Why do people say irrational numbers must contain any sequence of numbers?

Factual Questions
1

Consider the number 0.101001000100001000001… with an increasing number of 0s between the 1s.
[ol]
[li]Is this number irrational?[/li][li]It does not contain the number “2” or even the sequence “11”[/li][/ol]
Since this number does not contain the sequence “11”, even though it has all the sequence’s digits, doesn’t this show not all irrational numbers contain all sequences of numbers? How, then, do we know that pi, or e, or sqrt(2) contains my phone number?

For more such irrational numbers, you could count in base 8: 0.1234567101112131415161720… and it doesn’t contain the digits 8 or 9.

2

Yes.

You have correctly demonstrated that not all irrational numbers’ digit-expansions contain all finite digit-sequences, or even all single digits, for that matter.

Anyone who believes that every irrational number’s digit-expansion contains all digit-sequences is flatly mistaken.

However, may people believe that, in practice, most of the irrational numbers that happen to arise for genuine arithmetic interest do have the property of having digit-expansions containing all digit-sequences, in all bases, with asymptotically equal frequency, etc., etc.

For example, people believe this of pi, and e, and sqrt(2).

However, it’s never been proven for any of these; indeed, it’s never even been proven that each digit occurs infinitely often in each of these, and for all we know, we’ll eventually prove otherwise.

3

The term used to describe such numbers is normal. While all normal numbers must be irrational, not all irrational numbers, obviously, are normal.

4

The most interesting thing about normality of numbers is that almost every real number is normal, and yet it’s very difficult to show that any particular one is. So in some sense, all the numbers that we’re familiar with are very unusual.

5

I wouldn’t quite say it’s hard to show that any particular number is normal; it’s easy to show that, e.g., 0.01234567890001020304050607080910111213… is normal.

It’s just hard to show that any arithmetically interesting number (one which arises naturally out of interest as an actual quantity, rather than merely as a roundabout coding of a sequence of digits) is normal. Which is essentially because it’s hard to say anything about digit-sequence properties from arithmetic properties; digit-expansions aren’t actually generally a great way to think about arithmetic. (The one noteworthy connection I can think of between an arithmetic property and a digit-sequence property is the correspondence between rationality and eventual-periodicity; beyond that, I’ve got nothing)

So, yeah, almost all sequences of digits are normal, as a basic probabilistic fact. And there happens to be a (somewhat messy, not quite 1-to-1) correspondence between digit-sequences and “real numbers” in the interval [0, 1]. But the nature of the correspondence is such that the things which are most interesting as digit-sequences aren’t generally paired with the things which are most interesting as fractional quantities (beyond the rationality/periodicity correspondence), and so we shouldn’t expect that fractional quantities chosen for their interest as fractional quantities should happen to also have interesting properties as digit-sequences. The "interesting"ness measure on [0, 1] and the "interesting"ness measure on {0, …, 9}^N just don’t match up under the digit-expansion correspondence.

6

(Speaking informally for that last sentence, mind you…)

(Or, rather, the thing to say is, the "interesting"ness measure on [0, 1] just doesn’t match up with the uniform measure on [0, 1], so things which are common by the latter measure needn’t be common by the former measure.)

(Ah, I suppose the interesting observation that interesting numbers aren’t nearly a representative sample of the uniform distribution on [0, 1] was the point in the first place…)

7

Does that mean specifically that all but a countable subset of real numbers are normal?

8

No, only that the set of non-normal numbers has Lebesgue measure zero. All countable sets have that property, but there are also non-countable sets that have that property.

Specifically, the set of all numbers in [0, 1] whose decimal expansion does not contain the digit ‘4’ is a subset of the set of all non-normal numbers, one can clearly form a bijection between it and [0,1] by using number that has the same decimal expansion in base 9 using the digits (0,1,2,3,5,6,7,8,9), so it has to be uncountable.

9

Do we have any reason to believe pi is normal? I see it stated so often and confidently, and I don’t think mathematicians would say something like this with no reason.

10

My understanding is that there’s a lot of statistical evidence. People have analyzed the first 10[sup]N[/sup] digits of π and looked at the frequency of the digits and sequences of digits, and found no significant deviations (in a statistical sense) from what one would expect from a truly random distribution of digits. Of course, this isn’t a proof in the mathematical sense, but it’s still suggestive.

11

I will give an intuitive proof why most of the real numbers are normal… though there can be several argument this is what I thought…

Definition:
consider a real number in base b…
Simply normal: if each digit (from 0 to b-1) has equal probability of appearance in that number it is called simply normal to that base…

normal: if any finite pattern made of the digit (from 0 to b-1) has equal probability of appearance in that number it is called normal to that base…

absolutely normal: if a number is normal to any base it is called absolutely normal.

