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Just out of curiosity, is the number $0.112358132134...$ a rational or irrational number?

$...$ stands for Fibonacci sequence not repeating decimals!

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    $\begingroup$ It depends on what the dots stand for ;) If it's supposed to be the number you get from concatenating the decimal expansions of the Fibonacci numbers, I'm pretty sure it's irrational, but I don't see an easy proof of it. $\endgroup$ – Daniel Fischer Sep 25 '13 at 17:49
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    $\begingroup$ Assuming the Fibonacci numbers are the sequence, it is most likely irrational, and a possible proof might go along these lines: experiments indicate that somewhere in the Fibonacci sequence there is a number ending in any number of 9s - I have not yet proved this, but if so, that would show that it cannot be recurring. $\endgroup$ – Old John Sep 25 '13 at 17:58
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    $\begingroup$ Though you can now see proofs that it is irrational, there is an interesting and somewhat related article you might check out if you can find a copy: An Unanticipated Decimal Expansion. Allen Schwenk, Math Horizons , Vol. 20, No. 1 (September 2012), pp. 10-12, jstor.org/stable/10.4169/mathhorizons.20.1.10. $\endgroup$ – Benjamin Dickman Sep 25 '13 at 18:09
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    $\begingroup$ A follow-up question for you to consider, is the case where we don't concatenate, but actually do the addition to the previous term, so we get $0.11235954 ... $ where $9=8+1$ (1 comes from 13), $5=3+2$ (2 comes from 21), $4 = 1 + 3$ (3 comes from 34). $\endgroup$ – Calvin Lin Sep 25 '13 at 18:15
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For the sake of clarity, some definitions:
An immediately periodic sequence is a sequence $\{a_i\}$ such that $a_i = a_{i+p}$ for all positive integers $i$. (This is often what people mean by periodic.)
An eventually periodic sequence is a sequence $\{a_i\}$ along with an integer $N$, such that $a_i = a_{i+p}$ for all positive integers $i\geq N$. (Some people might consider this periodic.)

We will show that the number is irrational.

Claim: For all integers $n$, there is a squence of $m\geq n$ 0's followed by 1 1's which appear.

Proof: Recall that the Fibonacci numbers mod $k$ is an immediately periodic sequence for any $k$.
This claim follows by considering the Fibonacci sequence mod $10^{n+1}$, and remembering that the $F_{1}$ term is $1$. $_\square$

Corollary: The decimal concatenation of the Fibonacci numbers is never eventually periodic, hence it is not rational.

Corollary: The starting values of $F_1, F_2$ are not important, as long as they are non-zero.


For completeness, let's prove the fact that I asked you to recall.

First, we show that the Fibonacci numbers mod $k$ is eventually periodic mod $k$.
Simply consider all pairs $(F_i, F_{i+1}) \pmod{k}$ for $ i =1 $ to $k^2+1$. There are $k^2 + 1$ such pairs, but only $k^2$ distinct possibilities. Hence, by the Pigeonhole Principle, one of these pairs must repeat.

Suppose that $(F_i, F_{i+1} ) \equiv (F_j, F_{j+1} ) $. Then, for all integers $x$, we can inductively show that $F_{i + x} \equiv F_{j-x} \pmod{k}$. Hence the sequence is eventually periodic.

We can also show inductively that $F_{i-x} \equiv F_{j-x} \pmod{k}$. Hence, the sequence is immediately periodic.


This was the first proof, which has a slight hole as Nate pointed out. Easily patched, but I prefer the proof at the top.

Claim: For all $n$, there is a sequence of $m \geq n$ 9's that appear.

Proof: Recall that the Fibonacci numbers mod $k$ is an immediately periodic sequence for any $k$.
This claim follows by considering the Fibonacci sequence mod $10^n$, and remembering that the $F_{-2}$ term is $-1$.
So eventually, there is a (positive) term that is equal to $-1 \pmod{10^n}$. The term could have additional 9's, which is why I have $m \geq n$.$_\square$

Claim: For all $n$, there is a sequence of $m \geq n$ 0's that appear.

Proof: Same as above since $ F_0 = 0$. $_\square$

Hence, there exists exists an arbitrary string of 9's, the only way that it can be repeating is for it to be all 9's. But we showed that this is not possible, since arbitrary strings of 0's also appear. Hence, it is irrational.

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  • $\begingroup$ What is "an immediately repeating sequence"? $\endgroup$ – Did Sep 25 '13 at 18:02
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    $\begingroup$ I suppose to be complete, you should rule out the possibility that the number is of the form $0.11235\dots99999999\dots$, i.e. ends with infinitely many 9s, i.e. is a multiple of $10^{-n}$ for some $n$. $\endgroup$ – Nate Eldredge Sep 25 '13 at 18:03
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    $\begingroup$ @Nate: The sum of two numbers ending in $9$ doesn't end in $9$. $\endgroup$ – mjqxxxx Sep 25 '13 at 18:12
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    $\begingroup$ @BenjaminDickman No, apply a similar argument as above. I didn't need to take $F_{-2}$. In fact, I could have taken $F_1$, and showed there there must be $m$ 0's and then a 1. Hence, immediately not periodic. $\endgroup$ – Calvin Lin Sep 25 '13 at 18:18
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    $\begingroup$ @CommanderShepard Hm, its actually quite elementary, but I can see how it sounds advanced if you are 14. Do you agree with the following facts: A) a decimal is rational if and only if it is eventually periodic B) a sequence in which we know it has "01", "001", "0001", ... appearing somewhere in it is never eventually periodic C) The Fibonacci numbers modulo $k$ is immediately periodic (proof written above). $\endgroup$ – Calvin Lin Sep 25 '13 at 18:55
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Nonrigorous argument: As $n \to \infty$ $F_n \to \varphi F_{n-1}$. So given a very high N we can say $F_n \approx \lfloor(\varphi^{n-N}F_N)\rfloor$ $\forall n > N$. Since $\varphi$ is irrational, whenever $F_n$ has one more digit than $F_{n-1}$ there won't be any pattern to the extra digit, so $0.F_0F_1F_2...$ won't ever repeat.

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