Is the number $0.112358132134...$ rational or irrational?

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

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

Solution 1:

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.

Solution 2:

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.