Consider the block diagram above which descibes the evolution of a point

$(x,y)\in \mathbb{N}\times \mathbb{N}$. Suppose

$\mathrm{\Delta }$ is drawn from a random distribution with values spanning

$[0,2\pi )$. What happens to the value of the radius (r) over time?

First, I'll show the radius never increases (Lemma 1). Then, I'll show for all possible values of the radius, there exists

$\mathrm{\Delta }$ which results in the radius decreasing (Lemma 4). Thus, given sufficent time the radius goes to zero.

Lemma 1: The radius never increases.

$$\begin{array}{rl}{r}_n& =\sqrt{x_n^2+y_n^2}\\ & =\sqrt{(r_n\cos (\theta ){)}^2+(r_n\sin (\theta ){)}^2}\\ & =r_n\sqrt{\cos (\theta {)}^2+\sin (\theta {)}^2}\\ & =r_n\sqrt{\cos (\theta +\mathrm{\Delta }{)}^2+\sin (\theta +\mathrm{\Delta }{)}^2}\text{(by Pythagorean identity)}\\ & =\sqrt{(r_n\cos (\theta +\mathrm{\Delta }){)}^2+(r_n\sin (\theta +\mathrm{\Delta }){)}^2}\\ & \ge \sqrt{\lfloor r_n\cos (\theta +\mathrm{\Delta }){\rfloor }^2+\lfloor r_n\sin (\theta +\mathrm{\Delta }){\rfloor }^2}\\ & =r_{n+1}\end{array}$$

Lemma 2: If

$x\notin \mathbb{N}$, then

$\lfloor x^2\rfloor <x^2$.

Proof. If

$x\notin \mathbb{N}$, then

$x$ has a non-zero decimal component, so flooring

$x$ decreases its magnitude.

Lemma 3: If

$x$ is not irrational and

$k$ is irrational, then

$x\times k\notin \mathbb{N}$.

Proof. Because

$x$ is not irrational,

$x$ can be expressed as the ratio of two natural numbers

$x\equiv \frac{a}{b}$. Suppose

$x\times k\equiv m\in \mathbb{N}$, then

$k\equiv \frac{m}{x}\equiv \frac{m}{\frac{a}{b}}\equiv \frac{m\times b}{a}$, but then

$k$ is not irrational as it can be expressed as the ratio of two natural numbers.

Lemma 4: For all

$r_n$ there is a

$\mathrm{\Delta }$ for which

$r_n>r_{n+1}$.

Proof.

$r_n$ is irrational or not.

Case I:

$r_n$ is irrational. Setting

$\mathrm{\Delta }=-\theta$:

$$\begin{array}{rl}{r}_n& =\sqrt{(r_n\cos (\theta +\mathrm{\Delta }){)}^2+(r_n\sin (\theta +\mathrm{\Delta }){)}^2}\\ & =\sqrt{(r_n\cos (0){)}^2+(r_n\sin (0){)}^2}\\ & =\sqrt{r_n^2}\\ & >\sqrt{\lfloor r_n{\rfloor }^2}\text{(by Lemma 2)}\\ & =r_{n+1}\end{array}$$

Case II:

$r_n$ is not irrational. Setting

$\mathrm{\Delta }=\frac{\pi }{4}-\theta$:

$$\begin{array}{rl}{r}_n& =\sqrt{(r_n\cos (\theta +\mathrm{\Delta }){)}^2+(r_n\sin (\theta +\mathrm{\Delta }){)}^2}\\ & =\sqrt{(r_n\cos (\frac{\pi }{4}){)}^2+(r_n\sin (\frac{\pi }{4}){)}^2}\\ & =\sqrt{(r_n\sqrt{2}/2{)}^2+(r_n\sqrt{2}/2{)}^2}\text{(by Lemma 3, }{r}_n\frac{\sqrt{2}}{2}\notin \mathbb{N})\\ & >\sqrt{\lfloor r_n\sqrt{2}/2{\rfloor }^2+\lfloor r_n\sqrt{2}/2{\rfloor }^2}\text{(by Lemma 2)}\\ & =r_{n+1}\end{array}$$