The number system Simple question(or maybe not...): Its quite fascinating that with only 10 symbols (for base 10) we can represent any possible natural number. In particular, it seems that no matter how you arrange any two or more numbers, you always get a number. For example: "1233" + "56" + "0000122" = "1233560000122". The same does not apply for the Roman numeral system because of the 3-rule. You can't have "I" + "I" + "I" + "I" = "IIII" because that's and illegal statement. My question is, where can I find a formal proof that the number system we have is "unbreakable" or the opposite, which could be possible as well.
 A: This comes from the division remainder theorem, which tells that for any two integers a, b we can write a = k*b + r where r < b. Hence, for any chosen base we can call every number between 0 and h an algarism, and concatenate them to produce any natural number. Any string of digits will then represent a multiple of the base plus some remainder. This is simply not true for roman numerals, which do not have a fixed base. 
A: I think of decimals as a sort of weighted inclusion. To more easily see this, consider binary representations. Any natural number, then, can be written as a finite string of from the alphabet $0,1$. Let us define a sequence $(s_n)_{n\geq0}$ where $s_n \in \{0,1\}$ with the stipulation that for large enough $n, s_n =0$. Denote by $S$ the set of all such sequences. Then, there is a natural bijection between $\mathbf{N}$ and $S$. Since, $n = \sum_{k=0}^{\infty}s_k 2^k.$ To prove that this is a bijection is fairly routine; one would the division remainder theorem from Felipe Jacob's post.
However, if we write out the sequence $(s_n)$ to the left then we get something of the form $$\dots00000s_m s_{m-1} \dots s_3 s_2 s_1 s_0$$ Where $s_m =1$ is the last nonzero bit. Since the sequence contains infinitely many zeros at the end, we can refrain from writing them, and then have a finite string of symbols, which are more familiarly recognized as binary expansions. Hence, $5 = 101_2$. Where the $_2$ just reminds us that the RHS is a binary expansion. 
Thus, we can see that binary expansions are a sort of weighted inclusion, the element $s_k$ of the sequence $(s_n)$ tells you what the value of $2^k$ is in the related natural number. We know that $\infty$ is not a natural number, so each $(s_n)$ sequence can contain only finitely many $1$s else it would be infinite.
More  generally we see that nothing is really unique about $2$ here, and that any natural number $s$ can be used to form base-$s$ representation of natural numbers on the alphabet $\{0,1,\dots,s-1\}$.
Thus, we have for $10$, $T$ the set of all sequences $(t_n)_{n \geq 0}$ such that $t_n \in \{0,1,\dots,9\}$ with $t_n =0$ for large enough $n$. We can use a similar expansion, $n=\sum_{k=0}^{\infty} t_k 10^k$, to represent any number, but now notice our notion of weighted inclusion becomes more meaningful. For each $k$ there is a corresponding weight $t_k$ to which the number $10^k$ plays a role in representing $n$. Again, we can write out our strings (to the left) as
$$\dots0000t_m t_{m-1} \dots t_3 t_2 t_1 t_0$$
where now $t_m \neq 0$ is the last nonzero digit. Then, we lop off the zeros at the end, and have finite strings which are recognized as numbers in base-$10$.
Now, for your question, you've asked about concatenating numbers. Let us define $\circ$ as operation which takes two sequences $(s_n)$ and $(t_n)$ with the same base. Let $m_s, m_t$ be the place of the last  nonzero bits of $(s_n)$ and $(t_n)$. Now, define a new sequence $(u_n) = (s_n) \circ (t_n)$ by:
$$
  u_n = \left\{\def\arraystretch{1.2}%
  \begin{array}{@{}c@{\quad}l@{}}
    t_n & \text{if $n \leq m_t$}\\
    s_{n-m_t-1} & \text{if $n > m_t$}\\
  \end{array}\right.$$
For example, $24 \circ 12 =2412$. 
Now, Roman numerals follow their own patterns, which are sufficiently different from those of base-$k$ representations. Since Roman numerals have unique symbols for larger number, we have no hope of concatenation: e.g. XII $\circ$ MC $=$ XIIMC ?? It doesn't make sense, from a purely grammatical standpoint.  
I'll leave it to you to play around with this, if you'd like, but I hope what I wrote has helped or at least been interesting.
