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Topic:
Formal proof of the ambiguity of 0^0
Replies:
2
Last Post:
Oct 17, 2013 3:47 PM




Re: Formal proof of the ambiguity of 0^0
Posted:
Oct 17, 2013 2:54 PM


Dan Christensen <Dan_Christensen@sympatico.ca> writes:
> On Thursday, October 17, 2013 11:05:11 AM UTC4, Ben Bacarisse wrote: >> Dan Christensen <Dan_Christensen@sympatico.ca> writes: >> >> > On Thursday, October 17, 2013 7:13:01 AM UTC4, Ben Bacarisse wrote: >> >> Dan Christensen <Dan_Christensen@sympatico.ca> writes: >> >> <snip> >> >> > There exists infinitely many "exponentlike" functions ? one for each >> >> > natural number x0. >> >> <snip> >> >> >> >> > THEOREM 4: The Product of Powers Rule >> >> > >> >> > ALL(a):ALL(b):ALL(c):[a e n & b e n & c e n >> >> > => [~a=0 => a^b*a^c=a^(b+c)]] <snip other similar theorems>
>> >> Have you tried to do the same but with, say, 3^7=14 (leaving 0^0=1)? >> > >> > I'm not sure what you are getting at, but, using my definition, no >> > matter what value you assign to 0^0, you will not get 3^7 = 14. You >> > will only ever get 2,187. >> >> Just as you have shown that 0^0 can be anything you like, could you not >> show that 3^7 can be anything you like? > > No. As Theorem 2 shows, for all x, y in N, apart from the case of > x=y=0, x^y is uniquely determined. If you assumed any value for 3^7 > other than 2,187, you would get a contradiction.
I am not sure what I am failing to get across here. The new theorem 2 would say exactly the same about 3^7 as it currently says about 0^0. I'm asking why you can't do the whole thing you done to permit 0^0 to be any value one likes for some other pair of arguments.
>> For example theorem 4 would >> become >> >> ALL(a):ALL(b):ALL(c):[a e n & b e n & c e n >> => [~a=3 \/ (~b=7 /\ ~c=7 /\ ~b+c=7) => a^b*a^c=a^(b+c)]] >> >> By the way, your rules seem to be rather loose, but which I mean they >> decline to say things about many cases to which they might be applied. >> For example, your version of the above rule (number 4) says nothing >> about 0^1 * 0^1. It may be covered in other theorems, > > It can be shown that x^1 = x.
So? Your multiplication rule could cover that case, and others, but it doesn't. It just looks odd for it to say nothing about 0^9 * 0^2 when it could do so perfectly simply.
>> but surely it >> would be better to make the only exclusions be when a=0 and either b=0 >> or c=0? > > [snip] > > A matter of preference, I suppose, but that would work, too.
Your rule excludes an infinity of cases (no matter that other theorems may come to the rescue). A better rule would exclude only the one case you've decided to make special, and it's just a matter of preference?
 Ben.



