What is the Probability of Choosing an Irrational Number on the Real Line?

[mruncleramos]

Take the Interval [0,1] over the reals. Randomnly choosing a number, what is the probability that you will get an irrational number? A rational one?

 

[Hurkyl]

Is this homework? Have you had any thoughts about how to do this?

Anyways, you haven't given enough information: you can't really talk about randomness until you specify a probability distribution.

 

[EnumaElish]

Take the Interval [0,1] over the reals. Randomnly choosing a number, what is the probability that you will get an irrational number? A rational one?

With a continuous distribution, the probability of picking anyone number at all is zero, regardless of it being rational or irrational. So it has to be discrete. Now the tough question is, how do you specify a discrete distribution over a nonatomic domain (a domain without a "smallest unit")? May not be possible. E.g. the pseudorandom generators that are programmed into statistical software are defined on rationals only (for example, they will never return the number "pi," just a rational approximation to an arbitrary yet finite number of digits).

 

[matt grime]

it is possible to give a discrete distribution for this, but the probability of picking almost all numbers is zero.

whilst we do not have such a discrete dist. we do commonly abuse notation and appeal to measure theory for the answer

 

[Hurkyl]

You don't need a discrete distribution to ask about the probability of picking a rational number vs a nonrational number. You don't even need it to be discrete for one (or both) of those probabilities to be nonzero.

 

[EnumaElish]

You don't need a discrete distribution to ask about the probability of picking a rational number vs a nonrational number. You don't even need it to be discrete for one (or both) of those probabilities to be nonzero.

Can you give an example?

 

[Hurkyl]

Ack, I can't believe I said nonrational.

Let's set A = the set of all rationals in [0, 1], and B = the set of all irrationals in [0, 1].

Now, for ANY probability measure P, we have that P(A) + P(B) = P([0, 1]).

For example, if P is the uniform distribution on [0, 1], then we have:

P(A) + P(B) = 1

In particular, the uniform distribution is simply given by $P(R) = \int_R \, dx$, so we know P(A)=0 and P(B)=1.

 

[EnumaElish]

Sure, you can always get a set of numbers, but you'll never get a number from a continuous distribution (with positive probability, that is). Remember, the original quote said "randomly choosing a number."

 

[Hurkyl]

Sure, you can always get a set of numbers, but you'll never get a number from a continuous distribution (with positive probability, that is).

Incorrect. I get a number with probability 1.

The probability that I get any specific number, like 1/2, is zero¹, but that's not what the question asked.

1: well, there are technicalities...

 

[EnumaElish]

Incorrect. I get a number with probability 1.

Okay, you may be right. It may take a little more to sink in.

 

[HallsofIvy]

EnumElish: there is a difference between "getting a specific number" and "getting a number". If I "pick a number at random" then I have to get a number!

For example, suppose we pick a number with the uniform probability distribution on the interval [0, 1]. That means that every number is equally likely to be picked, the probability of picking a number in a given interval is equal to the length of that interval, and the probability of picking a number in a given measurable set is equal to the measure of that set.
In this case, the probability of picking any specific number (like $$\pi$$ or 1/2) is 0. The probability of getting some number is, of course, one. In fact, since the set of all irrational numbers between 0 and 1 has measure 1,the probability that the number chosen will be irrational is 1, the probability that the number chosen will be rational is 0 (which does not mean that it can't be rational! Probabilities of 0 and 1 in infinite spaces do not mean "impossible" or "certain".)

 

[EnumaElish]

EnumElish: there is a difference between "getting a specific number" and "getting a number". If I "pick a number at random" then I have to get a number!

That's what I realized yesterday, after Hurkyl's last post. Thanks, all.

 

[mruncleramos]

Wow, so many responses. I was just wondering and all. I have another question now. This is not homework, I was just thinking. Let's say we take a definite integral from 0 to 1 of an arbitrary real valued function, and answer is x. Clearly if we take away any random point from the interval 0 to 1, the integral will still equal to x. Same thing with 2, 3, 4, 5 etc. What happens if we take away all of the rational numbers? Will the integral still be the same? It seems to me as if it should. But when I think about taking away all irrational numbers...

