Well... I mentioned it in the chatterbox and many people answered... but not with base 0. so can anyway give me what base 0 is? or if anyone wanna comment on base 1... please... i dun like bases under binary.

I think that most n-based number systems can't have a value higher than n-1 in a single digit. Base 1 is forced to break this rule for non-zero values. It seems to me that since it already broke this rule, there is more than one way to notate any one value. For example, 3 could equal 111, 120, 300, 201, etc. Of course, for just one notation for each number, counting 1,11,111... will do.

I'm thinking that base 0 is impossible because each digit would be multiplied by 0^3,0^2,0^1,etc. and cannot equal anything other than 0. Also, for some reason or another, 0^0 is considered undefined. I'm not sure quite why, but I know that not everyone agrees on it. 0 raised to a negative power is obviously undefined because it is dividing by 0.

Tristan's point was that in base 1, then theoretically even 1 has no meaning, just as in base 2, 2 has no meaning. Personally I think we can stretch to allow the digit 1 in base 1, but putting higher digits in there is just stretching it too far.

I don't think Bases 1 or 0 are real bases, because in base 0 there are 0 digits. This means numbers don't exist!

In base 1, you have 0, 00, 000 and so on. This also doesn't work.

Also, to answer the question about 0^0 being undefined, some people say this argument: If going up 1 exponent is the same as multiplying the base (if 4^3 = 64, then 4^(3+1) = 64 * 4) then subtracting 1 must be the same as dividing by the base. (so if 4^3 = 64, then 4^(3-1) = 64 / 4)

When such an argument is reached, you can see that if 0^1 = 0, then 0^(1-1) = 0 / 0. But what is 0 / 0? :)

Tristan, l'Hopitals rule doesn't equate numbers, it equates limits, which is the reason that in some cases it appears that 0/0=1. In other cases 0/0=4 or 0/0=-7 etc. The driving factor in the differences here is that the zero in question here is not actually zero, it is in fact something approaching zero. I'm sure that I'm not really clearing things up for you, but I'll attempt to give an example. Due to many years since formal math instruction and a lack of reference texts, it'll be unfinished but the idea will be there.

Assuming that you understand the basic trig functions sin and cos;

sin(0)=0 and cos(0)=1

what happens when you take sin(0)/0?
how about (cos(0)-1)/0?

well, these both look like 0/0 cases to me, and so we use l'hopitals rule and create limits from the two equations. In the first we equate the answer to the limit of [sin(x)/x] as x approaches 0 (similarly in the secong we equate the answer to the limt of [(cos(x)-1/x) as x approaches 0). Now we see that the answer is ~approaching~ 0/0 (it is not equal to 0/0 - this is a property of limits that is very important). Since this is the case (as it would have also been if approaching inf./inf. or -inf./-inf) we can further evaluate using l'hopitals rule.

Now assuming that you're familiar with basic calculus, if you're not you'll have to take me at faith;

The original limits are described by the derivatives of the original numerators divided by the derivatives of the original denominators. This means that sin(0)/0 simplifies to inf/1, or just infinity, while the (cos(0)-1)/0 simplifies to 0/1, or zero. Two different cases of 0/0 providing different answer depending on the various initial condition - this shows that the term 0/0 is undefined (and this means something different fomr infinity).

I'll finalize with the clarification that there is likely some flaws in the math I've produced (I expected the sin(x)/x to reduce to 1 somehow...) and I'm sure that others will come along to clean it up a little, but hopefully you see the catch here is that l'hopitals rule doesn't deal with quantities, it derals with limits of equations, and these behave somewhat differently that the quantities they evaluate to.

To illustrate what Cory is saying, here's some spreadsheet output showing how, when x approaches zero, the corresponding quantities change:

         X(RAD)   SIN(X)  SIN(X)/X         COS(X)-1  (COS(X)-1)/X


               1 0.841471 0.841471          -0.4597 -0.4597


             0.1 0.099833 0.998334          -0.005  -0.04996


            0.01     0.01 0.999983          -.00005 -0.005


           0.001    0.001        1        -.0000005 -0.0005

The method of approach to (0,0) must be specified. If you consider the function y=0^x as x approaches 0, the answer will be 0, as zero raised to even a tiny amount above zero is zero. If you consider the function y=x^0 as x approaches 0, the answer will be 1, as any tiny number (above zero) raised to the zero power is 1.

I was reading recently about working with pure sets, that is, sets of sets of set etc of nothing. Would this count as a base-0 system?

Also, surely the easy way to explain base-1 is the word "tally"?

Peace,
Rawlyn.