I will prove most of the real numbers are normal in base 10…

How to generate a real number randomly between 0 and 1??
One possible way is to throw a 10-sided die again and again infinitely… Outcome will be the digits after decimal point… A perfect random number will be generated…
and that generated number will be normal(according to the definition) since we are playing randomly…

Now two questions arise here:

  1. how are we accounting rational number?

  2. how are we accounting irrational non-normal numbers?

  3. rational numbers(say of length finite n) will get generated in this process when all the outcome after nth throw will be zero… whose probability will be zero when we are generating the random number following the above rule(throwing the die infinitely)…

  4. every irrational non-normal number follow certain rules(simple or complex)… when we are randomly generating the numbers probability that the generated number will follow ‘ANY’ rule is also zero…
    So the probability of getting a non-normal number randomly is extremely small… That is a truly random pick from real numbers will yield a normal number with probability close to one…

Hence we can conclude most real numbers are normal

12

That idea can be used to show that the set of numbers in [0, 1] which are not normal in a particular base has measure zero (although you’ll need to do some work to make it rigorous), and you should be able to extend it to the entire line with a little more work, but I think you need a fundamentally different approach to prove anything about numbers that are normal in every base.

13

Probably unrelated, but since this is a minor zombie thread anyway, and sorta-kinda on the topic, I’ve got a question. I’m not a math wiz by any means, so this may have already been answered and I just didn’t understand it.

So, pi, 3.141… etc, continues infinitely without repeating, and contains every number sequence imaginable (Somewhere is your phone number, your SS# if you have one, the movie “Pluto Nash” in binary, etc. etc…), or so I understand current thinking to be.

If so, then shouldn’t pi itself be in there somewhere? If it is truly infinite, I mean.

To demonstrate: 3; 10% of all following digits, if randomly distributed, will be 3. 31 is the first two digits, ignoring the decimal. It’s easy to find another example of 31; it should happen, what, 1% of the time? Going on, 314 will appear, as will 3141, although with lesser and lesser frequency. 31415, the same thing. Continuing this line of thought, for any X digits of pi, that same series of numerals, minus the last one, will exist, such that it matches X-1 digits of the series you’re looking for. So if you look a match to the string 3141592653589793238 (the first 19 digits), you will find it, but the next digit may be a 7 instead of a 5. If all digits are random, AND the digits are infinite, this MUST happen, no?

Carrying this idea to the extreme, then if you were to figure out pi to, say, a billion trillion digits… somewhere, further on, is that exact same sequence, except with a different number at the end. And, that billion-trillion-digit sequence, PLUS the similar-except-one sequence, PLUS all the numbers between the two… that’s in there somewhere as well!

Doesn’t this mean that, for pie calculated to any given accuracy… it does, in fact, repeat itself after all?

14

Forgot this part:

To continue the point, consider that no number sequence can be ruled ineligible for repetition… you can’t say ‘this is the first, last, and only time the sequence 314159 can appear’, because if you do, eventually every sequence will have appeared at least once and there will be no series of numbers which can follow, ending the ‘pattern’ for good.

My brain hurts.

15

At least, it’s normal in base ten. Can it be proven that that number is universally normal? Or even that there is any other base at all in which it’s normal? Well, other than bases of the form ten[sup]n[/sup], of course.

Phnord Prefect, pi to any given precision will, presumably, show up elsewhere in the digits of pi, and in fact, will do so infinitely many times. But you’d be hard-pressed to call that a repetition, since the distance you have to go to find the “repeat” will be exponentially greater than the size of the repeat itself. That is to say, you can’t say “I found the sequence 314159, and that counts, because six digits is good enough”, because if six digits were good enough, you’d be far from reaching the place where you found that.

16

It’s always very easy to construct an irrational number that doesn’t contain a given sequence of digits. A good example of an obviously irrational number that at the same time obviously doesn’t contain all finite sequences of digits is 0.101001000100001… (co-incidentally it is algebraic). This number doesn’t even contain all the finite sequences of 1s and 0s (e.g. it doesn’t contain …010101…) let alone all the decimal sequences.

17

Given an artificially constructed abnormal number – in particular, the one you gave here – How do you demonstrate that it’s algebraic?

18

Sum of a geometric progression.

19

Only this one isn’t. I should look more clearly.
However, it is an infinite series. Some infinite series turn out to be algebraic.

20

I don’t know about the number 0.101001000100001… My guess is that it is not algebraic. But the number 0.110001000000000000000001…, the sum of all the 10^{n!} is definitely known to be transcendental. In fact I could even prove it. Unlike for e and pi, where I once went through the proof but could not tell you anything about the proof, which was one of the most complicated arguments I have ever followed in detail (it is one of the volumes of Herstein’s algebra text). The number above is studied and proved transcendental in Courant and Robbins, What is Mathematics?