# What would base 0 be? How would/could it work?

If I was trying to take the number $123$ in base $10$ and try and convert it into base zero I would do something like this:

$123 = 100 + 20 + 3$

$10^{\log_0(100)} + 10^{\log_0(20)} + 10^{\log_0(3)}$

But $\log_0(x)$ is the same thing as $\dfrac{\log(x)}{\log(0)}$ and the log of zero is undefined. So is there any other way to convert to base zero? Or does base zero simply not exist?

• What do you think you mean by the term "base zero"? Many questions are trivial so long as you know the meaning of the words you use. – anon Apr 19 '15 at 22:01
• @anon I am writing $10^{log_0(a)}$s down because it's the way I figured out how to convert bases. And by "base zero" I mean a number with a radix or base of zero. – Jamie Sanborn Apr 19 '15 at 22:03
• Do you know what you mean by that? One way I could give the words meaning is "a whole number is written in base $b$ if it is expressed as a polynomial in $b$ whose coefficients are integers $0\le a<b$." On that meaning, it's very obvious whole numbers can't be written in base zero. Maybe the words mean something different to you? I asked you what you meant by "base zero," and you come at me with "base of zero" - are you aware of how this comes across as not having any definition at all? It is critically important we know what we're talking about, which implies knowing what words mean. – anon Apr 19 '15 at 22:11
• This is 100% USDA certified nonsensical. – ncmathsadist Apr 19 '15 at 22:13
• As for writing down terms like $10^{\log a}$s, suppose you wanted to write eleven in base two: what good would the expression $10^{\log_2 10}+10^{\log_21}$ do us? How does that get us closer to the correct answer, which is $1011_2$? – anon Apr 19 '15 at 22:15

Base 0 does not make any mathematical sense.

Look at binary (base 2). There are two digits, 0 and 1. Thus, every other number you need to roll over the 1 back to a zero, and add 1 to the next column.

Now, look at base 1. Now, every number requires rolling over to the next row. This is essentially a tally system, where each '1' (in base ten) gets it's own column.

Now, if you think about base 0, that would mean every increase by '1' in any non-zero base represents an infinite amount of columns that need to be created to support the overflow. Thus, every number in base 0 would essentially be infinite, or even worse, every number would be the same number.

• @TrevorWilson thinking in another way, when you try to represent a number in base 1, for 0 you will get 0, and in other cases you have to carry infinite times to get infinite 0s yet still not able to represent the number. – Colliot Apr 20 '15 at 1:09
• @Aszune'sHeart Base 1 is sometimes used as name for unary numbers. Unary does not follow the rules of positional systems, but still have some properties in common with positional systems. Unary does have a use in computation theory. – kasperd Apr 20 '15 at 5:43
• @Aszune'sHeart Base 1 is essentially a tally system. '||||||' would be $6_{DEC}$ in base 1 – Jamie Sanborn Apr 20 '15 at 11:07
• @JamieSanborn why would a coefficient greater than 0 be allowed in base 1? – Colliot Apr 20 '15 at 12:34
• @Aszune'sHeart Tally marks are used in base one to give a better understanding. The '|' is just a placeholder. You could use zeros, having 000000 for $6_{DEC}$ in base 1 but people usually just use tally marks to make things clearer. – Jamie Sanborn Apr 20 '15 at 19:31

In base $10$, we use ten symbols.

In base $2$, we use two symbols.

In base $1$, we use one symbol (tally marks).

In base $0$, we'd use zero symbols. We can't express anything with zero symbols.

