Don H
Posts:
86
Registered:
12/13/04
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Re: 0^0=1
Posted:
Apr 22, 2012 3:11 PM
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"Bart Goddard" <goddardbe@netscape.net> wrote in message
news:XnsA03CAF9BF5D61goddardbenetscapenet@74.209.136.98...
> "DonH" <donlhumphries@bigpond.com> wrote in
> news:ZsDkr.6045$%E2.5976@viwinnwfe01.internal.bigpond.com:
>
>>
>> "Frederick Williams" <freddywilliams@btinternet.com> wrote in message
>> news:4F92FE5B.D2586615@btinternet.com...
>>> DonH wrote:
>>>>
>>>
>>>>
>>>> # If x^(a+0) = x^a, then the implication is that x^0 = 0; the x^0
>>>> is dedundant.
>>>
>>> x^a = x^(a+0) = x^a x^0 rather implies that x^0 = 1.
>>
>> # Does it? Maths' Order of Operations states that parenthesis is
>> evaluated first, which reduces (a+0) to a; leaving only x^a. Hence,
>> x^0 = 0.
>
> One wonders how you get from x^a = x^a to the conclusion that
> x^0 = 0.
>
>
> --
> Cheerfully resisting change since 1959.
# Maybe. But I'd claim x^0 is actually a nonsense, as it tries to use a
non-number (which is what zero is) in a calculation. Zero merely denotes
the absence of number.
x^a * x^b = x^(a+b)
x^a * x^0 = x^(a+0) = x^a
x^0 = 1
... is Webb's original statement. Is x^0 a true algebraic expression, or
invalid?
Perhaps the situation can be resolved by dealing with actual entities.
Say, there are six apples.
This can be represented by 1+1+1+1+1+1, or 2+2+2, or 3+3. Or, even 2^2
+2.
But does 6^0 = 1?
That is, you multiply all six by itself no times, and you get one apple?
Or, do you regard "all six" as an aggregate? In which case, "one" has
special significance?
6 x 1 = 6. But 6 x 0 = 0... unless the original six remain; then 6 x 0 =
6
On the other hand, if you start with no apples, then 0 x 6 = 000000 = 0.
0^0 = 1? If you have "a" nothing, then you still have the original
"nothing", as one "entity"?
Or 0^0 = 0. Fair enough. But serve you right for using 0 in calculating.
PS: If a computer says 0^0=1, lest it go crazy, then it is giving the wrong
answer- and your faith in computers is diminished.
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