# Is it irrational?

Suppose I generate a number $0 < x < 1$. In general, after the decimal point, the first digit is $1$, the second is $0$, the third is $1$, etc. However, every digit in position $n$ has a $1/n$ chance of being $2$ instead. The tenths place, for example, must have a $2$. The number might look like this:

$$x = 0.221210101020101010101010101010121010101010101010101010101010...$$

(There are an infinite number of possible values of $x$. My question applies to any number in this set, excluding numbers where the digits begin to repeat interminably at a finite point, which has 0 probability of happening. Ex. not interested in 0.222222...)

Such a number doesn't seem to be rational, because $2$ will continue to pop up, even as it becomes less frequent. However, as the position $n$ of a digit approaches infinity, the probability of that digit being $2$ approaches $0$. It seems that as $n$ goes to infinity (and there are an infinite number of digits), the digits approach rational behavior, for lack of a better phrase.

Would this number be irrational?

Clarification: What I am asking is whether a number would be considered irrational if its digits "approach repetition"; if they could be said to repeat interminably at infinity, but not at a finite point.

-
I think you should rephrase the Question to make it clearer that you are asking about the probability that this random variable will be a rational number. – hardmath Feb 19 at 13:07
This number could be rational and it could not. You didn't describe a single number. Your question is like saying "I generate my number by rolling a 6 sided die. Is the number even?". Well... it could be even, yes. – 5xum Feb 19 at 13:26
What does “considered irrational” mean? – Carsten S Feb 20 at 13:34
@CarstenS simply "be irrational". What else would you think? – user236182 Feb 20 at 13:35
@user236182, I don't know. It might be that Kanaya's concept of an irrational number is fuzzy. – Carsten S Feb 20 at 13:39

## 4 Answers

If you get infinitely many 2s, but with arbitrarily large gaps between them, then your number is irrational. This happens with probability one. (See the Borel-Cantelli theorem. I'm assuming your events with probability 1/n are independent, you didn't say that...)

But note if you use probabilities $1/n^2$ instead, then you get (with probability one) finitely many 2s, so your number is rational.

-
As I said, en.wikipedia.org/wiki/Borel–Cantelli_lemma/ – GEdgar Feb 19 at 14:00
This is the best answer, I think, because all of the others seem to imply that if there were probability $\frac1{n^2}$ for each it would also be irrational with probability $1$, or even not talk about probabilities at all. – Akiva Weinberger Feb 19 at 14:10
This is the kind of answer I was looking for; very interesting, thank you. And yes, the events are independent. – Kanaya Feb 19 at 14:38
If the exponent $a$ of the event probability $n^{-a}$ is any number less than $-1$, is the number rational? Or is it required that $a < -2$? – Trevor Alexander Feb 19 at 17:33

Assume that the digits do not happen to start repeating interminably at a finite position (0.2210101010101010...), since the probability of this happening is 0.

Well then it is irrational... A rational number has a decimal expansion that is eventually periodoc. If you assume that it isn't eventually periodic, then you're ruling out the case that it's rational.

I think what you're confused about is that it doesn't matter how you generated your number; either it is rational, or it isn't. It is possible that your process will generate a rational number, though the probability that it does is indeed zero.

And for that matter you cannot speak of the number generated by this process; many different number can be generated by this process, some are rational, some aren't. Every single number will have probability zero of occurring, and yet your process will always yield some number...

-

This isn't a number, it is a set of numbers. Probabilities don't make sense here, a number is part of your set or it isn't. The set is then $0.1010\ldots$ where some of the digits are replaced by $2$. One of the numbers in your set is $0.2222\ldots$, which is rational, another is $0.101010\ldots$, again rational. But it also contains the number where the $2$ show up only on positions that are perfect squares, i.e., $0.2012101012\ldots$, which isn't a repeating decimal and thus irrational.

Your set of numbers isn't countable, by the way, so it has to include transcendental numbers too (the set of algebraic numbers is countable).

-
I think this is the only correct answer. Moreover, I believe that many of the numbers in the set can't actually be constructed and therefore rely on the axiom of choice for their existence. – Kyle Strand Feb 20 at 0:21

That's not rational. And you don't write numbers by chance. Well, if a digit has certain 'probability' of appearing then it's not a fixed number.

You see one time the number may be 2 and another time it won't. And since you say that though the provability approaches 0, there is still the possibly, even if minute, that the number may be 2.

-