# Is $123456788910111121314\cdots$ a $p$-adic integer?

On the back of this question comes the natural question of whether the string $$1234567891011121314\!\cdots$$ is even a number at all. While that sort of question is vague, given the lack of generic definition for the word "number", I would feel comfortable answering this question in the affirmative if we knew that it were a $p$-adic number.

Now, my first impressions are that this number is $10$-adic integer (although it is not quite as easy to show this as I initially thought). However, it seems rather unlikely that it is $p$-adic for any prime $p$. Does anyone know how to show that it is or isn't $p$-adic, or if there are similar questions which have been answered— I mean, the $p$-adicity of strings like $2481632\!\cdots$ or $23571113\!\cdots$?

(Sidebar: I believe that if it is a $p$-adic number, it has to be a $p$-adic integer, but I admit that I could be mistaken here.)

EDIT: mixedmath's answer shows that if the question is interpreted with $1$ as the "leftmost digit" then the question makes no sense. However, KCd points out that this notation is frequently used when we intend $1$ to be the "rightmost digit". In this case, the question becomes formalizable (probably), and almost surely much more interesting.

Therefore, the question becomes whether or not the following $10$-adic integer is $p$-adic for some prime $p$:

$$\sum_{k=0}^\infty k\exp_{10}\left(k+\sum_{i=1}^k\lfloor\log(i)\rfloor\right)$$

where $\exp_{10}(x)$ is a [formal] power $10^x$. This probably still doesn't make perfect sense formally, but it is at least easy to imagine formalizing it. We might reasonably interpret it to be a statement about the existence of sequences $b_m\in\Bbb Z_p$ and $e_m,\, f_m,\, g_m\to\infty$ such that the finite sums agree up to a point:

$$\sum_{m=0}^{f_N} b_m p^m \equiv \sum_{k=0}^{g_N} k\exp_{10}\left(k+\textstyle\sum\lfloor\log(i)\rfloor\right) \qquad \text{mod} \exp_{10}(e_N)$$

for all $N\in\Bbb N$. Perhaps something regarding rearrangements would work a little bit better (for this formalism I worry about some inessential objections regarding instability of the ones digit).

If anyone would like to answer this other interpretation of the question, I would give a bounty for it.

• A string of digits is not any kind of definite "number" at all since it is not clearly defined unless you explain what your notation means. For instance, if I wrote .23232323... with the periodic repetition of 2 and 3, what is this? Well, if it means a number in base 10 then it is 23/99. If it means a number in base 11 it equals 23/120 (usual decimal notation). If it means a number in base 47 it equals 1/96 (usual decimal notation). If you want your expression to be a $p$-adic integer then it can be for any $p > 9$. There is no universal meaning once and for all of that string of digits. – KCd Jan 3 '15 at 4:02
• @KCd: If the string is a $10$-adic integer, then I find it hard to believe that there is ambiguity about what this string means (I admit I could be mistaken, but $\sum_{m\geq k} a_m 10^m$ seems pretty unambiguous to me). I posted the question under this assumption before realizing that I was not able to find an $a_m$ sequence that was obviously correct. So perhaps it is not $10$-adic, in which case I agree that the question doesn't make sense, and I would accept an answer along those lines. – Eric Stucky Jan 3 '15 at 4:05
• aaand two minutes later I get an answer along those lines :) – Eric Stucky Jan 3 '15 at 4:12
• If you meant for the string to be $1 + 2 \cdot 10 + 3\cdot 10^2 + \cdots + 9 \cdot 10^8 + 1 \cdot 10^9 + 0 \cdot 10^{10} + 1 \cdot 10^{11} + \cdots$, then yes it is a $10$-adic integer. But the same exact notation could mean an $11$-adic integer $1 + 2 \cdot 11 + 3 \cdot 11^2 + \cdots$ and so on. It is all a matter of context. In practice $10$-adic integers are very rarely used. – KCd Jan 3 '15 at 4:17

This is not an $n$-adic integer or a $p$-adic integer. A pivotal idea of a $p$-adic integer is that it can be well-represented in a sort-of-base-$p$, $$x = \sum_{k \geq \ell}a_kp^k,$$ and in particular it makes sense to speak of the first coefficient, the coefficient of $1$. This amounts to understanding what the number is mod $p$, which is very dependent on knowing what the "last digit" is. But your string does not have a well-defined last digit.
It is for this reason that you might see $p$-adic integers written in the form $$\cdots a_3a_2a_1,$$ where the expansion is on the left. In such a way, we have well-defined final digits, and so we can find any of the finite $p$-adic expansions.
• I think it is a fair interpretation to take the first digit to be $1$, the next to be $2$, and so on. It is common to write $p$-adic integers left to right also (try to write them in a computer algebra package :)). I agree that being more explicit about the built-in powers of 10 in the expression would take away any ambiguity. – KCd Jan 3 '15 at 4:20
• @KCd: ah, this is something I hadn't considered. Yes, if we consider the number as secretly being written left to right instead of right to left, then this is a $p$-adic number and integer, matching your comments to the OP above. – davidlowryduda Jan 3 '15 at 4:22