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

2 Answers
Mar 31, 2018

#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#

Mar 31, 2018

#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#