For irrationals, the problem with infinite sequence of digit 9 does not occur. So for any given irrational (given its definition), any digit is determined by th

For irrationals, the problem with infinite sequence of digit 9 does not occur. So for any given irrational (given its definition), any digit is determined by this definition, because any irrational number can be approximated with arbitrary precision by a rational number (which has unique decimal expansion). If you think otherwise, where is the problem? This is not affected by the fact that irrationals cannot be counted. Given the irrational, a rational close enough exists which has the same decimal expansion for the first n digits, for any n.