Science Forums 2

[Don Blazys]

To:Qfwfq.

In post#57, you ended your first paragraph with: "Let's look at the fallacies in your argument".

Well, I already know that there are no fallacies in my argument, so the only possible reply from my point of view was and is: "There are no fallacies in my argument."

I suppose that I could also have mentioned that "my grasp of fundamental logic" is quite good, and that from my point of view, it's your grasp of fundamental logic that needs to be "straightened out", but I prefer not to indulge in such childish gibberish.

Neither of us would bother to reply to the other if either of us was truly inept.

Moreover, I never ever considered you to be my "opponent".

I thought that we were exploring my proof together, as friends.

Now, here's where we differ:

I hold that the "blunt" equation:

(1)^(0/0)=1

is both true and correct, while you continue to assert that the expression

(1)^(0/0)

is both "undefined" and "meaningless".

Now, for positive integer variables, and co-prime terms, given the equations:

(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)) ___________________(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, ________________(2)

if we let z=1 and z=2 respectively, then the above becomes:

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

T(c/T)^((ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)) ________________________(3)

and

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

(T(c/T)^((ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)))^2 _____________________(4)

where the logarithms can now be "crossed out", so that all we have left is:

(T/T)a^x+(T/T)b^y=(T/T)c=T(c/T) ______________________________(5)

and

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c)^2=(T(c/T))^2 __________(6)

letting T=c in all of the above equations now results in:

(c/c)a^x+(c/c)b^y=(c/c)c=c(1)^(0/0), ___________________________(7)


((c/c)a^(x/2))^2+((c/c)b^(y/2))^2=((c/c)c)^2=(C(1)^(0/0))^2, ______(8)


(c/c)a^x+(c/c)b^y=(c/c)c=c(1) __________________________________(9)

and

((c/c)a^(x/2))^2+((c/c)b^(y/2))^2=((c/c)c)^2=(c(1))^2, ____________(10)

where it is now obvious that the equation:

(1)^(0/0)=1

is perfectly consistent with the cancellation of the logarithms.

If the expression:

(1)^(0/0)

was "undefined", as you claim it is, then both (7) and (8) would become:

a^x+b^y=c= "undefined"

and

a^x+b^y=c^2= "undefined",

and that would make no sense whatsoever, because clearly, defined co-prime equations do indeed exist!

The properties of logarithms can not render all co-prime equations "undefined"!

Thus, it must be the case that:

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

and

(1)^(0/0) is not "undefined".

The proof works!

Don.