(white)()`
Alternative method

An alternative method recognises that:

`0.bar(82) = 0.82 + 0.0082 + 0.000082 +...`

is a geometric series, with initial term `a = 0.82` and common ratio `r = 1/100`.

This has sum given by the formula:

`s_oo = a/(1-r) = 0.82/(1-1/100) = 0.82/(99/100) = (0.82*100)/99 = 82/99`