Actually that's not a problem, sm; take the contrapositive. His first mistake is somewhat later.

[QUOTE=Condor;258596]BTW - is it possible to edit posts? I don't see an icon for it. I didn't realize I had cut-and-pasted formating when I pulled that one in from the editor I prefer, and I agree it looks ugly. (Oh - is it that only the last post can be edited? This one did get an "edit" icon.)[/QUOTE]You have a time-limited window to edit a post (I believe it's an hour...)

Poor "Condor", "science man" and "CRNuthouse"!

They are so desperate, so frustrated, so obsessed and............. so stupid!

After all their compussive and incessant postings, they [B][I]still[/I][/B] dont realize that

[B][I]any[/I][/B] true equation, whether it be 2 + 3 = 5 or [U][COLOR="Navy"]httр://donblazys.com/03.рdf[/COLOR][/U]

is simply an actuality and that there is [B][COLOR=red]no lawyer-like argument to refute[/COLOR][/B].

Thus, the task of "refuting" this proof is utterly futile and truly Sisyphean !

(A fitting punishment for nincompoops!)

Don.

[QUOTE=Don Blazys;258768]Poor "Condor", "science man" and "CRNuthouse"!

They are so desperate, so frustrated, so obsessed and............. so stupid!

After all their compussive and incessant postings, they [B][I]still[/I][/B] dont realize that

[B][I]any[/I][/B] true equation, whether it be 2 + 3 = 5 or [U][COLOR="Navy"]httр://donblazys.com/03.рdf[/COLOR][/U]

is simply an actuality and that there is [B][COLOR=red]no lawyer-like argument to refute[/COLOR][/B].

Thus, the task of "refuting" this proof is utterly futile and truly Sisyphean !

(A fitting punishment for nincompoops!)

Don.[/QUOTE]

I almost fell for it:

[QUOTE]
[SIZE=7][TEX]\frac{T}{T}C^z = T \(\frac{C}{T}\)^{\frac {ln( \frac{C^z}{T})}{ln(\frac{C}{T})}}=T \(\frac{C}{T}\)^{\frac {\frac{ln( \frac{C^z}{T})}{ln(T)}}{\frac{ln(\frac{C}{T})}{ln(T)}}} =T \(\frac{C}{T}\)^{\frac {\frac{ln( \frac{C^z}{T})}{ln(T)}-\frac{ln(T)}{ln(T)}}{\frac{ln(\frac{C}{T})}{ln(T)}-\frac{ln(T)}{ln(T)}}}=T \(\frac{C}{T}\)^{\frac {\frac{{z}*{ln(C)}}{ln(T)}-1}{\frac{ln(C)}{ln(T)}-1}}[/TEX][/SIZE]
[/QUOTE]the last one has a different value for the exponent last I checked with T=6, C=3,and z=4, if these are within your bounds( haven't double checked) then it's done.

I'm sorry - I could not resist any longer once I ran out of my popcorn:
[url]http://www.youtube.com/watch?v=qLrnkK2YEcE[/url]

Can someone demonstrate any examples of widely used and commonly accepted proofs which would be rendered invalid when Don's reasonign is applied ?

Quoting science man:
[QUOTE]
The last one has a different value for the exponent last I checked with T=6, C=3,and z=4,
[/QUOTE]
You are wrong. (They all result in 81.)
Now you need to check how you checked! :smile:

Don.

Don Blazys,

refrain from posting personal insults.

[QUOTE=Don Blazys;258802]Quoting science man:

You are wrong. (They all result in 81.)
Now you need to check how you checked! :smile:

Don.[/QUOTE]

actually doing the [TEX]\frac{\frac{ln(c^z/t)}{ln(t)}}{\frac{ln(c/t)}{ln(t)}}[/TEX]

as written in you thing is a different value than [TEX]\frac{ln(c^z/t)}{ln(c/t)}[/TEX] I did it out in pari:

[CODE](17:42)>c=3;z=4;t=6;print((log(c^z/t)/log(t))/(log(c/t)/log(t)))
-3.754887502163468544361216832
(17:42)>c=3;z=4;t=6;print(log(c^z/t)/ln(t)/log(c/t)/log(t))
*** obsolete function.

