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An oft-used phrase describing irrational numbers is that their (decimal) expansions "never repeat".

The sense of "never repeating" intended is, of course, that their expansions don't repeat forever. And it's straight-forward to show that rational numbers do repeat forever.

It is also easy to show that irrational numbers never repeat forever because that would make them rational.

Now, the question: Suppose I define "never repeating" as simply meaning that the first N digits of the expansion (in whatever base you like) are not repeated. I.e. positions 1..N are not the same as positions (N+1)..2N. Clearly I can construct transcendental numbers with that property. But if we restrict ourselves to irrational algebraic numbers, can it be shown that there are any numbers in that set that "never repeat" in this sense?

It seems to me that even for a randomly chosen irrational algebraic number, the probably that it fulfils this property very very quickly becomes infinitessimally small. I.e. the first billions digits will not match the next billion digits. However, having looked at the first N digits, the next N digits could still be anything, and even though the probability shrinks as p^-N, there are sill infinitely many opportunities as N grows.

So, my question is: can it be shown that there exists a (real) irrational algebraic number for which it is never true that digits 1..N are the same as digits (N+1)..2N for any N?

Related:How to know that irrational numbers never repeat?

ThePopMachine
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  • Since any irrational algebraic number is an irrational number, therefore, whatever applies to irrational numbers applies to them as well. – vidyarthi Nov 19 '18 at 17:36
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    It's conjectured that every irrational algebraic number is [Normal](https://en.wikipedia.org/wiki/Normal_number), which would imply that no example of what you want can occur. Very little has been proven, though. – lulu Nov 19 '18 at 17:37
  • @vidyarthi: Irrational numbers include transcendentals, and it's clear that I can construct a transcendental with this property anytime I want. – ThePopMachine Nov 19 '18 at 17:38
  • @lulu: I didn't know the term "normal", but I sort of assumed it when I talked about the probability of this property. But I don't see how that leads us any close to showing what I asked. – ThePopMachine Nov 19 '18 at 17:39
  • oh sorry! I didnt see that you want the property for every number $n$. Maybe, as lulu said, such a number does not exist – vidyarthi Nov 19 '18 at 17:42
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    Study the definition. It says that any specified $N$ digits occurs infinitely often (and the density is specified). – lulu Nov 19 '18 at 17:42
  • @vidyarthi, no not, every number. Just any such number and any selection of the first N digits. – ThePopMachine Nov 19 '18 at 17:43
  • @ThePopMachine I meant for every natural number $n$ where it corresponds to the number of digits – vidyarthi Nov 19 '18 at 17:44
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    @lulu: I didn't ask if a particular string of N digits occurs again. I asked if the first N digits match the next N digits, for any N at all. – ThePopMachine Nov 19 '18 at 17:44
  • @vidyarthi: I don't follow what you mean. Can you give a complete sentence for me to agree or disagree with? – ThePopMachine Nov 19 '18 at 17:46
  • Somewhat surprisingly it is easy point at **rational** numbers that never repeat w.r.t. your definition. Say $1/6$. Or even $1/2$ :-) – Jyrki Lahtonen Nov 19 '18 at 17:46
  • I see. Right...in that case, I agree. I don't immediately see how normality settles the matter, even heuristically. – lulu Nov 19 '18 at 17:46
  • @JyrkiLahtonen: Heh heh-- yeah, true. That doesn't enlighten me about irrational algebraics at all :( – ThePopMachine Nov 19 '18 at 17:47
  • Incidentally, is there a term for algebraic numbers that are not rational? – ThePopMachine Nov 19 '18 at 17:48
  • $\sqrt 2$ repeats :-) – TonyK Nov 19 '18 at 17:49
  • @TonyK:. Yeah, some obviously do. – ThePopMachine Nov 19 '18 at 17:50
  • You know, I just realized that if the claim is true for a number for some string of digits in some base, then it's also true that the first two digits in some higher base are the same. I don't know if that helps to think about it. – ThePopMachine Nov 19 '18 at 17:51
  • @lulu: You know, thinking about this more, along with what I just said, instead of testing for longer and longer strings, we can just go to higher and higher bases. And this means that if we assume normality, then as we go to higher bases b, the probability of it "repeating" goes with only 1/b, not with (1/b)^N. So the probability of a number "not repeating" is Product over b=2..inf of (b-1)/b. Does this help? I don't know. – ThePopMachine Nov 19 '18 at 17:58
  • Yeah...maybe. I don't see anyway to get a handle on it, myself. – lulu Nov 19 '18 at 18:01
  • @JyrkiLahtonen: Actually, not for 1/6 and 1/2 because 1/6 starts with 0.00 in base 2 and 1/2 = 0.111... in base 3.... – ThePopMachine Nov 19 '18 at 19:07
  • Oh. I missed the part of this needing to happen in every base. Then there won't be any rational examples. They are purely periodic in any base coprime to the denominator. – Jyrki Lahtonen Nov 19 '18 at 19:19
  • @JyrkiLahtonen: Fundamentally, it seems silly ask a question this is partial to a particular base. – ThePopMachine Nov 19 '18 at 19:52
  • @ThePopMachine. I've been looking at this (rather belatedly) but I'm not altogether sure what's wanted. Are you asking for a positive real number whose first two digits are different in every number base? If so, my answer is that (unless we allow 1.0 and 2.0) there is no such number, whether rational or irrational, algebraic or transcendental. However, you say (Nov 19 at 17:38) you can produce a transcendental number that does have the required property, so maybe I've misunderstood the question. – Michael Behrend Nov 24 '18 at 17:45
  • @MichaelBehrend: I can produce a transcendental number that obeys this property in some base. The question is whether there can be a number that obeys it in no base. And is algebraic and irrational. You claim there can be no such transcendental. What's the rationale? – ThePopMachine Nov 26 '18 at 16:10

