As #3/11# is already in smallest form, we have to long divide #3# by #11# and as denominator has factors other than #2# and #5#, answer will have repeating numbers after decimal point.
#color(white)(xxxxxx)0.2 color(white)(x)7color(white)(x)2color(white)(x)7#
#color(white)(xx) 11| bar(3color(white)(x).0color(white)(x) 0)color(white)(x) 0color(white)(x) 0#
#color(white)(xxxxx)ul(2color(white)("x)2)color(red)(darr#
#color(white)(xxxxxxx)8 color(white)(x) "0"#
#color(white)(xxxxxxx)ul(7 color(white)(x) 7 )"#
#color(white)(xxxxxxxxx) 3 color(white)(x) "0"#
#color(white)(xxxxxxxxx)ul(2 color(white)(x) 2 )"#
#color(white)(xxxxxxxxxxx) 8 color(white)(x) "0"#
#color(white)(xxxxxxxxxxx)ul(7 color(white)(x) 7 )"#
#color(white)(xxxxxxxxxxxxx) 3 #
As the remainder continues to be #3#, the digit #27# will repeat endlessly and hence
#3/11=0.27272727......# or #0.bar(27)#
