What? No response? Have we finally put this to bed?

ahhh .. Jinx!

Quoting "rogue":
[QUOTE]
Let's take another look.
[/QUOTE]Before we do that,
do we now agree that my proof,
which can be found here:

[COLOR="Navy"][U]httр://unsolvedрroblems.org/S14b.рdf[/U][/COLOR]

does [B][I][U]not[/U][/I][/B] contain the expression (0/0)?

Don.

[QUOTE=firejuggler;316926]ahhh .. Jinx![/QUOTE]

Indeed! :smile:

[QUOTE]Before we do that,
do we now agree that my proof
does [B][I][U]not[/U][/I][/B] contain the expression (0/0)?
[/QUOTE]

Assuming you did not modify your proof as I suggested, then no. You need to modify the preconditions of your proof to state that c != T. If you do that, then I will agree with your statement.

[COLOR=black][FONT=Verdana][COLOR=black][FONT=Verdana]Quoting "rogue":[/FONT][/COLOR]
[COLOR=black][FONT=Verdana][QUOTE] [/FONT][/COLOR]
[COLOR=black][FONT=Verdana]You need to modify the preconditions of your proof to state that c != T. [/FONT][/COLOR]
[COLOR=black][FONT=Verdana][/FONT][/COLOR]
[COLOR=black][FONT=Verdana][/QUOTE][/FONT][/COLOR]
[COLOR=black][FONT=Verdana][/FONT][/COLOR]
[COLOR=black][FONT=Verdana]If c != T, then clearly, (T/T) != (c/c).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana] [/FONT][/COLOR]
[/FONT][/COLOR]

[QUOTE=Don Blazys;317516]If c != T, then clearly, (T/T) != (c/c).[/QUOTE]

Are you kidding me? I've stayed out of a thread I mostly don't understand, but that's clearly wrong. (T/T) = (c/c) unless T = 0 or c = 0. (I hope you're being sarcastic, but unfortunately, I'm really bad at it, especially on the blogotubes.)

[QUOTE=Don Blazys;317516]If c != T, then clearly, (T/T) != (c/c).
[/QUOTE]
If 2[TEX]\ne[/TEX]3, then clearly (3/3) [TEX]\ne[/TEX] (2/2).
Did I get this right?

The foundations of mathematics are its axioms,

which are defined as "self evident truths".

Consider the "symmetric axiom of equality"

which states that if [TEX]c=T[/TEX], then [TEX]T=c[/TEX]

and the "substitution axiom of equality"
which states that we can always substitute [TEX]\left(\frac{c}{c}\right)[/TEX] for [TEX]\left(\frac{T}{T}\right)[/TEX].

Well, if

[TEX]\left(\frac{c}{c}\right)*c^3= \left(\frac{T}{T}\right)*c^3 [/TEX] where [TEX]T=c[/TEX]

and the properties of logarithms allow the identity:

[TEX]
\left(\frac{T}{T}\right)*c^{3}=T*\left(\frac{c}{T}\right)^{\frac{\frac{3*\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}}[/TEX] where [TEX]T\not=c[/TEX]

then clearly, those so called

"symmetric and substitution axioms of equality"

are neither self evident, nor always true.

Think about it. If we can't always substitute [TEX]\left(\frac{c}{c}\right)[/TEX] for [TEX]\left(\frac{T}{T}\right)[/TEX],
then we must conclude that [TEX]\left(\frac{c}{c}\right)=\left(\frac{T}{T}\right)[/TEX] is not always true,
and that there do exist some identities in which [TEX]\left(\frac{c}{c}\right)\not=\left(\frac{T}{T}\right)[/TEX]
which of course shakes the very foundations of mathematics.

Don, something is really wrong with you: you forgot to say "don't you feel happy about it?" or something like that...

[QUOTE=LaurV;317530]Don, something is really wrong with you: you forgot to say "don't you feel happy about it?" or something like that...[/QUOTE]

I think he says "doesn't that make you happy?" :smile: