Question

How do I convert `-0.7bar3` to a fraction?

Answer 1

`-0.73bar3 = -11/15`

The bar over the last three means that it is repeated for ever.

Explanation:

First and foremost we must not forget that this is negative. They will be watching to see if you overlook that point.

set `x=-0.733333......`

So `10x=-7.3333..." "larr` one digit recurs, multiply by `10`

and `100x=-73.3333...`

`100x-10x=-73.333333...`
`color(white)("dddddddddddd") -ul(color(white)("d")7.333333...larr" Subtract" )`
`color(white)("dddddddddddd")-66.000000....`

`90x=-66`

#x=-66/90color(white)("d")->color(white)("d")-[ (66-:6)/(90-:6)]= - 11/15#

Answer 2

`-0.7bar(3) = -11/15`

Explanation:

Here's one method...

Given:

`-0.7bar(3)`

Note that `0.bar(3) = (0.bar(9))/3 = 1/3`.

So let's try subtracting that from `0.7bar(3)`:

`0.7bar(3) = 0.bar(3)+0.4 = 1/3+4/10 = 5/15+6/15 = 11/15`

So:

`-0.7bar(3) = -11/15`

Essentially, if you see a recurring decimal with a repeating tail that you recognise, you can try subtracting the fraction you know, then make sense of the remaining terminating decimal.

Answer 3

There are short cut rules you can apply.

Explanation:

The full method of converting recurring decimals to fractions is given in another answer. However, there are two short cut rules which are easy to learn and apply.

If all the digits recur

Write a fraction as follows:

`("the recurring digits")/("a 9 for each digit")" "larr` then simplify if possible

`0.55555.... =0.bar5 = 5/9`

`0.88888... = 0.bar8 = 8/9`

`5.737373.. = 5.bar(73) = 5 73/99`

`14.657657657... = 14.bar(657) = 657/999 = 73/111`

If only some of the digits recur

Write a fraction as:

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

`0.43777.... = 0.43bar7 = (437-43)/900 = 394/900 = 197/450`

`7.419191919.. = 7.4bar(19) = 7 (419-4)/990 = 7 83/198`

`17. 357616161... = 17.357bar(61) = 17 (35761-357)/99000 = 17 35404/99000 = 17 8851/24750`