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If I were to pick a number completely at random in the range [0,1), it seems to me that number would be irrational.

After all, there are a countably infinite number of rationals between zero and one, but an uncountably infinite number of irrationals, therefore, the odds of picking one of those very few rationals is immeasurably small, indistinguishable from zero.

I think the problem lies in the word "pick". It is (I suspect) impossible to pick an random number from the set of all real number; it would take certainly an infinite number of digits to represent the number.

Or am I wrong?

Michael Lorton
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    If by "randomly" picking a number you simply picked an infinite list of random integers between 0 and 9, then the chance of getting a rational number would tend to 0 because you would need to get an infinitely repeating pattern. – user138335 Apr 06 '14 at 22:06
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    Look up "probability space." – anon Apr 06 '14 at 22:06
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    The probability of picking any particular number is zero. Afterall, for any number you might pick, there are finitely many numbers equal to that number and there are uncountably many numbers that are not that number. – David H Apr 06 '14 at 22:09
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    There is a difference between "it will certainly be irrational" and "it will be irrational with probability $1$" – Henry Apr 06 '14 at 22:11
  • Assuming uniform probability, yes. See this question, showing it does have measure zero. (In your case, interpret "measure" to mean probability.) – Andrew Kelley Apr 06 '14 at 22:13
  • @Henry -- there is? – Michael Lorton Apr 06 '14 at 22:15
  • @AndrewKelley -- I've seen answers I didn't understand; I've seen questions I didn't understand; this is the first question I couldn't even pronounce. But I think it means the answer to my question is "yes". – Michael Lorton Apr 06 '14 at 22:18
  • @Malvolio: Yes, see Wikipedia on almost surely – Henry Apr 06 '14 at 22:19
  • @user138335: Note that by saying list you are implying countable... – Thomas Apr 06 '14 at 22:19
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    While you state a range, you don't state if you are looking at the reals, rationals, algebraic numbers or something else. This could be worth noting to some extent as if one limits the range to just rational numbers then the probability of selecting an irrational drops considerably. – JB King Apr 06 '14 at 22:47

2 Answers2

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Both of the other answers hinted at this: How do we choose a number "at random" in the unit interval [0,1)? The problem is that we would have to specify infinitely many digits. One standard way to "do this" is as follows:

Pick whether it is in the first half of the interval or the second half; define each to have probability 1/2. We have just picked the first binary digit. (It's 0 or 1.)

Then repeat the previous step infinitely many times, one for each binary digit (bit). Ah, there's the problem; if we really wanted to do this, it would take us for ever to be done. Let us now choose an option to continue: realistically or theoretically?

Realistically: Just stop after some large number of digits, and define the rest of the digits in some succinct, finite way. For example, define the rest to be zero. We now have a rational number. (We have repeating zeros.)

Theoretically: Imagine that we really do continue forever. (To deal with this rigorously is the so called product measure, but for us, it's fine to just imagine it.) As the first comment to the question reminds us, a number is rational if and only if it has a periodic decimal expansion (or binary expansion), i.e. it repeats a string from some point on. So, if we flip a coin forever, what's the probability that we repeat a pattern indefinitely from some point on? Zero.

So the answer to your question is...Yes, it will be irrational.

Andrew Kelley
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The process of picking out a number from $0$ to $1$ from a continuous distribution involves picking a number from smaller and smaller sub-intervals because the probability of picking any one number goes to zero as the size of the interval becomes arbitrarily small. The set of rationals can be covered by a set of intervals of arbitrarily small total length because they are countable, so for a given interval you get that the measure of the intersection of the interval with a set of intervals covering the rationals goes to zero as the length of the cover of the rationals is decreased. This implies that for any interval, the probability of choosing a number close to a rational in that interval goes to zero as the measure of closeness goes to zero. But the probability of choosing an arbitrary number in a given interval is positive. Hence the probability that you choose an irrational number should be identified as 1.

user2566092
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