Does radix 1 exists? Introduction
This probably sounds a bit weird to ask, but the question is, is there such a base that is base 1.
One thing we know is that this base will contain one symbol only, but will this symbol represent something i.e. a value, or nothing?
If we try to see a connection between other bases, for instance base 10, 4, 3, 2, we see following:


*

*base 10 {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

*base 4  {0, 1, 2, 3}

*base 3  {0, 1, 2}

*base 2  {0, 1}


So base 1 should have the symbol $0$, in order to fit this pattern.
Number representation
Let's say that we introduce a new symbol, which will be used in base 1, namely $\alpha$. We will probably think that if we list the first numbers in base 1, we will get something like:
$$\alpha,\alpha\alpha, \alpha\alpha\alpha, \alpha\alpha\alpha\alpha$$
Which is quite logical. But, if we observe other systems, such as base 10, we will see that 10 is actually built up of two symbols. Similarly, for radix 2, $10_2$ is equal to 2, which appears to be the base.
So, would it not now be logical to claim that the definition of 1 (base 10) in base 1 is $\alpha\alpha$, or with other words, two symbols?
Number conversion
If we would, for example, convert $\alpha\alpha\alpha\alpha$ to radix 10, we would get following:
$$(\alpha\alpha\alpha\alpha)_1=\alpha*1^3+\alpha*1^2+\alpha*1^1 + \alpha*1^0 = (4\alpha)_{10}$$
Sure enough, $$(\alpha\alpha)_1 = 1_{10} = (2\alpha)_{10}$$
So I suppose this statement is therefore true as well:
$$(\alpha\alpha\alpha\alpha)_1=(2\alpha)_{10} + (2\alpha)_{10} = 1 + 1 = 2$$
But the interesting part here is that our symbol $\alpha$ is nothing, i.e. does not have any value at all, which implies that all numbers do not exist as values.
Conclusion
I do not now how correct this is, so it would be great if someone could comment upon what is written here. The question is finally, is this plausible?
Thank you,
Artem
 A: After a minute of thought I realised a tiny mistake in what I wrote before. In this answer I will prove that $\alpha = 1$, and hence show that radix one consist of only one quantity.
So, let's change our previous idea a bit, and omit zero in each finite series of a base.(please note that I am avoiding writing out zero for simplicity) Therefore,


*

*base 10 {1, 2, 3, 4, 5, 6, 7, 8, 9 }

*base 4  {1, 2, 3}

*base 3  {1, 2}

*base 2  {1}

*base 1  { }


(considering 0 is a quantity that can be expressed in every base)
As it might be seen already, base 1 is an empty set, however, zero can still be used. In my question I defined this empty set as the symbol $\{ \}=\alpha$. Later on, I claimed that $(\alpha\alpha)_1=1_{10}$, because of my observations of other bases, where I found that $10_{10}=10;$ $ 10_2=2;$ $10_{\pi}=\pi$ (i.e. in any base, 10 is representing the value of that particular base.)
Now, it is clear that if we only are to use one symbol, $(\alpha\alpha)_1=1_{10}$. But, what if we try to construct a value of two similar symbols in any other integral base? For instance, $$11_{10}=11;$$ $$11_2=3_{10}$$
where the general trend is:
$$11_b=b+1$$
which means that the value of this number is one more than the base.
So the case for base 1 is following: $$(\alpha\alpha)_1=2_{10}$$
and by converting $$(\alpha\alpha)_1=(2\alpha)_{10}$$
and solving the equation $$2\alpha=2$$ $$\alpha =1$$
$$\therefore \{\} \to \alpha$$
From nothing, we have created something.
$\square$
A: All mathematical proofs aside, common sense defines base-1 as the old "hash-mark" system, where for each item counted you make a simple mark.  While this simple mark looks suspiciously like a base-10 1, and behaves much like the 1, it pre-dates the actual 1 by many years.  This system also predates the concept of zero, so it did not have one.  Zero could be represented by a clean slate with no marks on it, but was then indistinguishable from a blank slate.  You might benefit from writing your proof without using the Arabic number system at all. (I.E. 0123456789)  The confusion in the proofs comes from the fact that base-1 CANNOT reliably represent 0, so removing it from even the implied number set is required for a proof.
