Is there a way to map the concatenation operation over strings to the addition operation over $\mathbb{N}$ Given an alphabet, say $\Sigma = \{0,1\}$, I can make a one-to-one mapping from all possible strings $x \in \Sigma^*$ to $\mathbb{N}$. This could be done by ordering $\Sigma^*$ lexicographically and assigning the $i$th string $x_i$ to number $i \in \mathbb{N}$.
But given strings $x_i,x_j \in \Sigma^*$, is there any special mapping such that the concatenation operation $f:\Sigma^* \rightarrow \Sigma^* | (x_i,x_j) \rightarrow x_ix_j$ is also related to the usual addition performed over the corresponding indices $i,j \in \mathbb{N}$ to which $x_i$ and $x_j$ are mapped ?
For instance, if I assign the character $\{1\}$ to the number $1$, and string $x$ is assigned the number $10$, is there a mapping such that the string $x1$ is assigned the number $11$ ? (i.e. $10 + 1$)
Yes, if you don't want an injective mapping: just assign to all strings value 0.
No, if you want an injective mapping:

Empty string must correspond to value $0$. Otherwise,
$$value(\epsilon) = value(\epsilon \cdot \epsilon) = 2 value(\epsilon),$$
- contradiction when $value(\epsilon)\ne 0$.

Let &quot;0&quot; correspond to value $a$ and &quot;1&quot; correspond to value $b$ ($a,b &gt; 0$). Then
$$value(0^b) = b \cdot value(0) = ab = a \cdot value(1) = value(1^a)$$
- contradiction, since both $0^b$ and $1^a$ correspond to the same value.