Question

How do you convert `0.3bar4` (4 repeating) to a fraction?

Answer 1

`31/90`

Explanation:

`"we require 2 equations with the repeating part 4 being"`
`"after the decimal point"`

`bar4-=" the repeating 4"`

`"let "x=0.3bar4`

`rArr10x=3.bar4to(1)`

`rArr100x=34.bar4to(2)`

`"subtracting "(1)" from "(2)" eliminates the repeated 4"`

`rArr100x-10x=34.bar4-3.bar4`

`rArr90x=31rArrx=31/90`

Answer 2

`31/90`

Explanation:

While the answer can be worked out by a full process as explained by Jim G, there is a useful short cut which is quick to use.

If all the decimals recur:

Write the fraction as: `"the recurring digits"/(9 " for each digit"`

eg: `0.676767... = 0.bar(67) = 67/99" "larr` there are 2 recurring digits

eg: `7.394394394... = 7.bar(394) = 7 394/999" "larr` 3 recurring digits

If only some of the decimals recur

Write the fraction as:

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

eg: `0.23555... = 0.23bar5 = (235-23)/900 = 212/900`

eg: #3.4678678.. = 3.4bar(678) =3 (4678-4)/9990 = 3 4674/9990#

eg: `9.461565656.. = 9.461bar(56) = 9 (46156-461)/99000 = 45695/99000`

In your case we have: `0.3bar4`

The fraction is `(34-3)/90 = 31/90`