# Irrationals cannot behave as rationals in this sense, right?

Just thinking about decimal (or more general) representations of rationals and irrationals and we know that it can happen that some number can have two different decimal representations but all the examples that I know of are rationals.

Basically, digit-patterns in irrationals seem to prevent such thing from happening to them, so, if true, we would have another necessary and sufficient condition for rationality/irrationality, that is that the number is rational if and only if it can be written in two different ways in at least one base (maybe in all of them, I am tooo depressive now to say anything smart enough).

Not an elegant and mysterious necessary and sufficient condition but it does reveal something to us.

If you have a proof that irrationals cannot have two different decimal (or more general) representations then you can prove that to us here, if you do not have the proof then your opinion also qualifies as an answer.

Sorry for this bad style of a question, I really do not feel good at the moment.

Truly yours, Georg Cantor.

## 1 Answer

The only way for there to be ambiguity is for a number whose decimal expansion ends in an infinite tail of 9s, which can then also be represented as a number with an infinite tail of 0s. In this case the resulting number is necessarily rational (in the former case because an eventually periodic decimal representation exists, in the latter case because an infinite decimal representation exists).

In fact since a finite expansion exists, it is even a "decimal" number, i.e. a rational number which can be written with a power of $10$ as its denominator.

• In either case, the representation is eventually periodic; so a rational number. – Lubin Oct 17 '17 at 15:20
• It is even a decimal number. – Bernard Oct 17 '17 at 15:21