Question

How do we convert `0.4bar(82)` i.e. `0.4828282........` (`82` repeating endlessly) into a fraction?

Answer 1

`239/495`

Explanation:

For numbers with repeating decimals, there is a fairly simple way of algebraically retrieving their respective fractions:

Start by setting the value with repeating decimals equal to `x`:

`0.4bar 82=x`

The next step is to cancel out the repeating decimals to make the remaining values whole, and therefore writable as fractions.

Start by multiplying your value by a factor of 10 to place a whole set of the repeating part of the decimal on the left side of the decimal:

`1000x=482.bar82`

Then, create a similar value by multiplying by a factor of 10 to place the first set of the repeating part of the decimal on the right side of the decimal:

`10x=4.bar82`

Both of the right-side values now have never-ending decimals of `.bar82`, so if you subtract them they should give a whole number:

`1000x-10x=482.bar82-4.bar82`

Which simplifies to:

`990x=478`

Now to solve for x, divide both sides by 990:

`(990x)/990=478/990`

Which simplifies to:

`x=478/990=239/495`

Where `x` maintains its value of `0.4bar82`.

Answer 2

`(482-4)/990 = 478/990 = 239/495`

`239/495 = 0.48248482...`

Explanation:

The full method is explained elsewhere, but here is the short version:

If all the decimal digits recur....

Fraction = `("the recurring digits")/("a 9 for each digit")`

If only some of the digits recur:

Fraction: `("all the digits - non-recurring digits")/("9 for each recurring and 0 for each non-recurring digit")`

eg: `" "0.43434343.... = 43/99`

`" "0.157157157.... = 157/999`

eg: `" "0.347474747... = (347-3)/990 =344/990`

`" "0.625555.... = (625-62)/900 =563/900`

In this case you have `0.4828282...`

`(482-4)/990 = 478/990 = 239/495`