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Still not sure if we're talking about the same thing, but this is what I found.

Let $r$ be a non-negative real. If $r \ne 1, 2$ then there is a number base in which the first two digits of $r$ are the same. The proof below is partly by computer program.

If $r \ge 3$ the result is clear. Suppose $2 < r < 3$ and let $f$ be the fractional part of $r$. Then for a number base $b \ge 3$, the representation of $r$ begins $2.2$ iff $f \in [2/b, 3/b)$. Since $b > 2$ we have $3/(b + 1) > 2/b$, so the intervals for $b = 3, 4, \ldots$ cover the whole of $(0, 1).$

If $1 < r < 2$ the proof is similar, using bases $b \ge 2$.

The case $0 \le r < 1$ is the most interesting. If $b \ge 2$ is a number base and $d$ is a digit such that $0 \le d < b$, then the representation of $r$ in base $b$ begins $.dd$ iff $$r \in [(bd + d)/b^2, (bd + d + 1)/b^2).$$ So we have intervals $[0, 1/4)$, $[3/4, 1)$, then $[0, 1/9)$, $[4/9, 5/9)$, $[8/9, 1)$, etc. The question is, do these cover the interval $[0, 1)$? The answer isn't obvious, so I wrote a computer program to find out (using integer arithmetic throughout, to avoid rounding errors). The answer is that the intervals for $2 \le b \le 50$ cover $(0, 1)$, and $50$ is the least upper limit that will work.

In case this unexpected result was due a bug, I wrote programs to generate random reals and random rationals in $(0, 1)$ and look for a base in which the first two digits were the same. Such a base could always be found, and the maximum base required over $10^8$ trials was $50$.

Michael Behrend
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  • If I understand this, you are asserting that for any real number, there is a base in which the first two digits are the same. It order to answer my original question then, you just need to include my original observation in comments above. I.e. that if the first N digits match the next N digits in a particular base b, then this is equivalent to the first two digits in base b^N being the same. Maybe not too deep, but it brings it back to the original question. – ThePopMachine Nov 27 '18 at 00:28