So most p-adic notes inevitably beat us with a club with Ostrowski's theorem on absolute values.

However what if we forsake absolute values, and instead use a digit system where the value $v(x)$ of a number $x$ equals a tuple $(a,b)$. Here $a$ is the smallest negative exponent of some prime $p$ in the representation of $x$, while $b$ is the smallest positive exponent. For instance for $p=2$, $v(9/4)$ = $v(1/4 + 2)$ = $(2,1)$.

However we allow infinite expansions too:

We define a sequence as 'left-convergent' if for $\lim v_2(x_n - x_{n-1})$ the first tuple value approaches infinity, and likewise for 'right-convergent'.

In other words, high positive or negative powers of a prime both converge to different zeros. And therefore we can write unconditionally convergent expressions of the form $\sum_{-\infty}^{\infty} p^n a_n$ where $a_n$ is between $0$ and $p-1$.

So we basically obtain real (base $p$) plus p-adic 'combined' numbers.

The number zero can e.g. be represented as either:

$\cdots 0.0 \cdots$


$\cdots (p-1)(p-1)(p-1).(p-1)(p-1)(p-1) \cdots$

Since the latter is divisible by all factors of $p-1$, for $p>2$ we aren't dealing with a field (at least by the usual definition) as there are zero divisors. Though I don't know what happens with $p=2$?

Something like $\frac{1}{x}$ for $(x,p)=1$ will have multiple solutions. With the 'new' solutions corresponding to linear combinations of the real and p-adic solution.


$\frac{1}{3}$ for $p=2$ has representations:

$0.010101 \cdots$ (real)

$\cdots 0101011;$ (2-adic)

$\cdots 010101.1010101 \cdots$ (sum of both divided by two)

This sort of feels like 'field extension' since we are adding new solutions to an equation.

In these numbers, series like $\displaystyle \sum_{n=0} p^{(-1)^n n}$ converge to finite values. Also power series converge absurdly: $\sum_{n=-\infty}^{\infty} x^n = 0$ for all $x$ with a power of $p$ in the numerator or denominator.

How about products? (responding to Julian Rosen). In general we can use Cauchy products if at least one of the numbers has a finite representation. However multiplying an infinite p-adic and infinite real number together causes convergence issues. If we denote $\frac{1}{x}_p$ and $\frac{1}{x}$ for the p-adic and real representatives of a fraction. Then $\frac{1}{x}_p \cdot \frac{1}{x} = \frac{1}{x}_p \cdot \frac{x}{x^2} = \frac{1}{x^2} = \frac{x}{x^2}_p \cdot \frac{1}{x} = \frac{1}{x^2}_p$. Hence by implication there can't be a 'single' representative for this product.

My questions are:

  1. Does this approach make sense/does it have a name?
  2. Does it have any benefits or unique advantages?

E.g. tying p-adic results to real ones? For instance 'hypothetically' if we can demonstrate that the 'combined' expression (real + p-adic) is irrational and the real solution is rational, then the remaining term must be irrational.

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    $\begingroup$ What is the object / the number $x$ in: "...and instead use a digit system where the value of a number $v(x)$ equals a tuple..." ?! (This is just a first unclear point, appeared immediately after reading the line with Ostrowski's Theorem, which is not needed any longer...) $\endgroup$
    – dan_fulea
    Oct 18, 2021 at 17:16
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    $\begingroup$ If I wanted to "combine" $p$-adics and reals, my next construction is the direct product $\mathbb Q_p \times \mathbb R$. I can embed $\mathbb Q$ into it diagonally, i.e. represent $1/3$ as $(1/3, 1/3)$. I can also put some generalization of the absolute value on it via $(x,y) \mapsto (\lvert x\rvert_p, \lvert y \rvert_{\text{real}})$ and already catch more information than your construction there. It's up to you to show me how your construction beats my $\mathbb Q_p \times \mathbb R$ when solving some question. $\endgroup$ Oct 18, 2021 at 17:31
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    $\begingroup$ I think it's a quite interesting question based on the common motivations given for $p$-adic numbers as "base $p$ expansions that can go on forever in the wrong direction." Totally natural to ask: "why not both?" $\endgroup$
    – hunter
    Oct 18, 2021 at 18:31
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    $\begingroup$ You are probably fishing for the ring of Adeles. It does combine real, complex and p-adic numbers, and is quite useful in the theory of arithmetic groups. $\endgroup$ Oct 18, 2021 at 18:58
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    $\begingroup$ When are two expressions $\sum_n a_n p^n$ (with $a_n\in\{0,1,\ldots,p-1\}$) equal? Also, you speak of zero divisors, but is it clear how to multiply two combined numbers? $\endgroup$ Oct 19, 2021 at 4:18


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