• I agree with this answer except that if base $1$ is defined like all the other bases, then the one symbol is "$0$" and the things you can write with it ($0$, $00$, $000$, etc.) all have the same value (namely zero.) So although some people do call using tally marks "base $1$" I think this is not quite right. – Trevor Wilson Apr 20 '15 at 0:32
• @TrevorWilson I think the problem stems from treating "base 1" as a positional number system. The number two using tally marks can be represented as $\color{blue}{/}\color{green}{/}$, or $\color{green}{/}\color{blue}{/}$, so clearly the position of our tally-symbol does not matter. However, we are enumerating something, so in my opinion "base 1" is a valid number system -- just not a positional one. So when we use the symbol $0$ as our tally, we can't really compare it to other bases $\ge 2$. – Andrey Kaipov Apr 20 '15 at 1:05
• Sure, using tally marks is a valid number system. But if we want to call it "base $1$" then we have to relax the definition of "base $b$" in the case $b = 1$. We could say that digits $0,\ldots,b$ are allowed, rather than just digits $0,\ldots,b-1$. In this relaxed version of base $1$, digits $0$ and $1$ are allowed, but there is no point to using "$0$" except by itself, so we would count $0$, $1$, $11$, $111$, etc. I'm just pointing out that this is different from the usual definition of "base $b$". For example, "$222$" could mean fourteen in base $2$ but is usually disallowed. – Trevor Wilson Apr 20 '15 at 15:41

Base 0 unfortunately does not make any sense, for the very reason you specify.

Most digits in the number would be worth exactly zero, and the digit in the "ones position" would not even have a defined place value.

• If you see $123$ as $1\times 10^2+2\times 10^1+3 \times 10^0$, you might consider writing $123 = 123 \times 0^0$, using the convention $0^0=1$. – Henry Apr 19 '15 at 22:03
• @Henry Then 123 will have to be a separate symbol, as will EVERY number, and base-zero requires an infinitude of symbols, and that doesn't make any sense. – naslundx Apr 19 '15 at 22:08
• @Henry Writing $123 = 123 \times 7^0$ doesn't count as writing $123$ in base $7$ (for example) because the only symbols allowed in base $7$ are $0$, $1$, $2$, $3$, $4$, $5$, and $6$. Similarly, no symbols are allowed in base $0$. – Trevor Wilson Apr 20 '15 at 0:23

When you express a number in base $b$ you find it as a sum of various powers of $b$. For example to express $65$ in base 3 we first note $65= 27+27 + 9 +1 +1$ so $65=2\cdot 3^3 + 1\cdot 3^2 + 2\cdot 3^0$, hence $65=(212)_3$. Unfortunately all powers of zero are zero, and so sums of powers zero cannot be anything other than $0$, so zero is powerless to be the base for number system.

Base 0 would imply that each place holder in this theoretical base could take one of zeros values. Notice that's a contradiction.

Proof: assume it's possible to take a value from 0 possible values, then there is a possible value, and thus there was a value to begin with.

Using Henry's suggestion. $0^0$ could be interpreted as 1, in which case counting could be possible. However, this is not truly using noting to count, it's merely using ad hoc convention.

To possibly make this system work, you'd need to invent new math, however seeing as there is no motivating example, I doubt it would be worth the effort.

I don't get the hang up people have with zero. It is not a counting number. It is a place-holder between numbers. When I write 1230 in decimal, the 0 being there only increments the power of 10's up by 1 for 1, 2 and 3 but that is the only purpose the zero has here.

When counting in unary/base(1) there is no need for 0, why? Because unless we're dealing with bizarro stuff, 1 to any power is 1. So, in unary, 111 = 10101 = 1000000010001000. Unless the 0's now represent time or are code from something else non-numerical, they've lost their meaning. 0's don't count anything in unary, thus in base 1 we only use 1's or tally marks. And, in base 0, all we can use is 0, since the only value expressed in base 0 is ironically the value I said wasn't a real number, 0. The problem when base 0 is that writing anything aside from just 0^1, or 0 is either redundant or undefined.

So, it isn't that base zero or non-whole number bases don't exist - what's important to note are questions such as: does such a base offer anything useful in counting, and if not, does it provide us with anything else useful?

I know I rambled on, but, does that make sense?