On Saturday, October 19, 2019 at 8:09:28 PM UTC+3, Dan Christensen wrote:
> When you were first introduced to exponentiation in school, it was probably explained as repeated multiplication, usually starting with the exponent of 2, something like:
>
> 3 ^ 2 = 3 x 3 = 9
> 3 ^ 3 = 3 x 3 x 3 = 27
> 3 ^ 4 = 3 x 3 x 3 x 3 = 81
>
> and so on.
>
> With a bit of hand waving, working the pattern backwards (dividing by 3), we completed the picture for natural numbers
>
> 3 ^ 1 = 3
>
> 3 ^ 0 = 1
>
> Now, that worked for all non-zero bases (like 3 here), but the wheels come off with a zero base since you cannot work the pattern backwards, dividing by zero. By various other hand waving arguments, you were convinced that 0^n = 0 for non-zero n, but 0^0 should be left undefined. It worked, but it was hard to rigorously justify. I will attempt once again here to offer that a more rigorous justification.
>
> We will start by obtaining the general term in the initial exercise here.
>
> 3 ^ 2 = 3 x 3 = 9
> 3 ^ 3 = 3 x 3 x 3 = 27
> 3 ^ 4 = 3 x 3 x 3 x 3 = 81
> .
> .
> .
>
> 3 ^ (n+1) = 3^n * 3 where n >= 2 and 3^2 = 9
> .
> .
> .
>
> Generalizing the formula for any base m, we would have m^(n+1) = m^n * m where n >= 2 and 3^2 = 9.
>
> Now, we can ask, if there exists an exponent-like binary function for all combinations of base and exponent in N such that:
>
> 1. m^2 = m*m (turns out to be equivalent to m=/=0 => m^0 = 1)
> 2. m^(n+1) = m^n * m
>
> The bad news is that there are infinitely many such exponent-like functions. (Formal proof at https://www.dcproof.com/PowFunctionInfinitelyManyBase2..htm )
>
> The good news is that these functions differ ONLY for base = exponent = 0 suggesting that 0^0 be left undefined. (Formal proof at https://www.dcproof.com/PowFunctionEquivBases2.htm )
>
> We can prove the existence of a partial function for exponentiation that leaves 0^0 undefined as follows:
>
> ALL(x):ALL(y):[x in n & y in n => [~[x=0 & y=0] => exp(x,y) in n]]
> & exp(0,1)=0
> & ALL(a):[a in n => [~a=0 => exp(a,0)=1]]
> & ALL(a):ALL(b):[a in n & b in n => [~[a=0 & b=0] => exp(a,b+1)=exp(a,b)*a]]
>
> where n = the set of natural numbers.
>
> https://www.dcproof.com/PowPartialFunction.htm
>
>
> What about so-called "empty products?" This notion is often used as a justification for defining 0^0=1.
>
> The product operator is usually defined on a sequence (a_i) as follows:
>
> Prod(i=m to n): a_i = a_m * a_m+1 * a_m+2 * ... * a_n (for m <= n)
>
> For some reason that is not clear to me, many authors also adopt the convention that:
>
> Prod(i=m to n): a_i = 1 for m > n (an "empty product")
>
> IIUC this questionable convention is often used to justify simply defining 0^0=1 by analogy.
>
>
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

When human mathematickers reach a stage of stop, they simply turn to fictions like 0^0, for sure!

As if they want to travel in time

Ha ha ha 😂

BKK