Consider the block diagram above which descibes the evolution of a point $(x, y) \in \mathbb{N}\times\mathbb{N}$. Suppose $\Delta$ 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 $\Delta$ which results in the radius decreasing (Lemma 4). Thus, given sufficent time the radius goes to zero.

**Lemma 1**: The radius never increases.

$$
\begin{aligned}
r_n &= \sqrt{x_n^2+y_n^2} \\
&= \sqrt{(r_n\cos(\theta))^2 + (r_n\sin(\theta))^2} \\
&= r_n\sqrt{\cos(\theta)^2 + \sin(\theta)^2} \\
&= r_n\sqrt{\cos(\theta+\Delta)^2 + \sin(\theta+\Delta)^2} \quad\text{(by Pythagorean identity)} \\
&= \sqrt{(r_n\cos(\theta+\Delta))^2 + (r_n\sin(\theta+\Delta))^2} \\
&\geq \sqrt{\lfloor r_n\cos(\theta+\Delta)\rfloor^2 + \lfloor r_n\sin(\theta+\Delta)\rfloor^2} \\
&= r_{n+1}
\end{aligned}
$$

**Lemma 2**: If $x\not\in\mathbb{N}$, then $\lfloor x^2\rfloor\lt x^2$.

_Proof_. If $x\not\in\mathbb{N}$, 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 $x\times k \not\in \mathbb{N}$.

_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 $\Delta$ for which $r_n\gt r_{n+1}$.

_Proof_. $r_n$ is irrational or not.

**Case I**: $r_n$ is irrational. Setting $\Delta = -\theta$:

$$
\begin{aligned}
r_n &= \sqrt{(r_n\cos(\theta+\Delta))^2 + (r_n\sin(\theta+\Delta))^2} \\
&= \sqrt{(r_n\cos(0))^2 + (r_n\sin(0))^2} \\
&= \sqrt{r_n^2} \\
&\gt \sqrt{\lfloor r_n\rfloor^2} \quad\text{(by Lemma 2)} \\
&= r_{n+1}
\end{aligned}
$$

**Case II**: $r_n$ is not irrational. Setting $\Delta=\frac{\pi}{4}-\theta$:

$$
\begin{aligned}
r_n &= \sqrt{(r_n\cos(\theta+\Delta))^2 + (r_n\sin(\theta+\Delta))^2} \\
&= \sqrt{(r_n\cos(\frac{\pi}{4}))^2 + (r_n\sin(\frac{\pi}{4}))^2} \\
&= \sqrt{(r_n\sqrt2/2)^2+(r_n\sqrt2/2)^2} \quad\text{(by Lemma 3, } r_n\frac{\sqrt2}{2}\not\in\mathbb{N}) \\
&\gt \sqrt{\lfloor r_n\sqrt2/2\rfloor^2 + \lfloor r_n\sqrt2/2\rfloor^2} \quad\text{(by Lemma 2)} \\
&= r_{n+1}
\end{aligned}
$$