Quote:
Originally Posted by Don Blazys
0/0= N is "indeterminate" because: any number N*0=0.
Thus, if 0/0= N, where N can be "any number", then the expressions in my proof clearly work out to:
(1)^(0/0)=1, therefore, c(1)^(0/0)= c, and (c(1)^(0/0))^2=c^2.
In other words, we need not know the value of the "indeterminate form" (0/0) in order to determine that
(1)^(0/0)=1 because 1 raised to any power, (including any "indeterminate power" such as 0/0), still equals 1.
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Note Don that 0/0 is just as "undefined" as 6/0 is. The step from saying "any number N*0=0" to saying "0/0= N, where N can be 'any number'" is a paralogism; you should notice that the nexus involves dividing or multiplying by 0, so you draw your conclusions by an "as if" and quite circular argument.
The third assert is interesting because, while fraught with the formal woe of being an expression in which one term does not define a value, you justify the claim by the seemingly compelling argument that the whole expression is totally independent on the undefined term anyway, so it doesn't matter. Now, apart from the above difference between "can be any value" and "defines no value", even when loosely put in the former terms there remains the fact that "any value" can be
too. Now this is a very subtle matter because the indepence can, at first glance, seem so intrinsic that one may be convinced by the argument. However one may use the rules of logarithms to write:
which sheds a better light on the matter of independence; it is easier to interpret in the familiar terms of "any value" multiplied by zero. Now we know that
for any finite
a and that it can be for infinite "values" of
a too; notoriously
is also an indeterminate form, therefore so is
. So, re-casting the matter as follows:
where it is manifest that the rhs is
not independent on the exponent, one may understand that the exponential also isn't wholly independent on the blunt
form and there's no hook or crook in the equality because it relies only on the value of
and of
, no whatsoever singularities. Therefore the rhs can't be equated to 1 as an apodictical certainty and the same goes for the lhs.
Now, without the above manipulations, this was only less obvious than the more common case of multiplication by a factor such as
which, Don, is
not the same as multiplying by 1; this argument of yours totally forgets the very matter of the indeterminate form. It's the multiplication by an expression which defines a value of 1
only 
and defines no value for
. Even if you remember your own contention, that for
the expression "can be any value", you would have to admit that this isn't the same thing as multiplying by 1, or is it?
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