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

4 Answers

#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#.

May 2, 2016

#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#

May 2, 2016

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#

Jun 9, 2017

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#