Irrationals have a unique decimal representation in the mathematical sense: given a definition such as $x^2 = 2 $, any digit of the decimal expansion of $x$ can be determined.
> Irrationals have a unique decimal representation in the mathematical sense
Not all of them do. Actually, so many don't that mathematically, the number of them that do is zero.
Sure, there are exceptions like sqrt(2) and sqrt(3), but there is an uncountable infinity of irrationals between these two numbers that just don't have a representation.
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.
