I conjecture that for irrational numbers, there is generally no pattern in the appearance of digits when you write out the decimal expansion to an arbitrary number of terms. So, all digits must be equally likely. I vaguely remember hearing this about $\pi$. Is it true for all irrational numbers? If not, what about transcendental? If true, how might I go about proving this?

For my attempt, I'm not quite sure how to approach this, all I have are some experimental results to validate the conjecture. I started with $\sqrt2$. The occurrences of the various digits in the first 5,916 decimal terms are:

563, 581, 575, 579, 585, 608, 611, 565, 637, 612.

And here are the occurrences of the decimals in the first 1993 digits of $\pi$:

180, 212, 206, 188, 195, 204, 200, 196, 202, 210

Same for the first 9825 digits of $e$:

955, 971, 993, 991, 968, 974, 1056, 990, 975, 952

It does seem that the percentage representations of each digit is very close to 10% in all cases.

Edit: It's clear the conjecture is false (thanks for the answers). Still curious why all "naturally ocurring" irrational numbers (like the ones mentioned here) do appear to be normal. I know this is unproven, so feel free to provide conjectures.

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    $\begingroup$ Check out the Liouville number. It is transcendental but only consists of 0 and 1. $\endgroup$
    – Fabian
    Commented Apr 21, 2019 at 7:40
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    $\begingroup$ The digits of the Liouville number are highly patterned, just not patterned in a way that makes the number rational. $\endgroup$
    – Neil W
    Commented Apr 21, 2019 at 9:33
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    $\begingroup$ Try $0.1001000100001000001000000100.....$ $\endgroup$
    – BAI
    Commented Apr 21, 2019 at 10:12
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    $\begingroup$ Take the square root of 2 and remove all 1's from it. Are you claiming the result is now e.g. rational? $\endgroup$
    – user541686
    Commented Apr 22, 2019 at 6:47
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    $\begingroup$ @RohitPandey Or do they? $\endgroup$ Commented Apr 22, 2019 at 17:07

3 Answers 3


What you mention is not true for all irrational numbers, but for a special subset of them called Normal numbers.

From the wiki article:

While a general proof can be given that almost all real numbers are normal (in the sense that the set of exceptions has Lebesgue measure zero), this proof is not constructive and only very few specific numbers have been shown to be normal.


It is widely believed that the (computable) numbers $\sqrt{2}$, $\pi$, and $e$ are normal, but a proof remains elusive.

Note that there are infinitely many irrational numbers that are not normal. In 1909, Borel introduced the concept of a Normal number and proved (with a few gaps resolved later) the following theorem:

Almost all real numbers are normal, in the sense that the set of non-normal irrational numbers has Lebesgue measure zero.

Some additional points of interest: (added with thanks to @leonbloy)

  1. The number of non-normal irrational numbers is uncountable Theorem 4 of this reference.

  2. There is a subset of normal numbers called Abnormal numbers and Absolutely Abnormal numbers which are uncountable. Abnormal numbers are not normal to a given base $b$ while Absolutely Abnormal numbers are not normal to any base $b \ge 2$.

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    $\begingroup$ "almost all real numbers are normal" should be complemented by the known fact that "the set of non-normal numbers is uncountable" . So, there are relatively few non-normal numbers, but there are (uncountably infinite) many irrationals which are not normal. $\endgroup$
    – leonbloy
    Commented Apr 21, 2019 at 17:39
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    $\begingroup$ @leonbloy "there are uncountably infinite irrationals which are not normal". How? $\endgroup$ Commented Apr 21, 2019 at 18:25
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    $\begingroup$ Given the decimal expansion of a normal number $x$, you can construct a non-normal (or abnormal) number by taking any known abnormal number and replacing its decimals at positions $2^i$ with the $i$th decimal of $x$. The result follows from the uncountability of the normal numbers. $\endgroup$
    – Oliphaunt
    Commented Apr 21, 2019 at 20:36
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    $\begingroup$ @user1952500 Proof that non-normal numbers are uncountable: see eg core.ac.uk/download/pdf/56374383.pdf (page 8). It's even true that the set of all "absolute abnormal" numbers (not normal in any base) is uncountable maa.org/sites/default/files/pdf/upload_library/22/Ford/… $\endgroup$
    – leonbloy
    Commented Apr 21, 2019 at 21:01
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    $\begingroup$ @leonbloy in base 10 one can just note that the set of numbers whose decimal digits consist of only ones and zeros, is uncountable. It’s like a cantor set $\endgroup$
    – shalop
    Commented Apr 21, 2019 at 22:35

