Proving that $a_i,b_i \to 0$ where $a_{i+1} := |b_i - a_i|$ and $b_{i+1} := |a_i - a_{i+1}|$?

Question

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?

Answer 1

Prove, by induction that

$$a_n = \left(\sqrt{2} - 1\right)^n, ~~b_n = \left(\sqrt{2} - 1\right)^{n-1} - \left(\sqrt{2} - 1\right)^n. \tag1 $$

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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 $n \to \infty, b_n \to 0.$

Answer 2

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 $i,j$.