Proving that $a_i,b_i \to 0$ where $a_{i+1} := |b_i - a_i|$ and $b_{i+1} := |a_i - a_{i+1}|$? Let $a_0 := 1$ and $b_0 := \sqrt{2}$, and define
\begin{align*}
a_{i+1} &:= |b_i - a_i| \\
b_{i+1} &:= |a_i - a_{i+1}|
\end{align*}
Prove that $\lim_{i \to \infty}{a_i} = 0$ and $\lim_{i \to \infty}{b_i} = 0$.
Context: I am given the set $L := \{a+b\sqrt{2} : a,b\in \mathbb{Z}\} \subset \mathbb{R}$, and I want to prove it is NOT a lattice over $\mathbb{R}$, since it is not discrete.
My idea to solve it was to build a sequence converging to $0$ and I came up with this sequence.
Intuitively/empirically seems to be converging to $0$, but I am having trouble proving it formally. Does anyone have any idea or hint?
 A: Prove, by induction that
$$a_n = \left(\sqrt{2} - 1\right)^n, ~~b_n = \left(\sqrt{2} - 1\right)^{n-1} - \left(\sqrt{2} - 1\right)^n. \tag1 $$

Empirically true for $n=1$.
$$a_{n+1} = |b_n - a_n| = |\left(\sqrt{2} - 1\right)^{n-1} \times [ 1 - 2(\sqrt{2} - 1) ]|$$
$$ = |\left(\sqrt{2} - 1\right)^{n-1}  \times (3 - 2\sqrt{2})|$$
$$ = |\left(\sqrt{2} - 1\right)^{n-1} \times \left(\sqrt{2} 
- 1\right)^2| = \left(\sqrt{2} - 1\right)^{n+1}.$$
Therefore,
$$b_{n+1} = |a_{n+1} - a_{n}| = \left(\sqrt{2} - 1\right)^n - \left(\sqrt{2} - 1\right)^{n+1}.$$
So, (1) above is proven.
Further, it is immediate from (1) above that as $n \to \infty, a_n \to 0.$
Finally, $b_n$ may be re-written
$$\left(\sqrt{2 - 1}\right)^{n-1} [1 - (\sqrt{2} - 1)] = 
(2-\sqrt{2})a_{n-1}.$$
So, as $n \to \infty, b_n \to 0.$
A: If all you want is to show $L$ is discrete, it's much simpler than Dirichlet's theorem: the sequence $\{x_n=a_n+b_n\sqrt{2}| b_n = n, a=-\lfloor b\sqrt 2\rfloor\}_{n=0}^\infty$ is an infinite bounded subset of $[0, 1)$. So $L$ cannot be discrete. To be more precise, it contains a convergent subsequence $\{x_{n_i}\}$, then $|x_{n_i}-x_{n_j}|>0$ can be arbitrarily small for suitable $i,j$.
