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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.


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,


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marked as duplicate by MJD, Julian Kuelshammer, dfeuer, Davide Giraudo, Nick Peterson Oct 2 '13 at 18:50

This question was marked as an exact duplicate of an existing question.

You may wish to read the appropriate Wikipedia article:… – Brad Nov 3 '12 at 21:14
I was wondering how would we be able to convert a value such as $6$ in base $10$to corresponding value in base $1$? Since powers of 1 are all one, we could only write $6$ in base $10$ is =$\alpha\alpha\alpha\alpha\alpha\alpha$ - Is this so? – NoChance Nov 3 '12 at 21:20
Note, $\alpha$ is not a value, if we would follow this pattern. – Artem Nov 3 '12 at 21:31
There are countless ways to represent elements of any countable set with finite repetitions of just one symbol. The base-n pattern does not extend nicely to 1 though. – Karolis Juodelė Nov 3 '12 at 21:38
How can this question be a duplicate, if it was asked 6 months before What would base 1 be? – Artem Nov 2 '13 at 12:54
up vote 0 down vote accepted

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.

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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. $\square$

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I don't understand what you are trying to prove here. You said you are proving that $\alpha = 1$, but what is the definition of $\alpha$? What does it mean to "define the empty set" as $\alpha$? And what does it mean to write $\{\} \to \alpha$? – Trevor Wilson Dec 17 '12 at 20:37
I will need to think about a possible mistake with this assumption. It seems that $(xx)_b = x (mod b)$ breaks the whole thing. $\therefore (\alpha\alpha)_1 = \alpha (mod 1)$, $\implies 2\alpha=\alpha (mod 1)$ $\implies 2\alpha-\lfloor 2\alpha \rfloor = \alpha$ $\therefore \alpha = 0$. //sorry for confusing everything. – Artem Dec 17 '12 at 22:52
I'm afraid that I still don't understand your answer. Also, in your last comment, it is not correct that $2\alpha = \alpha \pmod{1}$ implies $\alpha = 0$. It is true for any integer $\alpha$. – Trevor Wilson Dec 18 '12 at 22:26
By the way, why is it not true? $\alpha$ is an integer, so I suppose that $2\alpha$ is an integer as well, so no fractional part, hence 0? – Artem Dec 18 '12 at 22:46
The three dots means "therefore", right? If $\alpha$ is, say, 1, then $2\alpha = \alpha \pmod{1}$ but $\alpha \ne 0$. So the implication fails. – Trevor Wilson Dec 18 '12 at 22:55

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