Irrational fractional parts - patterns Some fractional-part list plots are:


$\text{listplot of }|[\pi x]-\pi x|\text{, for }x \in \mathbb{Z} \text{ and } \text{listplot of }|[ex]-ex|\text{, for }x \in \mathbb{Z}$


$\text{listplot of }|[\sqrt{2} \  x]-\sqrt{2} \  x|\text{, for }x \in \mathbb{Z}\text{ and } \text{listplot of }|[x/300]-x/300|\text{, for }x \in \mathbb{Z}$
Any rational multiple of $x$ seems to yield a zigzag pattern as in last plot, whereas irrational multiples yield different patterns. Not only this, but irrational multiples of a square yield fairly chaotic plots as shown below.

$\text{listplot of }|[x^2\sqrt{2}]-x^2\sqrt{2}|\text{, for }x \in \mathbb{Z}$
Is this always the case, or are there exceptions to the rule? What is the explanation for these patterns, and is this a very loose test for irrationality?
NB Mathematica code:
ListPlot[Abs[Round[Table[N[ m Sqrt[2]], {m, 1, 1000}]] - 
Table[N[m Sqrt[2]], {m, 1, 1000}]]]

Update
Crisscross $\pi$ pattern is replicable by rational multiples (eg $12414/238$) - but doesn't stand up to the $x^2$ test.
 A: I have very little idea about what happens in the $x^2$-plots, but I have a few thoughts on the other plots. For the remainder of this answer we define for each fixed number $t$ the $t$-plot function:
$$
f_t(x)=|\text{round}(t\cdot x)-t\cdot x|,\quad x\in\mathbb Z
$$
Only fractional parts matter
Like stated in the comments we may define an equivalence relation by $a\sim b\iff a-b\in\mathbb Z$. Suppose now that $a=m+q$ with $m\in\mathbb Z$ and fractional part $q\in[0,1)$. Then for $x\in\mathbb Z$ we get
$$
x\cdot a =\underbrace{x\cdot m}_{\text{in }\mathbb Z}+x\cdot q\sim x\cdot q
$$
showing that $x\cdot a$ has the same fractional part as $x\cdot q$. Since the $t$-plots depend only on the fractional parts of the multiples of $t$, this shows that all numbers equivalent to $q$ will produce the exact same $t$-plot as $q$.
Irrationals resemble rationals arbitrarily well
Suppose the plot has dimensions $[0,n]\times[0,0.5)$. Each pixel on the screen has a height $h$ corresponding to a rational subsection of the interval $[0,0.5)$. Assuming that $t$ is irrational it is evident that $f_t(x)$ must be so too. Then for each integer $x\in[0,n]$ the value of $f_t(x)$ will be plotted in the $k$-th row from the bottom where $k$ is determined by the property that
$$
f_t(x)\in(k\cdot h,k\cdot h+h)
$$
Now, fixing a particular $x$, we may find an $\varepsilon_x>0$ such that $f_{t-\varepsilon_x}(x)$ and $f_{t+\varepsilon_x}(x)$ are both contained this interval. Thus defining $\varepsilon=\min_x(\varepsilon_x)$ for all the integers $x\in[0,n]$ we have found an $\varepsilon>0$ having this property for all $x$-values in our plot. Thus we may choose a rational number $t'\in(t-\varepsilon,t+\varepsilon)$ which is then bound to have function values $f_{t'}(x)$ in the same $h$-intervals as $f_t(x)$ did thus producing an identical plot to that of $f_t(x)$ on the screen.
The short story is that for a given number of points $n$ and a given resolution given by the pixel height $h$ it is possible to find a rational number $t'$ close enough to the irrational number $t$ so that they produce the exact same plots on the screen.
More will follow soon...
