Question

How do you convert 0.16 (6 repeating) to a fraction?

Answer 1

`1/6`

Explanation:

y = 0.16666.
Multiply by 10: 10y = 1.66666.. = 1 + 0.6666...= `1 + 2/3=5/3`.
Divide by 10: `y=5/3xx1/10= 5/30=1/6`.

Answer 2

`0.1bar(6) = 1/6`

Explanation:

First multiply by `10(10-1) = (100-10)` to get an integer.

The first multiplier of `10` is to shift the decimal representation one place to the left, so the repeating section begins just after the decimal point. Then the `(10-1)` multiplier is used to shift the digits one more place to the left (the length of the repeating pattern) and subtract the original to cancel the repeating tail.

`(100-10) * 0.1bar(6) = 16.bar(6) - 1.bar(6) = 15`

Then divide both ends by `(100-10)` and simplify:

`0.1bar(6) = 15/(100-10) = 15/90 = (1*color(red)(cancel(color(black)(15))))/(6*color(red)(cancel(color(black)(15)))) = 1/6`

Answer 3

A `underline("very slightly")` different way of writing the solution

`0.16bar6" "->" " 1/6`

Explanation:

Note: if the 6 is repeating then a way of showing this is to put a bar over the last 6 you write: `->" "0.16bar6`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let `x=0.16bar6`

Then `10x=1.6bar6`
Also `100x=16.6bar6`

`100x-10x->" "16.6bar6`
`color(white)(ggggggg2222222)underline(color(white)(.m)1.6bar6)" Subtract"`
`color(white)(222vvvvvvvvv22)underline(" "15.00" ")" "`

So `90x =15`

Divide both sides by 90
`" "x=15/90`

But `15/90->(15-:15)/(90-:15) = 1/6`

`=>x=0.16bar6" "->" " 1/6`

Answer 4

Short cut methods for finding the fraction:

Explanation:

The details of how to convert a recurring decimal into a fraction are shown in the other answers.

However, sometimes you just want a quick method.
Here is the short cut.

If all the digits after the decimal point recur:

Write down the digits (without repeating) as the numerator.

Write a `color(magenta)(9)` for each digit in the denominator. Simplify if possible.

`0.77777.. = 0.barcolor(magenta)(7) = color(magenta)(7/9)" "(larr"one digit recurs")/(larr"one 9")`

`0.613613613... = 0.bar(color(magenta)(613)) = color(magenta)(613/999)" "(larr"three digits recur")/(larr"three 9s")`

`6.412941294129.... = 6.bar(4129) = 6 4129/9999`

If only some digits recur

Numerator: write down all the digits - non-recurring digits
Denominator: a `9` for each recurring and a `0` for each non-recurring digit

In `0color(red)(.524)color(blue)(666...)`, only the `color(blue)(6)` recurs while the `color(red)(524)` do not.

`0color(red)(.524)color(blue)(666...) = 0color(red)(.524)color(blue)(bar6) = (5246-color(red)(524))/(color(blue)(9)color(red)(000)) = 4722/(color(blue)(9)color(red)(000))`

`0.1353535... = 0.1bar(35) = (135-1)/990 = 134/990`

`4.23861861861.. = 4.23bar(861) = 4 (23861-23)/99900 =23838/99900`