How Are Rational and Irrational Numbers Distributed Along the Real Line?

[Square1]

Hey.

I've been wondering, from -infinity to positive infinity, what does the distribution of rational vs. irrational numbers look like? Are there any notable or interesting patters/points to make about this relationship?

 

[Filip Larsen]

If I remember correctly, both rational and irrational numbers are dense in real numbers meaning that no matter how small an (open) interval from the real numbers you look at, there will always be both rational and irrational number present, or similarly, that no matter how close you select two unequal rational numbers there will always be an irrational number in between, and vice versa.

If the above holds true it is tempting to propose that there are "equally many" rationals as there are irrationals and that both sets are equally "smeared out" over the reals. Unfortunately I haven't done that kind of rigorous math needed to support such claims in a long time, so perhaps someone here with a more fresh knowledge subject matter can confirm or reject that claim.

 

[lurflurf]

There are many more irrationals than rationals.

 

[fleem]

Both are infinitely dense and there are an infinite number of each. Even if we try to normalize by rewording the question, "What is the probability that a randomly chosen number will be rational as opposed to irrational?", we find that there are infinite integers that can be used for the numerator and denominator of a rational number, which means we can express any randomly chosen number as a ratio of integers. If we then try to normalize THAT by rewording the qurestion again to, "If we choose a finite set of random integers, what is the probability a randomly chosen number can be expressed as a ratio of two of those integers?", we find that the probability is non-zero that that set will contain multiple infinitely-large integers. I think this line of renormalization goes on forever. However, if your definition of "rational number" excludes numbers expressed as ratios of infinitely-large integers, then you are obliged to specify a finite limit to the set of integers you allow (you can't say the set includes an infinite number of integers where each has a finite value, and if you say the set has a finite number of finite integers, you can't know whether that set will tell you anything about irrational number density vs. rational number density).

 

[Square1]

Hmm alrighty then.

 

[Deveno]

Both are infinitely dense and there are an infinite number of each. Even if we try to normalize by rewording the question, "What is the probability that a randomly chosen number will be rational as opposed to irrational?", we find that there are infinite integers that can be used for the numerator and denominator of a rational number, which means we can express any randomly chosen number as a ratio of integers. If we then try to normalize THAT by rewording the qurestion again to, "If we choose a finite set of random integers, what is the probability a randomly chosen number can be expressed as a ratio of two of those integers?", we find that the probability is non-zero that that set will contain multiple infinitely-large integers. I think this line of renormalization goes on forever. However, if your definition of "rational number" excludes numbers expressed as ratios of infinitely-large integers, then you are obliged to specify a finite limit to the set of integers you allow (you can't say the set includes an infinite number of integers where each has a finite value, and if you say the set has a finite number of finite integers, you can't know whether that set will tell you anything about irrational number density vs. rational number density).

i don't believe the bolded statement is true. in fact, the odds that a randomly chosen real number will be rational is equivalent to saying a randomly generated infinite decimal will be periodic, and the odds of that are smaller than any known positive number (which we can call 0, unless you believe in infinitesimals).

in the usual view, the rationals are evenly distributed among the reals.

however, there are certain ways of looking at the rationals, which do indeed lead to clustering, which you can find here:

http://linas.org/math/chap-rat/chap-rat.html

the catch here, is that this is closure with respect to a different metric than the usual one.

 

[TylerH]

I can't really speak on how the rationals are dispersed within the reals and irrationals other than to say the rationals are dense within the reals(http://en.wikipedia.org/wiki/Dense_set). This was explained before in other terms, that in any open interval of the reals, there is a rational.

As to the probability of randomly choosing a rational from the reals, it is VERY unlikely. We can say that the probability of choosing a rational from the reals is equal to the size of the set of rationals over the size of the set of reals. When considering the sizes of sets, or cardinality, there arise different levels of infinity(http://en.wikipedia.org/wiki/Cardinal_number). The rationals are known as being "countable" and thus equal in size to the natural numbers, these are of size apleph-null or aleph 0, whereas the reals and irrationals are commonly assumed (by the continuum hypothesis, http://en.wikipedia.org/wiki/Continuum_hypothesis) to be of size aleph 1. Aleph 1 is infinitely larger than aleph-null, so the probability is infinitesimal.

You might want to study infinite sets(http://en.wikipedia.org/wiki/Infinite_sets). The general concept of infinite sets is prerequisite to (and more approachable than) considering distributions of an infinite set within another. One of my favorite exercises from my proof class was to prove the rationals as being countable (equal in size to the natural numbers) by constructing a one to one function(http://en.wikipedia.org/wiki/Bijection) between them. You want learn much about distribution by studying infinite sets and the like, but that's mostly because distribution-related stuff is reserved for the upper levels of proof-based math. Specifically, AFAIK, it is first formalized in real analysis with the concept of neighborhoods(http://en.wikipedia.org/wiki/Neighbourhood_(mathematics)) but not actually generalized to general sets until topology. (Not to be pretentious... This is just my understanding. I've never taken anything above an intro to proof class, so take what I say with a very large amount of salt. )

 

[SteveL27]

Between every two irrationals there's a rational. And between every two rationals there's an irrational.

Yet, there are way more irrationals than rationals, in two senses:

* The rationals are countable -- that means they can be put into 1-1 correspondence with the counting numbers 1, 2, 3...

The irrationals are uncountable -- they can NOT be put into 1-1 correspondence with the counting numbers.

* The rationals have measure zero. That means that the "probability" of randomly choosing a rational from the reals is zero.

What measure zero means, in simplified technical terms, is that you can cover the reals with a collection of intervals, the sum of whose length is arbitrarily small. For example, given some tiny number epsilon > 0, enumerate the rationals r1, r2, r3 ... (They're countable, so we can do this). Then cover the first rational r1 with an interval of length epsilon/2. Cover r2 with an interval of length epsilon/4. In general, cover r_n with an interval of length epsilon/2^n. Adding up all these intervals, we've covered the rationals with a collection of intervals whose sum is epsilon. But epsilon was an arbitrarily small number. So by definition, the rationals have measure 0.

That proof is wild, if you think about it. Imagine the number line with the rationals labeled r1, r2 ... as above. Even though the rationals are dense in the reals, you can put an interval around every one of the rationals and the vast majority ("almost all" is the technical term) of the reals will NOT be in any of the intervals. Now that's literally impossible to visualize, but it's true.

The rationals inside the reals are very strange indeed.

 

[pwsnafu]

That proof is wild, if you think about it. Imagine the number line with the rationals labeled r1, r2 ... as above. Even though the rationals are dense in the reals, you can put an interval around every one of the rationals and the vast majority ("almost all" is the technical term) of the reals will NOT be in any of the intervals. Now that's literally impossible to visualize, but it's true

Close, but no cigar.

A point in a set X is "almost all" if the complement, X^c,

1. is finite, or
2. is countable, or
3. has measure zero.

In this context we want the latter.

It is true that there always exists an open interval cover of the rationals which has measure less than any specific ε > 0. But each individual open cover will have non-zero measure because it is a countable union of open intervals (you used the word "around"). In other words, once you pick an open cover (which you have done), the excluded points are not almost all.

 

[SteveL27]

It is true that there always exists an open interval cover of the rationals which has measure less than any specific ε > 0. But each individual open cover will have non-zero measure because it is a countable union of open intervals (you used the word "around"). In other words, once you pick an open cover (which you have done), the excluded points are not almost all.

Yes, sorry about that, thanks for the correction.