y = 0.16666. Main point by 10: 10y = 1.66666.. = 1 + 0.6666...= #1 + 2/3=5/3#.Divide by 10: #y=5/3xx1/10= 5/30=1/6#.

You are watching: 0.16 (6 repeating as a fraction)


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First main point by #10(10-1) = (100-10)# to get an integer.

The very first multiplier of #10# is to transition the decimal representation one ar to the left, so the repeating section starts just ~ the decimal point. Then the #(10-1)# multiplier is provided to transition the number one an ext place to the left (the size of the repeating pattern) and also subtract the initial to cancel the repeating tail.

#(100-10) * 0.1bar(6) = 16.bar(6) - 1.bar(6) = 15#

Then division both end 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#


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Tony B
might 2, 2016

A #underline("very slightly")# different means of writing the solution

#0.16bar6" "->" " 1/6#


Explanation:

Note: if the 6 is repeating climate a way of mirroring this is to put a bar end the last 6 friend 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#


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EZ together pi
Jun 9, 2017

Short reduced methods because that finding the fraction:


Explanation:

The details of exactly how to transform a recurring decimal right into a portion are presented in the various other answers.

However, occasionally you simply want a fast method.Here is the brief cut.

If all the number after the decimal point recur:

Write under the number (without repeating) together the numerator.

Write a #color(magenta)(9)# for each digit in the denominator. Leveling if possible.

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#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 number recur")/(larr"three 9s")#

#6.412941294129.... = 6.bar(4129) = 6 4129/9999#

If just some number recur

Numerator: create down every the number - non-recurring number Denominator: a #9# because that each recurring and a #0# for each non-recurring digit

In #0color(red)(.524)color(blue)(666...)#, just the #color(blue)(6)# recurs when 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))#