**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