How do you convert the recurring decimal #0.bar(32)# to a fraction?

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Answer 1 · 2018-04-24T08:37:33

#x = 32/99#

#x = 0.bar(32)#

#2# digits are recurring :
#100x = 100xx0.bar(32)#
#100x = 32.bar(32)#

#=> x = 0.bar(32)# and #100x = 32.bar(32):#

#100x - x = 32.bar(32) - 0.bar(32)#
#99x = 32#
#x = 32/99#

Answer 2 · 2018-04-24T09:04:30

#0.bar(32) = 32/99#

There is a nifty short cut method to change recurring decimals into fractions:

If all the digits recur

Write a fraction as :

#("the recurring digit(s)")/( 9" for each recurring digit")#

Then simplify if possible to get simplest form.

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

#0.272727... = 0.bar(27)= 27/99 = 3/11#

#0.bar(32) = 32/99#

#3.bar(732) = 3 732/999= 3 244/333#

If only some digits recur

Write a fraction as:

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

#0.654444... = 0.65bar4 = (654-65)/900 = 589/900#

#0.85bar(271) = (85271-85)/99900 = 85186/99900 = 42593/49950#

#4.167bar(4) = 4 (1673-167)/9000 = 4 1506/9000= 4 251/1500#