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Informally speaking, taking the limit of two's complement as the number of bits goes to $\infty$, the integers are just the eventually constant binary sequences (which are naturally represented by finite binary sequences). For this to work, said sequences must start with the least significant bit, i.e., $1001011\overline{0}$ is interpreted as $2^0+2^3+2^5+2^6$ and $1001010\overline{1}$ is interpreted as $2^0+2^3+2^5-2^7$. The arithmetic and ordering of these strings is natural (and efficient for microprocessors when we restrict from $\mathbb{Z}$ to, say, $\{-2^{63},\ldots,2^{63}-1\}$).

The above can be reinterpreted as the following less direct construction. If $R$ is the inverse limit of rings $\lim_{\infty\leftarrow n}\mathbb{Z}/2^n\mathbb{Z}$, then the diagonal map $\Delta\colon\mathbb{Z}\rightarrow R$ given by $m\mapsto \lim_{\infty\leftarrow n}(m\mod 2^n)$ is an injective ring homomorphism. [Edit: The image is characterized as the set of $\vec x\in R$ for which the truth value of $x(n+1)=x(n)$ is eventually constant.] Moreover, the ordering of $\mathbb{Z}$ is coded via $m\geq 0\Leftrightarrow(m\mod 2^n: n\in\mathbb{N})$ is eventually constant.

Update: I couldn't resist the temptation to write a functional programming implementation.

2 added 130 characters in body

Informally speaking, taking the limit of two's complement as the number of bits goes to $\infty$, the integers are just the eventually constant binary sequences (which are naturally represented by finite binary sequences). For this to work, said sequences must start with the least significant bit, i.e., $1001011\overline{0}$ is interpreted as $2^0+2^3+2^5+2^6$ and $1001010\overline{1}$ is interpreted as $2^0+2^3+2^5-2^7$. The arithmetic and ordering of these strings is natural (and efficient for microprocessors when we restrict from $\mathbb{Z}$ to, say, $\{-2^{63},\ldots,2^{63}-1\}$).

The above can be reinterpreted as the following less direct construction. If $R$ is the inverse limit of rings $\lim_{\infty\leftarrow n}\mathbb{Z}/2^n\mathbb{Z}$, then the diagonal map $\Delta\colon\mathbb{Z}\rightarrow R$ given by $m\mapsto \lim_{\infty\leftarrow n}(m\mod 2^n)$ is an injective ring homomorphism. [Edit: The image is characterized as the set of $\vec x\in R$ for which the truth value of $x(n+1)=x(n)$ is eventually constant.] Moreover, the ordering of $\mathbb{Z}$ is coded via $m\geq 0\Leftrightarrow(m\mod 2^n: n\in\mathbb{N})$ is eventually constant.

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Informally speaking, taking the limit of two's complement as the number of bits goes to $\infty$, the integers are just the eventually constant binary sequences (which are naturally represented by finite binary sequences). For this to work, said sequences must start with the least significant bit, i.e., $1001011\overline{0}$ is interpreted as $2^0+2^3+2^5+2^6$ and $1001010\overline{1}$ is interpreted as $2^0+2^3+2^5-2^7$. The arithmetic and ordering of these strings is natural (and efficient for microprocessors when we restrict from $\mathbb{Z}$ to, say, $\{-2^{63},\ldots,2^{63}-1\}$).

The above can be reinterpreted as the following less direct construction. If $R$ is the inverse limit of rings $\lim_{\infty\leftarrow n}\mathbb{Z}/2^n\mathbb{Z}$, then the diagonal map $\Delta\colon\mathbb{Z}\rightarrow R$ given by $m\mapsto \lim_{\infty\leftarrow n}(m\mod 2^n)$ is an injective ring homomorphism. Moreover, the ordering of $\mathbb{Z}$ is coded via $m\geq 0\Leftrightarrow(m\mod 2^n: n\in\mathbb{N})$ is eventually constant.
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