No, the different $p$-adic number systems are not in any way compatible with one another.
A $p$-adic number is a not a number that is $p$-adic; it is a $p$-adic number. Similarly, a real number is not a number that is real, it is a real number. There is not some unified notion of "number" that these are all subsets of; they are entirely separate things, though there may be ways of identifying bits of them in some cases (e.g., all of them contain a copy of the rational numbers).
Now, someone here is bound to point out that if we take the algebraic closure of some $\mathbb{Q}_p$, the result will be algebraically isomorphic to $\mathbb{C}$. But when we talk about $p$-adic numbers we are not just talking about their algebra, but also their absolute value, or at least their topology; and once you account for this they are truly different. (And even if you just want algebraic isomorphism, this requires the axiom of choice; you can't actually identify a specific isomorphism, and there's certainly not any natural way to do so.)
How can we see that they are truly different? Well, first let's look at the algebra. The $5$-adics, for instance, contain a square root of $-1$, while the $3$-adics do not. So if you write down a $5$-adic number which squares to $-1$, there cannot be any corresponding $3$-adic number.
But above I claimed something stronger -- that once you account for the topology, there is no way to piece the various $p$-adic number systems together, which the above does not rule out. How can we see this? Well, let's look at the topology when we look at the rational numbers, the various $p$-adic topologies on $\mathbb{Q}$. These topologies are not only distinct -- any finite set of them is independent, meaning that if we let $\mathbb{Q}_i$ be $\mathbb{Q}$ with the $i$'th topology we're considering, then the diagonal is dense in $\mathbb{Q}_1 \times \ldots \times \mathbb{Q}_n$.
Put another way -- since these topologies all come from metrics -- this means that for any $c_1,\ldots,c_n\in\mathbb{Q}$, there exists a sequence of rational numbers $a_1,a_2,\ldots$ such that in topology number 1, this converges to $c_1$, but in topology number two, it converges to $c_2$, and so forth. (In fact, more generally, given any finite set of inequivalent absolute values on a field, the resulting topologies will be independent.)
So even on $\mathbb{Q}$, the different topologies utterly fail to match up, so there is no way they can be pieced together by passing to some larger setting.
I have done something like this with ordinary degree students. I will just show you what I have done it may give you some ideas.
I think the better students enjoyed the projects but some of them might have been too long.
The project titles were:
- Nowhere-Continuous Functions
- Intermediate Value Theorem
- Fixing Nasty Functions and making them Nice
- Continuity
- Leaving Cert Questions
- L’Hopital’s Rule
- Linear Algebra
- Differentiation
- Extrema of Functions of Several Variables with an Application to Statistics
- Riemann Sums
- Further Techniques of Trigonometric Integration
- The Natural Exponential Function as a Power Series
and The Most Beautiful Formula in Mathematics
- The Area & Volume of familiar Shapes & Solids
- Euler’s Method of Numerical Solution of Differential
Equations
- The Length of the Monza Circuit
- Dynamical Systems
You can find projects 1-9 here, projects 10-15 here and project 16 is question 3 in here.
Best Answer
Here is a very concrete question that can be explained by appealing to the $p$-adic continuity of addition and multiplication. We don't even need completions: the problem takes place entirely in the rational numbers and is not a fake problem in any sense.
We can form binomial coefficients $\binom{r}{n}$ when $r$ is not necessarily an integer, and this is important because they occur in the coefficients of the power series for $(1+x)^r$ in calculus. Their formula, for $n \geq 1$, is $$ \binom{r}{n} = \frac{r(r-1)\cdots(r-n+1)}{n!}. $$ If you look at the expansion for $\sqrt{1+x}$ and for $\sqrt[3]{1+x}$, corresponding to $r = 1/2$ and $r = 1/3$, the series start off as $$ 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \frac{7}{256}x^5 + \cdots $$ and $$ 1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^3 - \frac{10}{243}x^4 + \frac{22}{729}x^5 + \cdots $$ The surprise is that the denominators are entirely powers of 2 in the first case and 3 in the second case. Think about that: $\binom{1/3}{5}$ involves division by $5!$, but the 2 and 5 factors cancel out. As a more extreme example, $\binom{-3/22}{7} = -\frac{1071892575}{39909726208}$ and $39909726208 = 2^{11}11^7$. Even though the definition of $\binom{-3/22}{7}$ involves division by $7!$, the primes that survive in the denominator seem to have nothing to do with $7!$ and everything to do with the denominator of $-3/22$.
Claim: For $n \geq 1$ and nonzero rational $r$, if a prime $p$ is in the denominator of $\binom{r}{n}$ then $p$ is in the denominator of $r$.
Proof: We show the contrapositive. If $p$ is not in the denominator of $r$ then $|r|_p \leq 1$, so the denominator of $r$ is invertible modulo any power of $p$, and therefore $r$ is a $p$-adic limit of positive integers, say $r = \lim_{k \rightarrow \infty} a_k$ with $a_k \in {\mathbf Z}^+$. That is a $p$-adic limit. By $p$-adic continuity of addition and multiplication (and division), we get $\binom{r}{n} = \lim_{k \rightarrow \infty} \binom{a_k}{n}$, another $p$-adic limit. By combinatorics we know $\binom{a_k}{n}$ is a positive integer, so $|\binom{a_k}{n}|_p \leq 1$. The $p$-adic absolute value on ${\mathbf Q}$ is $p$-adically continuous, so $|\binom{r}{n}|_p = \lim_{k \rightarrow \infty} |\binom{a_k}{n}|_p \leq 1$. Thus $p$ is not in the denominator of $\binom{r}{n}$. QED
The special case $r = 1/2$ can be explained in terms of Catalan numbers: $\binom{1/2}{n} = (-1)^{n-1}C_{n-1}/2^{2n-1}$, where $C_{n-1}$ is the $(n-1)$th Catalan number (a positive integer). Therefore the denominator of $\binom{1/2}{n}$ is a power of 2. For the general case, I know of no argument that explains why primes in the denominator of $\binom{r}{n}$ must be primes in the denominator of $r$ in such a clean way as this $p$-adic method.
The converse of the claim is also true: for $n \geq 1$ and nonzero rational $r$, if a prime $p$ is in the denominator of $r$ then $p$ is in the denominator of $\binom{r}{n}$. That is, if $|r|_p > 1$ then $|\binom{r}{n}|_p > 1$. More precisely, if $|r|_p > 1$ then $|\binom{r}{n}|_p \geq |r|_p^n$, so in fact $|\binom{r}{n}|_p \rightarrow \infty$ as $n \rightarrow \infty$. Let's leave that as an exercise. (The data for the coefficients of $\sqrt{1+x}$ and $\sqrt[3]{1+x}$ suggest that perhaps the sequence $|\binom{r}{n}|_p$ is monotonically increasing if $|r|_p > 1$, and that too can be proved in general by looking at the $p$-adic absolute value of the ratio $\binom{r}{n+1}/\binom{r}{n}$.) In particular, for $n \geq 1$ the denominator of $\binom{1/2}{n}$ is a power of $2$ other than $1$.