# Non-existence of irrational numbers?

I realize the title of my question will probably cause the raising of some eyebrows, so let me explain. Not sure whether to file this under "math" or "philosophy". This also might be able to be explained away due to some possible misconceptions I have about irrational/transcendental numbers, so please correct me where I'm wrong, since I'm quite certain I am...but this is nagging me, so here it is:

Ok, so imagine a non-repeating decimal with an infinite number of digits in it. The numbers are random, but for the sake of argument, let's say the number starts with "0.2368210236..." and continues from there.

So I look at the first decimal digit - 2 - and think to myself "in the rest of this decimal, there are an infinite number of times that '2' shows up."

So I look at the first TWO decimal digits - 23 - and again, think to myself "due to the randomness and infinity of this decimal, there is statistically a 100% certainty that the numbers "23" show up, in that order, somewhere in the rest of this decimal.

Continuing this line of thought, I reach the conclusion that if I choose ANY number of digits (variable: x) that start off this infinite decimal number, then a) there is statistically a 100% certainty that particular sequence shows up somewhere else in the decimal, and b) that it shows up an infinite number of times.

If such is the case, then what happens as x approaches infinity? Would that not mean that, given enough length in the decimal, there would eventually come a point where the decimal starts repeating itself, ad infinitum? Would that not mean that what was originally thought of as an irrational number is actually rational? Would this line of thinking not apply to ALL irrational numbers?

I guess what I'm asking is this: could it be that there ARE no irrational numbers, just rational numbers whose scope and vastness outstrip our ability to see their rationality?

Alright, mathphiles; have at me!

• Can you formalize your new math in an axiomatic and self-consistent way? – Ferdinand.kraft Apr 18 '13 at 4:20
• Not repeating is not the same as saying that digits and digit sequences don't repeat. A rational number has a period repetition - there is some $N$ and $d$ such that for all $n\geq N$ the the $n$th digit will be the same as the $n+d$th digit. Knowing that finite sequence repeats once is irrelevant, and there is no way to "go to a limit" with those finite repetitions. – Thomas Andrews Apr 18 '13 at 4:30
• Hmmm...formal math isn't particularly my greatest strength, if you get my drift, but let me ponder this some more and I'll give it a shot at least. – Steve Beresh Apr 18 '13 at 4:31
• By the way, there are plenty of irrationals that don't have "random digits." For example, there are irrationals with only $0$ and $1$ as digits, base $10$. We know $\pi$ is irrational, but we don't even know if there are infinitely many $3$s in the expansion of $\pi$ - or at least we didn't last I'd heard. – Thomas Andrews Apr 18 '13 at 4:42
• Consider the non-repeating decimal $0.11010010001\ldots$ (i.e., 1 separated by 0, 1, 2, ... zeroes). This number isn't rational (as it doesn't repeat), and e.g. 11 doesn't ever appear again either. Infinite structures are very subtle beasts... – vonbrand Apr 18 '13 at 16:40

No. If you choose the digits independently and randomly, the resulting number is transcendental almost surely. (This follows from the fact that there are only countably many algebraic numbers, let alone rational numbers, and a countable subset of the reals has measure zero.)

You are confused about several things. Mathematicians have a precise and completely formal way of defining numbers that, in particular, does not depend on decimal expansions. Using these definitions, there are also precise and completely formal proofs that the resulting numbers are irrational or even transcendental.

• It's that "almost" that keeps nagging at me; if we're dealing with an infinitely long decimal, it would seem to follow that there's a 100% certainty of ANY sequence of digits occuring...even if that sequence is an infinitely repeating pattern. – Steve Beresh Apr 18 '13 at 4:29
• @Steve: any finite sequence of digits occurs almost surely, but an infinite repetition occurs almost never. (Just so we're clear, "almost surely" has a precise mathematical meaning; it means "except on a set of measure zero.") – Qiaochu Yuan Apr 18 '13 at 4:30
• Ah, ok. I'm a lay-mathematician; I don't have any formal training in upper-upper-level math. I appreciate the comments; I knew I was getting confused about SOMETHING; just couldn't put my finger on what it was. Thanks! :) – Steve Beresh Apr 18 '13 at 4:32

This is the problem Pythagoras faced. We once believed all numbers could be expressed as a ratio of two integers, hence the term rational number. The diagonal of a unit square is $\sqrt 2$ which is irrational. This is easy to see. Take two unit squares and cut them along their diagonals. You now have four right triangles whose legs are each equal to $1$. Arrange them into a square by putting the four right angles together. Since this one square is made from two unit squares, the whole thing has area equal to $2$. Therefore, the sides of the square which correspond to the diagonals of our original unit squares have length of $\sqrt{2}$. Therefore, this number is real. You may like neither limits nor non-repeating decimals, but irrational numbers exist.

