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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. $10 + 1$)

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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 "0" correspond to value $a$ and "1" correspond to value $b$ ($a,b > 0$). 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.

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  • $\begingroup$ Thanks for the comment! My mistake, I've never used "one-to-one" when talking about functions, and I decided that it means "bijective" without checking it. Let me fix it. $\endgroup$ – Dmitry Jul 18 '20 at 20:19
  • $\begingroup$ It became even simpler) $\endgroup$ – Dmitry Jul 18 '20 at 20:24

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