This definitely does not hold for all irrational or transcendental numbers. As noted in the comments by Fabian, various Liouville numbers come to mind as examples where the digits of the number are not at all uniformly distributed, yet these same numbers were constructed with the specific intention of being transcendental.

The property you refer to - this "equidistribution of digits" and such - is what defines the so-called simply normal numbers in base $10$. If you have heard $\pi$ exhibits this very property, it's technically wrong because so far $\pi$ has not been proven to be simply normal in any base. It is suspected to be simply normal in every base, and even (absolutely) normal in every base, but it remains an open problem.

Simple normality in base $b$ means that the frequency of each digit in the first $n$ digits tends to $1/b$ as $n$ tends to infinity. Normality in base $b$ means that for each finite digit sequence of length $k$, its frequency in the first $n$ digits tends to $1/b^k$ as $n$ tends to infinity.

In fact, it seems only rather contrived numbers, such as $0.123456789101112...$ (Champernowne's number, obtained by concatenation of the naturals) among others in the article, are known for sure to be simply normal in base $10$, and in fact is normal in base $10$ (but not even known to be simply normal in other bases that are not powers of $10$). Nothing is known about a lot of the more "natural" numbers - like $\pi,e,\sqrt 2$. But I suppose Chaitin's constant could be considered a "natural" number, and it is normal in every base.

Trivially, we do know that almost all real numbers are normal in every base, equivalently a real number drawn uniformly randomly from $[0,1]$ is normal with probability $1$.

As for how to prove it for common numbers? Well, the proof for Chaitin's constant (overview in this Math Overflow post) relies on its algorithmic randomness, which is in fact just a much stronger form of normality. Roughly, normality in base $b$ says that each finite digit sequence of length $k$ has frequency in the first $n$ digits tending to $b^{-k}$ as $n$ tends to infinity, whereas algorithmic randomness says that there is some constant $c$ such that for every $n$ the shortest program (in some prefix-free encoding) that outputs the first $n$ bits has bit-length at least $n-c$, which intuitively means incompressible up to a constant. Note that if a number was not normal, it can be compressed using arithmetic encoding. On the other hand, Champernowne's constant is a very clear example of a highly compressible (so not algorithmically random) but normal number, since the first $n$ bits can obviously be output by a fixed program run on $n$ (which can be stored in $O(\log n)$ bits in prefix-free encoding).

Since $\pi$'s digits do not follow some nice pattern like Champernowne's number, nor are they algorithmically random, it is unlikely that normality proofs for known normal numbers would give much clue for $\pi$. So far, at least.

Of course this raises the question of "why do we conjecture them to be normal then?" Empirical evidence based on the first trillions of digits of $\pi$ do 'support' it, but of course that is nowhere near proof. It is just like if you toss a coin $1000000$ times and observe $500469$ heads and $499531$ tails, and conclude that you do not have evidence that it is not a fair memoryless coin, since the number of heads for a fair memoryless coin would be in the range $[499500,500500]$ with likelihood about $1/2$. So does your observation count as empirical evidence that it is a fair memoryless coin? Not really... Lack of evidence against is not really evidence for. Similarly, that is all we have for the question of $\pi$'s normality, so far.

Also, such empirical evidence is notoriously hard to interpret. Again take the coin example, and suppose you observe exactly $500000$ heads and $500000$ tails. Would you think it is a memoryless coin? No! How about $500001$ heads and $499999$ tails?

