Standard crank division by zero thread - Page 51

xilman · 2012-11-10, 08:20

Quote:

Originally Posted by rogue

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.

rogue · 2012-11-10, 14:34

Quote:

Originally Posted by xilman

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.

Don Blazys · 2012-11-10, 15:20

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.

LaurV · 2012-11-10, 17:12

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?

xilman · 2012-11-10, 17:13

Quote:

Originally Posted by rogue

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.

rogue · 2012-11-10, 17:15

Don, respond to post 533. You have conveniently been ignoring it.

rogue · 2012-11-17, 16:35

Quote:

Originally Posted by rogue

Don, respond to post 533. You have conveniently been ignoring it.

gd_barnes · 2012-11-26, 16:56

Quote:

Originally Posted by rogue

Don, respond to post 533. You have conveniently been ignoring it.

I think we finally stumped Don.

axn · 2012-11-28, 10:35

http://lmgtfy.com/?q=don+blazys+is+wrong

Flatlander · 2012-12-10, 01:07

Quote:

Originally Posted by Don Blazys

... 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.

chappy · 2012-12-10, 01:50

Quote:

Originally Posted by axn

http://lmgtfy.com/?q=don+blazys+is+wrong

led to the Straight Dope smack down, which led to: http://www.correctpi.com/ All these years wasted! I've been doing it wrong!

2012-11-10, 17:12	#554
LaurV Romulan Interpreter Jun 2011 Thailand 5×1,931 Posts	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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2012-11-10, 15:20	#553
Don Blazys Feb 2011 10100011₂ Posts	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.

2012-11-10, 17:15	#556
rogue "Mark" Apr 2003 Between here and the 2×3²×353 Posts	Don, respond to post 533. You have conveniently been ignoring it.

2012-11-28, 10:35	#559
axn Jun 2003 2×3×7×11² Posts	http://lmgtfy.com/?q=don+blazys+is+wrong