0^0=1

It does.

No.

Yes.

Question at parsifal, it fails because the proof for n^0=1 would require you to divide by zero?

No.

Upon further reflection, I'm willing to concede that it certainly does not not equal 1.  But I shall budge no further.

Actually, I changed my mind. I've given the matter some more thought, and 0⁰ = 1.

it does it called the mulitplicity of no numbers.

Eh?

Generally when adding two zeros together you get 0 (e.g. 1+0+0 = 4). Thats simple.

Multiplying is a litle bit more complcated because,
0x0 = 0, agreed?
yet x^0 = 1.

This is because we generally say that the product of no numbers is one - thats kind of a stock phrase. This means that if you multiply no numbers together you get 1. The same thing happens if you do 0!. You can get the answer by putting 0 into rules for either factorials or indices. Of course this is dangerously close to a proof by definition, I'm not sure there is a much better proof than by using the definitions and seeing what happens if you put a 0 in.

In some ways its more interesting in words. zero times zero is zero, as you are multiplying together two numbers, both of thm zero. But you can write this with zero because you multiply no times at all, if you follow, that is the subtle difference. The former of multiplying by zero gives 0. The later, not multiplying anything at all gives one.

Ohh ok, sure, I get that.

But how does 1+0+0=4?

It doesnt I presumeably meant to type 1+0+0=1 or 4+0+0=4. I can't be sure which one of those I meant. I'm sure someone will now appear with one of those bizarre proofs for how 1=2 and whatnot.

Of course this is dangerously close to a proof by definition, I'm not sure there is a much better proof than by using the definitions and seeing what happens if you put a 0 in.

Binomial expansion of (1-1)ⁿ for n=0, anyone?

1. Am i missing something?

1. Am i missing something?

You said it was "dangerously close to a proof by definition". I was pointing out that there is a more solid proof available.

Ah I see what your getting at. It is another example though i'm not sure its more fundamental as its expression for binomial coefficients is derived from factorials. Its a more complex example involving elements of both x⁰ and 0!. Maybe its more fundamental, its certainly further reason to be confident in the hypothesis.

according to Bronstein's "Handbook of Mathematics" 0^0 is not defined. there is no reason given though. additionally 0^0=exp(0*ln(0))=exp(-0*infinity) seems a bit odd, too.

Ah I see what your getting at. It is another example though i'm not sure its more fundamental as its expression for binomial coefficients is derived from factorials. Its a more complex example involving elements of both x⁰ and 0!. Maybe its more fundamental, its certainly further reason to be confident in the hypothesis.

I think it's more fundamental because (⁰₀) can also be interpreted as the number of permutations of an empty set which are themselves empty sets. If you place your hand into an empty bag and pull out zero marbles, there's only one possible configuration.

What happens if you define 0^0 as 0? Will it return 0 as the answer?

according to Bronstein's "Handbook of Mathematics" 0^0 is not defined. there is no reason given though. additionally 0^0=exp(0*ln(0))=exp(-0*infinity) seems a bit odd, too.

4 / 0 is also not defined.  Say 4 / 0 = X.  Then multiply both sides by zero.  Then 4 = 0.  Not defined.  0 ^ 0 is not defined, but it makes a nice emoticon.
Google calculator says it's one, but a good discussion is here:
http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

The left and right hand limits of 4/x as x goes to 0 are non convergent. One goes to negative infinity while the other goes to positive infinity.

0/0 is indeterminate, while 4/0 does not exist.