Challenge: "Dividing" a number above 0 and ending up with the same, or a greater number (creative task) Here's a question/challenge for those of you who know quite a bit about math, or enjoy to be creative with what you do know (just for reference: I'm virtually illiterate when it comes to any math more advanced than the basic basics, and you're probably way more educated than me about this subject :9 ):
By breaking something in half, how can I end up with the two halves to each have the same value as the one I started with (before breaking it up)?
Our goal is take a number with a value higher than 0, divide it into parts (in halves, in thirds, in quarters, it really doesn't matter), and end up with each part to have the same value as, or higher than, the number we started with. 
So, for example, imagine starting with 1. 
Then you divide it by two, and instead of ending up with 1/2, you'll have 1.
Now, conceiving and believing that this is possible, what would you do to achieve the desired result?
If division doesn't allow this, then what kind of mathematical method, practice or theory would you use?
In advance: Thank you very much for your input! 
It is greatly appreciated :)
 A: Since the word "creative" appears in the title, I'll assume some "creative license" :)


*

*If you take the roman numeral one (simplified) $$\mathrm{I}$$ and divide it in half (cutting it horizontally) you get two small $\mathrm{I}$'s.

*We can define a set-theoretic "operator" by $aS:=\{ax:x\in S\}$ which multiplies all elements in the set by $a$. Then, for example, $$\frac{\mathbb{Q}}{2}=\mathbb{Q}$$ where $\mathbb{Q}=\{a/b:a,b\ne 0\in \mathbb{Z}\}$ is the rationals.

*In computer science, many algorithms are analyzed for efficiency using the Big O notation, which measures the time that a process will take asymptotically. It holds that $${1\over 2}\mathcal{O}(n)=\mathcal{O(n)}$$

*In logic, and especially Boolean Algebra, the "sum" of two Truth values obeys certain properties: $$T+T=T$$ where $T$ is "true". This meets your condition in a sense.

*Take the infinite series $$S=2+2+2+\cdots$$ Then $$\begin{align}S/2&=1+1+1+\cdots \\ &=(1+1)+(1+1)+\cdots \\ &=2+2+2+\cdots \\ &=S\end{align}$$
That should give you some ideas to investigate! (Note: some of the claims above are a bit sketchy... they are not necessarily wrong, but you may be interested in learning about more rigorous definitions such as the idea of convergence. Cheers!)
A: If you consider the number $x > 0$
as a point on the number line,
then you can linearly map
both the intervals
$[0, x/2)$
and
$[x/2, x)$
1-1 and onto the interval
$[0, x)$.
A: Perhaps something like this might fit in with the concept.

(source)
Note that this is more of an illusion than a real trick. You're shortening the bar a little bit when you do this.
A: This is impossible within the fields or rational ($\mathbb{Q}$), real ($\mathbb{R}$) or complex ($\mathbb{C}$) numbers. It comes down to finding a solution $x \neq 0$ of $$
   \frac{x}{n} = x
$$
which implies $n = 1$.
In finite fields of numbers - those are finite sets of numbers which obey the usual rules of arithmetic, including the possibility of dividing by arbitrary non-zero numbers - we can find numbers that behave somewhat as you describe. These finite fields have the property that there always exists some $n$ such that $\underbrace{1 + \cdots + 1}_{n\text{ times}} = 0$. The simplest such field contains only the numbers $0$ and $1$, and is governed by the rules $$
  0 + 0 = 0,\, 0 + 1 = 1 + 0 = 1,\, 1 + 1 = 0,\, 0\cdot 0 = 0\cdot 1 = 1\cdot 0 = 0,\, 1\cdot 1 = 1 \text{.}
$$
This implies that $$
 1 + 1 + 1 = 1
$$
meaning that we have, in a way, split $1$ into the three parts $1$, $1$ and $1$, whose sum are again $1$.
