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

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 Ben Bacarisse Posts: 1,972 Registered: 7/4/07
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 UTC-4, Ben Bacarisse wrote:
>> Dan Christensen <Dan_Christensen@sympatico.ca> writes:
>>

>> > On Thursday, October 17, 2013 7:13:01 AM UTC-4, Ben Bacarisse wrote:
>> >> Dan Christensen <Dan_Christensen@sympatico.ca> writes:
>> >> <snip>

>> >> > There exists infinitely many "exponent-like" 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.

Date Subject Author
10/17/13 Ben Bacarisse
10/17/13 Dan Christensen