[Math] Is Cantor’s diagonal argument dependent on the base used

binarydiscrete mathematicselementary-set-theoryfractionsreal numbers

Applying Cantor's diagonal argument to irrational numbers represented in binary, one and only one irrational number can be generated that is not on the list.

Wikipedia image:


But if you change the base from 2 to 3 or higher, including base-10, there are an infinite number of irrational numbers that can be generated that are not on the list, through various combinations of digits at tenths, hundredths, thousandths, etc places.

How can this be possible? Bases are just representations of a number, not a number in themselves. But binary produces fewer number of uncountable irrationals than ternary or quaternary.

Please correct me if I've gone wrong somewhere.

Best Answer

Cantor's diagonal argument proves (in any base, with some care) that any list of reals between $0$ and $1$ (or any other bounds, or no bounds at all) misses at least one real number. It does not mean that only one real is missing. In fact, any list of reals misses almost all reals. Cantor's argument is not meant to be a machine that produces reals not in your list. It's an argument by contradiction to show that the cardinality of the reals (or reals bounded between some two reals) is strictly larger than countable. It does so by exhibiting one real not in a purported list of all reals. The base does not matter. The number produced by cantor's argument depends on the order of the list, and the base chosen. In base $2$, you have no freedom in choosing the digits, so each ordering of the list produces in this way one real guaranteed not to be in the list. In other bases you have more freedom in choosing the digits, but this is irrelevant.