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Every vector $v$ in a finite-dimensional vector space space $V$ of dimension $n$ over a field $F$ has a unique representation in terms of a basis ${\frak B} \subseteq V$, where a basis for $V$ is a set of $n$ linearly independents vectors in $V$. Intuitively, a vector $v \in V$ doesn't have an "absolute representation" (except perhaps as the symbol "$v$"), but only "relative representations" in terms of other vectors in $V$ together with scalars in $F$.

A vector $v \in V$ and a basis $\frak B := \{b_0, b_1, \cdots, b_{n-1}\}$ yield a unique $n$-nuple of scalars $(s_0, s_1, \cdots, s_{n-1}) \subseteq F^n$ with the property that $v$ is the linear combination

$$ v = s_0 b_0 + s_1 b_1 + \cdots s_{n-1} b_{n-1}. $$

The $n$-tuple $(s_0, s_1, \cdots, s_{n-1})$ is sometimes called the coordinate vector of $v \in V$, although I like to think of it as $(s_0, s_1, \cdots, s_{n-1})_{\frak B}$ to emphasize that its only relationship to $v$ is through the basis $\frak B$.


Every integer $d$ in the commutative ring of integers ${\mathbb Z}$ has a unique representation in terms of a base $b \in {\mathbb Z_{>1}}$ where a base for ${\mathbb Z}$ is an integer larger than $1$. Intuitively, an integer $d \in {\mathbb Z}$ doesn't have an "absolute representation" (except perhaps as the symbol $d$), but only "relative representations" in terms of other integers in ${\mathbb Z}_{>1}$ together with integers in ${\mathbb Z}$.

An integer $d$ and a base $b$ yield a unique $n$-tuple of integers $(s_0, s_1, \cdots, s_{n-1})$ with the property that $d$ is the "non-linear combination"

$$ d = s_0 b^0 + s_1 b^1 + s_2 b^2 + \cdots + s_{n-1} b^{n-1}. $$

The integer $d$ is then written down as $s_{n-1} \ldots s_2s_1s_0$, or $s_{n-1} \ldots s_2s_1s{_0}_{b}$ to clarify that $b$ is the base.


Now, I know that thinking of the integers as a vector space (or an algebra, or a curve, or something) over a field (or ring) is absurd (since they don't satisfy the definitions), but are these similarities just a coincidence, or is there something to say about it?

(Wikipedia says that it'd be really convenient if the integers could be constructed as a curve over $F_{un}$.)

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    $\begingroup$ I think base expansions of integers doesn't have anything to do with bases of vector spaces. It may just be a manifestation of $\mathbb Z$ being a Euclidean domain (en.wikipedia.org/wiki/Euclidean_domain) $\endgroup$ – Qfwfq Jul 18 '18 at 10:11
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    $\begingroup$ The coefficients of the expansion are only unique up to a choice of representatives for $\mathbb{Z}/b\mathbb{Z}$, and the base expansions themselves are a result of the fact that $\mathbb{Z}$ is a discrete valuation ring, for various valuations. $\endgroup$ – Somatic Custard Jul 18 '18 at 12:13
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    $\begingroup$ This blog post on dyadic models may be relevant. $\endgroup$ – MTyson Jul 18 '18 at 15:50

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