Is sqrt(2) a normal number?

In a paper from 2018, Pierpaolo Uberti writes:

In this paper we study the property of normality of a number in base 2. A simple rule that associates a vector to a number is presented and the property of normality is stated for the vector associated to the number. The problem of testing a number for normality is shown to be equivalent to the test of geometrical properties of the associated vector. The paper provides a general approach for normality testing and then applies the proposed methodology to the study of particular numbers. The main result of the paper is to prove that an infinite class of numbers is normal in base 2. As a further result we prove that the irrational number √2 is normal in base 2.

And here’s the background:

Given an integer b ≥ 2, a b-normal number (or a normal number) is a number whose b-ary expansion is such that any preassigned sequence of length k ≥ 1 occurs at the expected frequency 1. . . .

The interest in studying normal numbers lies not only in their randomness but also in the fact that they are extremely difficult to identify and obscure in many other aspects. Despite the appeal of the concept, its trivial interpretation and the proof that almost all real numbers are normal, the proof for given irrational numbers to be normal in some base is still elusive. . . .

For many years, ever since I heard about the idea of the distribution of the decimal (or, in this case, binary) expansion, I’ve wondered if sqrt(2) is a normal number. Indeed, I’ve always felt that if there was one mathematical theorem more than any other I’d like to prove, it’s this.

I was reminded of this topic yesterday when reading this comment by X, so I googled *Is sqrt(2) a normal number?* and came across the above-linked paper.

The question is, is Uberti correct? I guess the answer is no, he’s just bullshitting. Or, to put it another way, he’s probably correct that sqrt(2) is a normal number, but he’s incorrect that he’s proved it. I say this because:

1. The article is from 2018 on Arxiv but it still hasn’t been published anywhere, which would seem unlikely if it’s really a proof of this longstanding conjecture.

2. Another quick google search led to this short online discussion with complete skepticism. Nobody there goes to the trouble of completely dismantling Uberti’s argument but only because they don’t seem to feel it’s worth the time (in contrast, to, say, really bad statistical arguments which are worth shooting down because of their real-world implications).

3. The author is not a mathematician and he seems to have published nothing in pure math.

So . . . too bad! We still don’t know if sqrt(2) is a normal number.

I’m bummed. But, I guess, my bad for getting fooled by that just for a moment. We always have to remember: dead-on-arrival papers don’t just get published by Statistical Science, Journal of Personality and Social Psychology, and SSRN. They also appear on Arxiv.

As to the answer to the question posed in the title of this post: I think it’s gotta be Yes! But it sounds like the result may never be proved.

21 thoughts on “Is sqrt(2) a normal number?

  1. I found Wikipedia helpful:
    “In mathematics, a real number is said to be simply normal in an integer base b[1] if its infinite sequence of digits is distributed uniformly in the sense that each of the b digit values has the same natural density 1/b. A number is said to be normal in base b if, for every positive integer n, all possible strings n digits long have density b−n.”
    https://en.wikipedia.org/wiki/Normal_number

    But that raises an obvious question: Why not use the following definition instead?
    “In mathematics, a real number is said to be simply normal in an integer base b[1] if its infinite sequence of digits is distributed *according to the Gaussian distribution*.”

  2. As a mathematician I have mixed feelings about this. On the one hand it’s clear the author doesn’t really know what they are doing. For instance, to answer the question of where he goes wrong it’s in proposition 3 where he tries to claim that if in the limit the number of ones in x equals the number of zeroes in x then x is normal in base 2. This is confused in two ways. First, both sides will be infinite provided x’s binary expansion isn’t eventually all 0 or 1 (eg provided x isn’t of the form n/2^k for some natural numbers n,k). Moreover, even if you moved the equality under the limit the claim is trivially refuted by considering 1/3= (in binary) .01010101…. which is clearly non-normal but has the property indicated (once you decode all his inadvertently obfuscatory notation).

    On the other I’m happy to see people trying to understand and explore mathematical problems and we all start somewhere. This is very much the kind of mistake a student could make in an introduction to proofs course (most ppl find the shift to thinking rigorously about math quite hard) and (as you note) it poses no danger to legitimate mathematics. Any mathematician or math adjacent PhD should be able to tell in a couple minutes it’s not worth their time.

    The more I think about it the more I want to encourage the author to keep trying but suggest he spend a bit more time working through problems in a real math textbook first. If he is reading this I can’t promise I’ll have the time to respond but if he wants to email me at my last name at invariant.org I’d be happy to take a few minutes to explain in more detail where he goes wrong and suggest a book or two for him to work through.

    • What a good response!

      I’ve learned a lot of math from trying (and failing) to prove soemthing I thought the real pro’s had missed. Even if my conjectures were just fantasy, they gave me a motivation to learn the definitions, etc. carefully that I would never have had solving end-of-chapter problems or just reading without an objective.

    • he tries to claim that if in the limit the number of ones in x equals the number of zeroes in x then x is normal in base 2.

      What does it mean that sqrt(2) is not “normal”? That means there is information to be gained by computing it to enough decimal points? Obviously I am thinking Carl Sagan’s Contact book, and where the burden of proof should be.

    • As someone who enjoys toying with math problems from time to time, I appreciate this response. Yes, it’s good to have math amateurs; at the very least, we help dispel the notion that math is inaccessible or uninteresting. In addition, digging into a problem helps you understand its difficulties. I only recommend that amateurs clearly identify themselves as such, so as not to cause confusion. And that they take the time to work through problems in a good textbook, so as to become familiar with a base of knowledge.

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