Marshall
Posts:
1,928
Registered:
8/9/06
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Re: 0^0=1
Posted:
May 7, 2012 6:00 PM
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On Monday, May 7, 2012 10:27:58 AM UTC-7, Jussi Piitulainen wrote:
> Dan Christensen writes:
>
> > So, why not 0^0 = 0? Can this be ruled out using only natural number
> > arithmetic?
>
> Not directly, as far as I can see, but almost nothing supports it
> while much requires x^0 = 1, either for all x or for all natural x.
Using only a very simple definition for the natural numbers,
and adding iterated operators to that, we easily and
naturally come to the conclusion that 0^0 = 1.
Define a natural number as either zero or the successor to
a natural number:
nat := zero | succ nat;
This gives us successor as well, so we can count. Now define
addition as iterated successor. In other words, give a count
for how many times to apply successor:
x + zero := x;
x + succ y := succ x + y;
Now we have addition. Define multiplication as iterated
addition:
x * zero := zero;
x * succ y := x + (x * y);
Define iterated multiplication:
x ^ zero := succ zero;
x ^ succ y := x * (x ^ y);
With a little massaging you can put this in your
favorite functional language and execute it directly.
This produces the result that 0^0 = 1.
After I went to the trouble of typing this in I found:
http://www.willamette.edu/~fruehr/LLC/final/lecs/nat/nat5.html
Marshall
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