• Your example may not actually be perfectly correct. The issue involves the notion of a surface and what it means to "cut". You would require an beyond infinitely-sharp cutters, and they cannot be conceived. For this issue of irrational vs. indeterminate rational rests on the distinction of the real vs. idealism. – Marcos Jun 7 '13 at 4:52
• It's a little bit like the question of whether the area of a surface is different when you bend it: Is the area larger on the convex side? One thinks of the surface as infinitely thin, but such a notion is useless for edge-cases like the above. If there isn't a difference, then there isn't even a notion of "curved surface". – Marcos Jun 7 '13 at 5:29
• @MarkJ Geometry doesn't depend upon the thickness of a geometer's pencil. – John Douma Jun 7 '13 at 13:52

Let me answer your question with another question. What rational number contains ever sequence in its decimal expansion? It turns out none. Either the decimal terminates or it has some point where it becomes periodic (a terminating decimal could be seen as 0 repeating).

Just because a string will repeat (even infinitely often), doesn't mean that the entire string must be a repeating sequence.

You can create such a sequence as follows:

Concateate the sequence with itself and add a 0. This gives us 110.
Concatenate the sequence with itself and add a 0. This gives us 1101100.
Concatenate the sequence with itself and add a 0. This gives us 110110011011000.
Repeat to infinity.

Due to the concatenation, you can see why given any sequence, we can find it again later on. In fact, we can find it infinitely many times.

You can also see why this will never repeat itself.

• But given the reiterations of the pattern in your example, could we extrapolate that to mean that any irrational number will at some point become "fractal-esque"? – Steve Beresh Apr 18 '13 at 4:24
• @Steve: what do you mean by that? – Qiaochu Yuan Apr 18 '13 at 4:32
• @SteveBeresh You can't really ask about "all" numbers having a property if you can't define that property. "Fractal-esque" is not a meaningful word. Most of the examples people will give here might look fractal, but that's only because most simple computable numbers look fractal. – Thomas Andrews Apr 18 '13 at 4:36
• What I mean by "fractal-esque" is the property demonstrated in your example: there's a recurring pattern there, that could in theory go on infinitely - just keep concatenating the sequence and adding a 0. While there would never be a finite sequence of numbers that would repeat, there would definitely be a recurring pattern, just as in a fractal. – Steve Beresh Apr 18 '13 at 4:51

I am answering this question 4 years late simply because someone showed me this link, and I think the existing answers ignore aspects of your question.

It seems you are actually talking about randomly generating a real number $R$ in the interval $[0,1]$ by writing $$R=0.A_1A_2A_3… = \sum_{n=1}^{\infty} A_n 10^{-n}$$ and by independently selecting the $A_n$ digits uniformly over $\{0, 1, …, 9\}$. That is fine. So let us first agree that your conclusions should be restricted to such randomly generated numbers in the interval $[0,1]$ (not all numbers in the interval $[0,1]$).

You correctly infer that (with probability 1) the decimal expansion of our number $R$ has the property that any particular sequence of a given length $x$ (where $x$ is a positive integer) occurs an infinite number of times. This property can be formally proven about such randomly generated numbers. Clearly this property does not hold for all real numbers since the real number 0 = 0.0000000 does not even have any “1” digits.

It indeed follows that the decimal expansion of this random number $R$ will (with probability 1), contain all possible finite length sequences, and that each of these finite length sequences will be repeated (at randomly placed locations) an infinite number of times.

