Question

How do you convert 0.916 (6 repeating) to a fraction?

Answer 1

`0.91bar(6) = 11/12`

Explanation:

In case you have not encountered it, you can indicate a repeating sequence of digits in a decimal expansion by placing a bar over it.

So:

`0.91666... = 0.91bar(6)`

`color(white)()`
Method 1

Multiply by `100(10-1) = 1000-100` to get an integer:

`(1000-100) 0.91bar(6) = 916.bar(6) - 91.bar(6) = 825`

Divide both ends by `1000-100` to find:

`0.91bar(6) = 825/(1000-100) = 825/900 = (color(red)(cancel(color(black)(75)))*11)/(color(red)(cancel(color(black)(75)))*12) = 11/12`

Why `100(10-1)` ?

The factor `100` shifts the given number two places left, leaving the repeating section starting just after the decimal point. The factor `(10-1)` shifts the number a further `1` place - the length of the repeating pattern - then subtracts the original to cancel out the repeating tail.

`color(white)()`
Method 2

Given:

`0.91bar(6)`

Recognise the repeating `6` tail as the result of dividing by `3`, so multiply by `color(blue)(3)` to find:

`color(blue)(3) * 0.91bar(6) = 2.75`

Notice that `2.75` ends with a `5`, so we can attempt to simplify the decimal by multiplying by `color(blue)(2)`:

`color(blue)(2) * 2.75 = 5.5`

Notice that `5.5` ends with a `5`, so we can attempt to simplify by multiplying by `color(blue)(2)` again:

`color(blue)(2) * 5.5 = 11`

Having arrived at an integer, we can divide by the numbers we multiplied by to get a fraction:

`0.91bar(6) = 11/(2*2*3) = 11/12`

Answer 2

`0.916bar6=11/12`

Explanation:

Given `0.9166666...` written as `0.91bar6`

Let `x=0.91bar6`

Then `100x=91.6bar6`

Also `1000x=916.6bar6`

So:

`1000x-100x=916.6bar6`
`" "ul(color(white)(9)91.6bar6)larr" subtract"`
`" "825.0`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
But `1000x-100x` is the same as `900x`

`=>900x=825`

Divide both sides by 900

`x=825/900=11/12`