There is a technique for changing recurring decimals to fractions.
Let #x = 0.21919191919....... # to infinity
#10x = 2.19191919191919...#to infinity.
#1000x = 219.1919191919 # to infinity
Do you see what will happen if we subtract #1000x -10x?#
#1000x = 219.1919191919 # to infinity
#ul(-10x)" " ul(-2.19191919191919...)#to infinity.
#990x = 217.0000000000000...# to infinity
Solve for #x#
#x = 217/990#
Notice the following:
Not all the decimals recur.
The recurring starts after 3 decimal places. This was the reason for multiplying by #1000#. The digits must line up so that when you subtract there will be zero's to infinity.
This method can be summarised as follows:
#x = (219-2)990 = 217/990#
#291: " " # write down all the digits until they start to recur.
#-2: " "#subtract the digits that do NOT recur
#div 990:" "# divide by a 9 for each recurring digit followed by a 0 for each non-recurring digit.
