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?
 A: 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.
A: 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.
A: 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.
A: 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.
A: 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.
A: 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?