However, it is a mistake to conclude this means the decimal expansion will eventually be periodic. As Paul's answer notes, it means it will not be eventually periodic. An eventually periodic sequence has some finite transient and some finite period, of the form: $$0.\underbrace{b_1 b_2 ... b_m}_{\mbox{transient}} c_1 c_2 ... c_p c_1 c_2 ... c_p c_1 c_2 ... c_p...$$ where $m$ is the length of the transient and $p$ is the length of the period. A number is rational if and only if its decimal expansion has this eventually periodic structure. It can be shown that this eventually periodic structure prevents the expansion from containing every possible finite length sequence: Indeed, if $c_1 \neq 0$, then the above does not contain the all-zero sequence $000..000$ of length $(m+p+1)$. This is because $m+p+1>m$ and $m+p+1>p$. So if you take any $(m+p+1)$ consecutive numbers in the above sequence, at least one of them must be $c_1$ (which is nonzero). On the other hand, if $c_1=0$, then the above sequence does not contain the all-1 sequence $111...111$ of length $(m+p+1)$. Thus, with probability 1, our randomly generated number $R$ will not have an eventually periodic decimal expansion, and so it will be irrational.

• I think that if you assume that all real numbers where where for all n, the first n digits repeat later are rational, then it's easy to prove from that that every single real number is rational. I think the author made a mistake somewhere else as described in my answer. – Timothy Jul 28 '18 at 14:47
• @Timothy What precisely do you mean by "the first $n$ digits repeat later"? (do you mean repeat periodically, or just show up again later?) How does such property relate to something in my answer? What aspect of my answer do you disagree with? Are you disagreeing with the fact that a randomly chosen number (uniform over $[0,1]$) will, with prob 1, contain all finite sequences an infinite number of times? (in scattered locations with random gaps between reoccurences) – Michael Jul 28 '18 at 15:11
• No, I think the author made a mistake and that wrong fact can be derived from their mistake. – Timothy Jul 28 '18 at 17:06

None of the other answers answer this question by figuring out where the author probably made the mistake and pointing out that mistake. Some of them prove that there are irrational numbers and that might enable the author to catch their own mistake but I prefer to explain the mistake.

Even if all random number are rational, that doesn't necessarily mean all real numbers are rational but I'm guessing you were probably thinking that once you prove that all random numbers are rational, it's easy to show that all real numbers are rational. I think you made a mistake in thinking that all random numbers are rational by using the following argument:

"For any natural number n, there exists a natural number m such that for all x from 1 to n, the m + x$^{th}$ digit is equal to the x$^{th}$ digit. Therefore there exists a natural number m such that for all natural numbers n, for any x from 1 to n, the m + x$^{th}$ digit is equal to the x$^{th}$ digit. From that, it's easy to show that for some natural number n, the first n digits keep repeating over and over."

The problem with that argument is that the second sentence doesn't follow from the first.

• Your answer has an unusual parallel with mine, in that you follow my outline of: (1) first mention you will add something that prior answers leave out, (2) distinguish between all real numbers and randomly generated numbers, (3) clarify some mistake. The difference is that you leave out the agreement that the asker correctly concluded randomly generated numbers will have all finite length sequences an infinite number of times. Another difference is that you apparently "quote" the asker but that quote is nowhere in the asker's writing (and I cannot follow the point of that quote). – Michael Jul 28 '18 at 15:27
• Anyway I suspect the point of your quote is to conclude that if a number has a decimal expansion that contains all possible finite sequences an infinite number of times, then it cannot be eventually repeating (and hence cannot be rational) which would be a 4th point of similarity between our answers (and I made that point by adding detail to a prior comment that Paul already made). What then is your answer adding that is new? – Michael Jul 28 '18 at 15:29
• @Michael I didn't read through all the existing answers before I wrote my answer but I probably should have. I independently thought of something similar to your answer. After I read all of the answers, I edited mine to say that none of the answers answer the question because I saw that they all missed something. – Timothy Jul 28 '18 at 17:04
• I see. But I am reading your answer because this morning you posted a comment under my answer. I don't know the reason for that comment. – Michael Jul 28 '18 at 18:39
• @Michael I thought that the Author correctly figured out that assuming that every number such that for every positive integer n, the string of the first n digits of that number appears again later, that number is a rational number, it's easy to show that all numbers are rational so to prove that all numbers are rational, it suffices to prove that assumption so I tried to figure out the mistake the author probably made in proving that assumption. – Timothy Jul 28 '18 at 18:47