How to decide whether a language is decidable when not involving turing machines?
It might come as a surprise to you, but the language $$L=\{k\in\mathbb N :\text{the binary expansion of} \sqrt2\text{ contains $k$ consecutive $1$s}\}$$ is decidable, even if we could not construct an algorithm that can demonstrate whether the binary expansion of $\sqrt2$ contains $k$ consecutive $1$s or not given arbitrary $k$.

As Yonatan N indicated, let us take a closer look at $L$. There are two disjoint cases.


$L$ is $\mathbb N$. Then $L$ is trivially decidable. For any given $k$, just return "Yes".
$L$ is not $\mathbb N$. Then there are some numbers in $\mathbb N$ that is not in $L$. Let $m$ be the minimum of such number. Then $L$ must be $\{1, 2, 3, \cdots, m-1\}$. Why?


Since $m&gt;2$ is the minimum number that is not in $L$, $m-1$, which is a number smaller than $m$, must be in $L$. That is, $\text{bin}(\sqrt 2)$ contains $m-1$ consecutive $1$s. So it also contain $k$ consecutive $1$s for all $k\le m-1$.
Since $\text{bin}(\sqrt 2)$ does not contains $m$ consecutive $1$s, it does not contain $k$ consecutive $1$s for any $k\gt m$. That is, $k\not\in L$.


Now we can construct an algorithm that decides $L$. For any given $k$, check whether $k\lt m$. If yes, return "Yes". Otherwise, return "No".


In either case, there exists an algorithm that decides $L$. Hence $L$ is decidable.



Most people, I believe, felt a bit disoriented the first time when this kind of proof/conclusion was encountered. Or at least myself.

The essential point is we do not have to identify/construct/bind to one algorithm that decides $L$.  We do not have to understand fully what is $L$. All we need is there exists an algorithm that decides $L$, whatever $L$ turns out to be. This deviates from "the naive sense of decidability" that you might have even before you encountered the theory of computation/decidability/computability. 

In particular, I do not know whether $L$ is $\mathbb N$ or not. Some people claimed to have proved that $L=\mathbb N$. I have not checked their proof, yet. Although it is certainly interesting to find whether $L=\mathbb N$, its result will not change the fact that $L$ is decidable.