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Old 10-28-2008   #58 (permalink)
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Don Blazys
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Re: A Mathematical Emergency.

To: Qfwfq

There is no fallacy in my argument.

First of all, let's keep in mind that we are discussing the meanings of the expressions:

(1)^(0/0) and

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

as they relate to my proof of the BC.

Thus, if all variables are positive integers and all terms assumed co-prime, then:

(T/T)a^x+(T/T)b^y=(T/T)c^z=

T(c/T)^((zln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1))

and:

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c^(z/2))^2=

(T(c/T)^(((z/2)ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)))^2,

where letting z=1 and z=2 respectively allows us to "cross out" the logarithms, which results in:

(T/T)c=T(c/T), and

((T/T)c)^2=(T(c/T))^2.

Now, my questions to you are:

(1.) Can we now let T=c? (I say yes.)

(2.) Can we "cross out" the logarithms and let T=c if z>2? (I say no.)

(3.) Did letting z=1 and z=2, then "crossing out" the logarithms eliminate the possibility of encountering those pesky "indeterminate forms"? (I say yes.)

(4.) Do those pesky "indeterminate forms" occur if and only if fail to "cross out" the logarithms at z=1 and z=2 before we let T=c? (I say yes.)

(5.) Are those pesky "indeterminate forms" trivial? (I say yes.)

(6.) Do those pesky "indeterminate forms" somehow prevent us from letting T=c at z=1 and z=2? (I say no.)

(7.) Does the possibility of "division by zero" prevent us from letting T=c at z>2? (I say yes.)

(8.) Can we re-define T as "any positive real number other than unity", then show that at T=1 and T=2, the "limit as T approaches c" is c and c^2 respectively? (I say yes.)

Please let me know what your answers to these questions are.

By the way, I disagree that:

(0/0)=N and (6/0)=(infinity)

is a "fair comparison", because while any number:

N*0=0,

there is no number called "infinity" such that:

(infinity)*0=6.

Clearly, if limits are not introduced, then the "blunt" equation:

(0/0)=N

still belongs in the realm of numbers, while:

(6/0)=(infinity)

remains entirely nonsensical.

I also can't see how you can possibly claim that the implication:

51/3=2 (implies) N*0=0

is "formally true". The equation 51/3=2 is false, while the equation N*0=0 is true. How can something false imply, "formally" or otherwise, something true?

Don.
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