Show given property for a sequence $a_{n}$ The sequence $a_n$ is defined as 
$ a_0$ is an arbitrary real number,
$ a_{n+1}$ = $\lfloor a_{n}\rfloor$ ($a_{n} - \lfloor{a_{n}}\rfloor$)
Show that for every $ a_0$:
$$\exists m\geq0, \forall n \geq m, a_{n+2}= a_n$$
Floor function $\lfloor x \rfloor$, example, $\lfloor 3.2 \rfloor = 3$ and $\lfloor -3.2 \rfloor = -4$
Here is my attempt: [link]
What I have noticed is that due to the floor function denoted as $\lfloor x \rfloor$ all of these sequences will approach zero. I am not sure if this sequences has a divergent property of periodically switching between a few particular elements, but maybe.
Though why I think it approaches zero:
Let $ a_0= 16.2 \Rightarrow 
a_1 = 16 (16.2 - 16) = 16 \cdot 0.2 = 3.2 \Rightarrow 
a_2 = 3 (3.2 - 3) = 3 \cdot 0.2 = 0.6 \Rightarrow 
a_3 = 0 (0.6 \cdot 0) = 0$
 A: The case $a_n \geq 0$ (i.e. $a_0 \geq 0$) is straightforward as $a_{n+1} = \lfloor a_n\rfloor(a_n-\lfloor a_n\rfloor) \leq \lfloor a_n\rfloor \leq a_n$ so $a_n$ is decreasing and is bounded below by $0$ so by the monotone convergence theorem the sequence converges to the only possible (positive) limit-point $a_n = 0$. By the definition of convergence there is a $N$ such that $0 < a_n < 1$ for $n\geq N$. But then $a_{N+1} = 0$ and $a_n = 0$ for all $n > N$.
The tricky case is $a_0 < 0$. First note that we can write $a_n = -m - \delta$ where $\delta \in [0,1)$ and $m \in\mathbb{N}$. Depending on the value of $\delta$ relative to $m$ we can have different behavior: 


*

*If at some point $\delta = 0$ then $a_n = 0$ from that point onward. Likewise if $\delta = \frac{1}{m+2}$ then
$$a_{n+1} = -(m+1)\left(1-\frac{1}{m+2}\right) = - m - \frac{1}{m+2} = a_n$$
and we have $a_n = -m-\frac{1}{m+2}$ from that point onward. 

*If $\delta \in \left(\frac{1}{m+2},1\right)$ then $$|a_{n+1}| = |m+1|(1-\delta) < m + \delta = |a_n|$$ so the sequence is smaller in absolute value at the next step.

*If $0 < \delta < \frac{1}{m+2}$ then 
$$a_{n+1} = -(m+1)(1 - \delta) = -m - \delta'~~~\text{where}~~~\delta' = 1 - (m+1)\delta~~\text{so}~~~\delta' \in \left(\frac{1}{m+2},1\right)$$
and it follows that
$$|a_{n+2}| = (m+1)^2\delta < m+\delta = |a_n|~~~\text{since}~~~\delta < \frac{1}{m+2}$$
so the sequence is smaller in absolute value after two steps.

*Finally if $-1 \leq a_n  < 0$ then $a_{n+1} = -1 - a_n$ and $a_{k+2} = a_k$ holds for all $k\geq n$.
Combinding the results above we can construct a subsequence $\{a_{n_k}\}_{k=1}^\infty$ that is decreasing in absolute value. Take $n_1 = 0$ and for $k\geq 1$ we take $n_{k+1} = n_k + 1$ if $|a_{n_k+1}| \leq |a_{n_k}|$ and $n_{k+1} = n_k + 2$ otherwise. 
If this subsequence at some point has $-1 < a_{n_k} < 0$ then it will start to alternate so $a_{n+2} = a_n$ holds for all $n\geq n_k$. If this does not happen then by the monotone convergence theorem this subsequence must converge. There are two options: first if $a_{n_k}$ ever becomes an integer then the sequence converges to $0$ on the next step (and it follows that $a_n = 0$ for all $n > n_k$). Otherwise the subsequence must converge to one of the fixpoints on the form $-m-\frac{1}{m+2}$ for some $m\in\mathbb{N}$. As we will show below this is impossible as these fixpoints are repelling. 
Let $e_n = a_n - \left(-m-\frac{1}{m+2}\right)$ and assume that $e_n\to 0$ which guarantees the existence of a $N$ such that $|e_n| < \epsilon = \frac{1}{m+2}$ for $n\geq N$. This choice of $\epsilon$ guarantees that $\lfloor a_n \rfloor = -(m+1)$ for all $n\geq N$ and the recursion gives us $e_{n+1} = -(m+1)e_n$ and by induction $e_{n+k} = (-1)^k (m+1)^k e_n$ for all $k\geq 1$. Taking $k\to \infty$ we get a contradiction unless $e_n = 0$ which is only possible if $a_0 = -m -\frac{1}{m+2}$ a case we already have covered.
In conclusion we see that $a_{n+2} = a_n$ holds for all sufficiently large $n$.
A: If your function is $a_{n+1} = \lfloor a_n \rfloor (a_n - \lfloor a_n \rfloor)$ (on the photo, but not the question)
Then
If $a_0>0$ 
$\lfloor a_n \rfloor < a_n$ and $ (a_n - \lfloor a_n \rfloor)<1$
$0 \le a_{n+1} < \lfloor a_n\rfloor] < a_n$ and $\lfloor a_{n+1}\rfloor \le \lfloor a_n \rfloor - 1$
The integer component is falling by at least 1 with every iteration.
eventually it must fall below 0. And which point your sequence equals 0 thereafter.
If $a_0 < 0$ the pattern is more complicated
it is possible for $a_{n+1}$ to jump to an integer and then to $0.$
if $1<a_n<0$ 
$a_{n+1} = -1 - a_n$ which is in $(-1,0)$
$a_{n+2} = a_n$
And a stable oscillation emerges.
$a_n < -1$ while it is possible that $|a_{n+1}| \ge |a_n|$ but in these cases $\lfloor a_{n+1} \rfloor = \lfloor a_{n} \rfloor$ and $a_{n+1} - \lfloor a_{n+1} \rfloor < a_{n} - \lfloor a_{n} \rfloor$ increasing the likelihood that $|a_{n+2}|<|a_{n+1}| - 1$ 
The sequence either heads to $0$, or it falls into a regular oscillation.
