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I observed a strange anomaly in the sequence of bases of a number system.

For every $n$-base, $n \geq 2$, it holds that its number symbols are written with numerals which are all less than $n$. For example, a binary ($2$-ary) base has number symbols which consist only of $1$s and $0$s.

The anomaly manifests when we investigate $n = 1$. By the above pattern, its number symbols should consist only of $0$s. But if this were so, the only number we could express would be $0$. Hence, we have to break the pattern and allow $1$ to appear in our number symbols. We have to break another pattern, namely that the number symbols of a $n$-base consist precisely of $n$ different types of numerals, if we want $0$ to be representable in our system. That is, if we allow only $1$s in our number symbols, we cannot represent $0$.

For clarity, here is a sample definition of the value of number symbols, and an example from my imagined unary system. $$ (k_m k_{m-1} \dots k_1 k_0)_n = \sum_{i=0}^m n^i k_i $$ Then, the number symbols of the unary system are the same as that of the binary, with the difference that in the unary system, they denote a natural number equal to the number of $1$s which occur in them. For example, $(1101)_1$ is $3$.

I find it surprising that the elegant pattern described above completely breaks down at $n = 1$. Do we have analogous examples, or a deeper reason that this happens? Thank you for your input! :)

God bless
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    Technically you don't need the digit $0$, as the number $0$ can be written as an empty string of digits. That is, $0 = ()_1$ – eyeballfrog Jul 09 '23 at 12:56
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    @eyeballfrog Good idea, though my definition of valuation does not allow for that. It would also be hard to write empty strings on paper. :) Is this trick ever used in serious mathematics? – God bless Jul 09 '23 at 13:00
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    It does sort of show up in computational complexity theory. There unary is used to make inputs exponentially longer, artificially lowering the complexity of an algorithm, which can be helpful in some proofs. In that context, the empty input is indeed used for $0$, though complexity theory is more concerned with behavior under very large inputs rather than small ones. – eyeballfrog Jul 09 '23 at 13:10
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    I don't understand your question. Base-$n$ system is only defined for $n\ge 2$. The "unary" system is not a Base-$1$ system, it is somewhat different. So no wonder the "pattern" breaks for $n=1$: we've already broken the pattern by defining it differently than for any other $n$'s. Had we not broken the pattern, all we could represent would be the number $0$. And all of this is already known to you. It is like breaking a chair and then asking "why is this chair broken?". –  Jul 09 '23 at 13:35
  • @StinkingBishop You can word the issue this way, but my question remains: why is it the case that this nice property for number systems only holds for $n \geq 2$? – God bless Jul 09 '23 at 13:46
  • How should the unary base system be useful in proofs ? It is an utterly artificial thing which I would not even call number system. – Peter Jul 09 '23 at 13:53
  • I feel like "if this were so, the only number we could express would be $0$" is the answer to "why is it the case that this nice property for number systems only holds for $n\ge2$?". If you think it would bring closure for me to write an answer saying that and that for an integer $n$ to have positive integers below it, we need $n\ge2$, I could. And if this doesn't answer your question, maybe it can help you clarify what you would like to know. – Mark S. Jul 09 '23 at 15:39
  • Base 1 is not a positional numeration system, therefore cannot be compared with them. 2) Besides,"a strange anomaly" is not very different from an "anomaly" (without adjective) :)
    • – Jean Marie Jul 09 '23 at 15:54
  • @Godbless This is off topic but also relevant because both bring up the notion of paraconsistent math, where something is allowed to be both true and not true. Several years ago I was trying to make a list of axioms for ultrafinite math, which is where there is a largest number. Let's call it $K$. Working in unary, it made sense to have an axiom where $K + 1 = $. I really could not think of any way of making a system of axioms without incomplete expressions. – Teg Louis Jul 09 '23 at 19:42