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Old 2012-11-10, 08:20   #551
xilman
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Quote:
Originally Posted by rogue View Post
Don didn't ask for it, but I will. What commentary do you have for your answer?
Because no conditions were placed on either c or T. If either are zero, the corresponding ratio is undefined.
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Old 2012-11-10, 14:34   #552
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Quote:
Originally Posted by xilman View Post
Because no conditions were placed on either c or T. If either are zero, the corresponding ratio is undefined.
That is true, but in his proof he establishes the condition that neither c nor T are zero.
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Old 2012-11-10, 15:20   #553
Don Blazys
 
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Your teachers have conditioned your mind
to use badly flawed axioms and have thus
destroyed your ability to think rationally.

Of course, they will never admit to that,
because they don't want to be held accountable,

But trust me, a lot of those so called
"institutions of higher learning"
have known about my equations for years
and are doing their best to keep them a secret.

You see, instead of revising their textbooks,
they would rather program you with nonsense.

Perhaps you should find a "personal injury lawyer" and
sue your school for having turned your brain into mush.
If you need a witness to testify to that,
then I will be more than happy to oblige.

Now, here is the question that should convince both
you and everyone else here, that those so called
symmetric and substiturion axioms of equality
are really nothing but a load of rubbish.

Given the identity:

\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}}
can we substitute \left(\frac{c}{c}\right) for \left(\frac{T}{T}\right) ?

Just try to answer this simple yes or no question
without going into one of your hand waving
song and dances. After all, you are neither Michael Jackson,
nor a geisha, and a simple yes or no question about
a "self evident truth" should not require any commentary.
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Old 2012-11-10, 17:12   #554
LaurV
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Yes.

\left(\frac{c}{c}\right)*c^{3}=T*\left(\frac{c}{T}\right)^{\frac{\frac{3*\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}}



Why not?

Last fiddled with by LaurV on 2012-11-10 at 17:12
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Old 2012-11-10, 17:13   #555
xilman
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Quote:
Originally Posted by rogue View Post
That is true, but in his proof he establishes the condition that neither c nor T are zero.
Not so.

He states without proof that (T/T)*c^1 is identically equal to T*(c/T). He doesn't give a proof of that statement. And, indeed, this is not an identity if T=0.

Ah, I see what you may be saying. IFF it's an identity, then T is non-zero.

In that case, then I agree with the admissibility of the original substitution.
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Old 2012-11-10, 17:15   #556
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Don, respond to post 533. You have conveniently been ignoring it.
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Old 2012-11-17, 16:35   #557
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Quote:
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Don, respond to post 533. You have conveniently been ignoring it.
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Old 2012-11-26, 16:56   #558
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Quote:
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Don, respond to post 533. You have conveniently been ignoring it.
I think we finally stumped Don.
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Old 2012-11-28, 10:35   #559
axn
 
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http://lmgtfy.com/?q=don+blazys+is+wrong
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Old 2012-12-10, 01:07   #560
Flatlander
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Quote:
Originally Posted by Don Blazys View Post
... shakes the very foundations of mathematics.
Sounds like something http://en.wikipedia.org/wiki/BBC_Horizon would say.
"The very nature of reality itself." etc

Makes me cringe.
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Old 2012-12-10, 01:50   #561
chappy
 
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Quote:
Originally Posted by axn View Post
led to the Straight Dope smack down, which led to: http://www.correctpi.com/ All these years wasted! I've been doing it wrong!
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