 

[EnumaElish]

My guess is, you'll still end up with the same integral value. I guess this will follow from Cantor's taxonomy of infinities: uncountable infinity (irrationals) + countable infinity (rationals) = uncountable infinity (reals). Therefore, U.I. = U.I. - C.I.

BTW, the integral you are describing is identical with the uniform probability distribution for 0 < x < 1. Under that measure, Hurkyl has suggested that P(rationals) = 0 and P(irrationals) = 1. Obviously P(reals) = 1. This imples P(irrationals) = 1 = P(reals). When you remove all rationals you are removing a set with measure zero.

 

[matt grime]

Removign a countable set doesn't affect the integral. Countable sets have measure zero. Let us prove it using some measure theory (as i know no measure theory this is hand wavy). First let us suppose that the function f is bounded and positive , ie 0<= f(x) <M for x in [0,1] and some M.

Now take any countable set of points

x_1, x_2, x_3,...

we want to show that this countable set of points conttributes "nothing" to the integral. Let us estimate the contribution.

Let d be some real positive number. About the point x_n put a little open set of width d2^{-n}. Then the x_i's can contribute no more than

$$\sum_n Md2^{-n}$$=

that is they contribute less than the rectangle of width d_2^{-n} and height M to the integral.

Now we can work out that geometric sumand it is Md. but d was arbitrary so the contribution must be less than Md for any d in the positive reals (and is not negative) so it must be zero.
Yes this is surprisig and no it isn't rigorous.

Your idea of throwing away ione point at a time shows that we can remove any finite number of points. it doesn't then follow that we can remove an infinite number of points. We see that we can always remove a countable number of points, but not an uncountable number (we could remove all the points and that would certainly affect the integral).

 

[HallsofIvy]

To make sense of that question you have to, as Matt Grime did, change from the Riemann integral to the Lebesque integral.

 

[mruncleramos]

Measure Theory. That sounds pretty cool.

 

[EnumaElish]

Your idea of throwing away ione point at a time shows that we can remove any finite number of points. it doesn't then follow that we can remove an infinite number of points. We see that we can always remove a countable number of points, but not an uncountable number (we could remove all the points and that would certainly affect the integral).

I think what you mean is "we can remove a finite, and at most a countably infinite, number of points and still get the same sum."

 

[matt grime]

and what was the point of that post?

 

[Hurkyl]

For the sake of precision, I would like to point out that there are some uncountable sets of points we can remove without affecting the integral. (e.g. a Cantor set -- uncountable, yet with measure zero)

 

[EnumaElish]

there are some uncountable sets of points

Do you mean "sets containing an uncountable number of points"?

 

[EnumaElish]

and what was the point of that post?

You juggled finite vs. infinite, and then countable vs. uncountable. I was just paraphrasing you in a way that (I thought) is marginally more organized.

 

[matt grime]

you are merely restating in more words what is already explicit.

i "an uncountable set of poitnts" and "a set containing an uncountable number of poitns" are indeed synonymous though the former is perhaps better as it does not imply that we have defined what it means for a number to be uncountalby large.

 

[DaTario]

Take the Interval [0,1] over the reals. Randomnly choosing a number, what is the probability that you will get an irrational number? A rational one?

As there are infinitely more irrational numbers than rational in this interval, the probabilily of getting a rational must be 0 while irrationals are obtained with prob = 1.

 

[matt grime]

"infintely more" isn't very rigorous is it?

 

[HallsofIvy]

Depends upon which integral you are using. Matt Grime's answer- that removing a countable set of numbers (or even an uncountable set with measure 0) does not change the integral- is for the Lesbesque integral. If you remove a countable number of points from an interval, the resulting function would not be Riemann integrable.

 

[DaTario]

"infintely more" isn't very rigorous is it?

Ok, I should have said perhaps:

as the the cardinality of the set of irrational number is higher than that of rationals, it follows that picking up a rational number has probability zero, while picking up an irrational has probability 1.