Multiply by #10(100-1) = 1000-10# to get an integer:
#(1000-10)0.0bar(65)=65.bar(65) - 0.bar(65)=65#
Then divide both ends by #(1000-10)#:
#0.0bar(65) = 65/(1000-10) = 65/990 = (color(red)(cancel(color(black)(5)))*13)/(color(red)(cancel(color(black)(5)))*198) = 13/198#
The factor of #10# in our multiplier is chosen to shift the original number until the repeating section starts just after the decimal point.
The factor of #(100-1)# shifts the number a further #2# places to the left and subtracts the original, in order to cancel out the #2# digit repeating pattern tail, leaving an integer.
Once we get to a division of integers in the form #65/990# we can then simplify by cancelling out any common factors. In more difficult examples you might take the trouble to factor the numerator and denominator completely into prime factors to check for common factors, but in our example here it's fairly easy to spot that the only common factor is #5#.
