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Old 10-25-2008   #56 (permalink)
Don Blazys's Avatar
Don Blazys
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Re: A Mathematical Emergency.

To:Qfwfq.

As mathematicians, we must distinguish between "reasonable implications" and "unreasonable implications".

Now, if N is a non-negative integer, then the "blunt form":

(0/0)=N

implies that any non-negative integer N multiplied by 0 equals 0.

That's a "reasonable implication" because N*0=0 is a "true statement".

However, the "blunt form":

(6/0)=N

implies that any non-negative integer N multiplied by 0 equals 6,

and that's an "unreasonable implication" because N*0=6 is a "false statement".

Thus, the result: (1)^(0/0)=(1)^N=1

is predicated on a "reasonable implication" and a "true statement", so we need not define the exact value of either (0/0) or N.

In fact, if it were somehow concievable that the "blunt form" (0/0)=N defined a "particular value", then that would be a most stupid and ugly thing because it would assign a completely unwarranted significance to that "particular value" and would thereby falsely imply that for the equation:

(1)^(0/0)= (1)^N=1,

some particular value of (0/0) or N is "more true" than the rest!

By contrast, the "result":

(1)^(6/0)=(1)^N

is predicated on an "unreasonable implication" and a "false statement" so we need not even consider it.

Please keep in mind that indeterminate forms such as (0/0) are often initially encountered in their "blunt forms", and that in some cases (such as my proof), they should remain that way, while in other cases, they should be evaluated by introducing more subtle concepts such as limits.

Speaking of limits, it would be easy to introduce the "limit as T approaches c" in my proof, and thereby show that:

z=1 and z=2 results in (1)^(1),

while z>2 results in (1)^(infinity)="indeterminate",

but it would also be quite unnecessary, and would, in my opinion, diminish some of it's generality, simplicity, power, and of course, beauty.

Don.

Last edited by Don Blazys; 10-25-2008 at 12:24 AM..
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