For full compatibility with GP 1.39.15, type "default(compatible,3)", or set "compatible = 3" in your GPRC file.

New syntax: ln(x) ===> log(x)

log(x): natural logarithm of x.


(17:46)>c=3;z=4;t=6;print(log(c^z/t)/log(t)/log(c/t)/log(t))
-1.169600413701046825003995111
(17:46)>c=3;z=4;t=6;print(log(c^z/t)/log(c/t))
-3.754887502163468544361216832[/CODE]

[QUOTE=science_man_88;258772]I almost fell for it:



the last one has a different value for the exponent last I checked with T=6, C=3,and z=4, if these are within your bounds( haven't double checked) then it's done.[/QUOTE]
You should double check - the "identity" is indeed valid, as it is merely an expansion and re-arrangement of the terms. You need to add some parentheses to your code.


But the problem is the part inside the [] below:
[INDENT][TEX]\frac{T}{T}C^z = T \(\frac{C}{T}\)^{\frac {ln( \frac{C^z}{T})}{\[ln(\frac{C}{T})\]}}[/TEX][/INDENT]
Anything to the right of this in Don's derivation is no longer part of a "true equation" if T=C. Actually, the term becomes indeterminate; but Don will deny that.

It seems sad that a person who obviously has the capability to imagine combinations in new and interesting ways can so delude himself about what they mean. And about how he applies a double standard (and is using "lawyer-like" arguments) when he insists on letting C=T because of the "truth" inherent in his "identity," yet insists the same argument doesn't apply to numbers that would allow him to see that it is indeterminate.

[QUOTE=Condor;258875]You should double check - the "identity" is indeed valid, as it is merely an expansion and re-arrangement of the terms. You need to add some parentheses to your code.


But the problem is the part inside the [] below:
[INDENT][TEX]\frac{T}{T}C^z = T \(\frac{C}{T}\)^{\frac {ln( \frac{C^z}{T})}{\[ln(\frac{C}{T})\]}}[/TEX][/INDENT]
Anything to the right of this in Don's derivation is no longer part of a "true equation" if T=C. Actually, the term becomes indeterminate; but Don will deny that.

It seems sad that a person who obviously has the capability to imagine combinations in new and interesting ways can so delude himself about what they mean. And about how he applies a double standard (and is using "lawyer-like" arguments) when he insists on letting C=T because of the "truth" inherent in his "identity," yet insists the same argument doesn't apply to numbers that would allow him to see that it is indeterminate.[/QUOTE]

what I was pointing out is without more parentheses that second exponent is clearly different that the first. an example to go on:

3/4/2/4 or (3/4)/(2/4) first one is 3/(4*2*4) = 3/32 the second is (3/[TEX]\strike {4}[/TEX])/(2/[TEX]\strike {4}[/TEX]) =3/2 = 1.5

Quoting Condor:
[QUOTE]
Anything to the right of this in Don's derivation is no longer part of a "true equation" if T = c.
[/QUOTE]

That's not true. [B][COLOR=red]It all depends on the value of[/COLOR][/B] z. Please follow this carefully.

If z=1,

then (T/T)*c^z = T*(c/T)^((z*ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1))

becomes (T/T)*c^1 = T*(c/T)^1

where clearly, we [B][U][COLOR=red]can[/COLOR][/U][/B] let T = c because doing so gives us the "true equation"

(c/c)*c^1 = c*(c/c)^1

Quoting Condor:
[QUOTE]Actually, the term becomes indeterminate; but Don will deny that.
[/QUOTE]
As I just demonstrated, indeterminate forms are [B][COLOR=red]not an issue[/COLOR][/B] in my proof.

They are "removable singularities" that are easily avoided and don't even exist if we
do the algebra [B]correctly[/B] and evaluate the exponents at z = 1 [B][I]before[/I][/B] we let T = c.

Don.