##0.0bar(9) = 1/10## or ##0.bar(09) = 1/11##
We will be using the notation that ##0.099999… = 0.0bar(9)## This avoids any ambiguity as to which digits are repeating.
There are lots of ways to show that ##0.bar(9) = 1##. Once we have that, then we simply divide by sides by 10, and we are done.
##0.0bar(9) = 1/10*0.bar(9) = 1/10*1 = 1/10##
. . .
Now, in this case we used our prior knowledge that ##0.bar(9)=1## to show what we wanted, however, what if there is a different repeating decimal? For example, how would we find the fraction representation of ##0.bar(21)##? Here’s one way:
Let ##x = 0.bar(21)##
Then, multiply each side by ##100## (note that this is because there are ##2## digits in ##21## and ##100 = 10^2##).
##100x = 21.bar(21)##
Note that despite shifting the decimal two places, because there were infinite ##21##s after the decimal originally, there are still infinite ##21##s after the decimal.
Now, subtract ##x## from both sides, noting that ##x = 0.bar(21)##
##100x – x = 21.bar(21) – 0.bar(21)##
##=> 99x = 21##
##=> x = 21/99 = 7/33##
This trick will work for any repeating decimal, so long as you multiply by ##10^k## where ##k## is the number of digits that repeat. The idea is to shift the decimal without changing the value afterwards, thereby allowing the subtraction trick shown above. This trick, incidentally, is one of the methods shown in the above link demonstrating that ##0.bar(9) = 1##.
As a final note, if the initial question was referring to ##0.bar(09)## and not ##0.0bar(9)## then applying the same trick we have
##x = 0.bar(09)##
##=> 100x = 9.bar(09)##
##=> 100x – x = 9.bar(09) – 0.bar(09)##
##=> 99x = 9##
##=> x = 9/99 = 1/11##