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    $\begingroup$ "So in a twisted sort of sense, you could say "almost all" real numbers are normal." Given that the rationals are also a dense subset of the reals, but allmost all reals are irrational, this doesn't follow at all. user1952500 states that there is an -albeit constructive- proof that almost all reals are normal, but that is actually quite a remarkable result. $\endgroup$
    – Chieron
    Commented Apr 21, 2019 at 9:16
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    $\begingroup$ Wouldn't 123456789/9999999999 be both normal and rational? $\endgroup$
    – EvilSnack
    Commented Apr 21, 2019 at 13:26
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    $\begingroup$ @EvilSnack No. Normal numbers must contain each finite string of digits the right fraction of the time. Your example never contains $13$. $\endgroup$ Commented Apr 21, 2019 at 15:00
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    $\begingroup$ @Chieron was correct to criticize your post. The Cantor set is uncountable but has measure zero. Also, there are other errors in your answer. Chaitin's constant being normal is unrelated to its mere uncomputability; it is trivial to find uncomputable numbers that are not normal. Rather, it's because of its algorithmic randomness. Thirdly, the distributions should not become closer and closer to being equal, by the central limit theorem. Next time, please don't make claims and arguments that you can't justify. $\endgroup$
    – user21820
    Commented Apr 22, 2019 at 6:16
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    $\begingroup$ I didn't say you deliberately got things wrong; I just said that you should refrain (next time) from posting answers with claims that you aren't able to justify. This is necessary because most who read your answer lacked the mathematical background to even realize there were multiple errors in it. Concerning your third error, if you flip a fair coin $n$ times, the number of heads and tails will not tend to become closer and closer to being equal as $n → ∞$. Likewise for the digits of normal numbers. This is a common fallacy, related to the gambler's fallacy. $\endgroup$
    – user21820
    Commented Apr 22, 2019 at 7:23

As user1952500 says, most (in a quite precise sense) numbers are normal and hence as you expect. However, it is very simple to create exceptions: numbers which are irrational but are not normal. The rational numbers have decimal expansions which, after a while, terminate or are periodic so just create a sequence that does not terminate or repeat but also does not have a uniform distribution of digits. A very simple way is to omit some digits. Fabian mentions the Liouville numbers which are an example of this style. There is nothing very special about the many zeros, you could swap the digits for others, e.g. $3$ and $7$. Another way would be to write an irrational number in a base less than $10$ and then regard it as a base $10$ number. It would not be normal as it would have no $9$s.

You can't prove normality by checking calculated digits as that will always only be a finite subset. Maybe the first quadrillion digits of $\pi$ behave as expected and after that $9$ never appears.

However, I would guess that irrational numbers which have not been artificially constructed are probably normal but it is very hard to prove. It has not been achieved for $\pi$ yet.

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    $\begingroup$ "very hard" is an understatement, I think. Noone has the slightest idea, how a proof of the normality of , lets say , $\pi$ could look like. It cannot even be ruled out that eventually, only digits $0$ and $1$ appear. It is very likely that such a proof is completely out of reach. $\endgroup$
    – Peter
    Commented Apr 21, 2019 at 8:27
  • $\begingroup$ Good answer (+1) $\endgroup$
    – Peter
    Commented Apr 21, 2019 at 8:28
  • $\begingroup$ @Peter Indeed but do we have a scale of hardness? So, you are suspecting that eventually only $0$ and $1$ appear. I was only expecting the weaker condition that $9$ ceases to appear. $\endgroup$
    – badjohn
    Commented Apr 21, 2019 at 8:36
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    $\begingroup$ @Peter I think the basic problem is that the definition of normality is not a property of a number, it is a property of a particular representation of a number. Can it even be proved that if a number is normal represented in one base it is normal in every base? (And if that conjecture is false, the whole topic is going to fall apart in a mass of special cases, and seems more like numerology than math.) $\endgroup$
    – alephzero
    Commented Apr 21, 2019 at 15:32
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    $\begingroup$ @alephzero Indeed, I don't find properties dependent on the decimal representation, or any other particular one, as interesting as properties independent of the representation. I think that it is possible for a number to be normal in one base but not another or normal in all bases but I need to check. $\endgroup$
    – badjohn
    Commented Apr 21, 2019 at 15:50

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