Standard crank division by zero thread - Page 52

Don Blazys · 2012-12-11, 10:49

Quoting "rogue":

Quote:

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

Why should I respond to anything that you write when you can't
even answer the simple yes or no question in this post!

Sorry "rogue", but your argument that removable singularities are
not removable was simply wrong, and unfortunately, since you are
either too stupid to see that (or not man enough to admit it),
I will no longer teach you anything.

Face it... you lost.

Now get lost and stop posting on my very popular "crank" thread!

Quoting "Flatlander":

Quote:

Makes me cringe.

You see, not only have your teachers destroyed your ability to
reason by teaching you axioms that are badly flawed, but sadly,
they have also turned you into a wimp who would rather cringe
in fear than face the truth.

Here again is the question that should convince all of you who are
cringing behind your fake names that those so called symmetric and
substitution axioms of equality are 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 $\left(\frac{T}{T}\right)$ ?

LaurV says:

Quote:

Yes.

xilman says:

Quote:

No.

Now, can the rest of you fake named wonders muster up the courage
to do the same and simply answer yes or no without any commentary
whatsoever?

rajula · 2012-12-11, 10:58

Quote:

Originally Posted by Don Blazys

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 $\left(\frac{T}{T}\right)$ ?

Yes or no.

Don Blazys · 2012-12-11, 11:01

Which is it?

rajula · 2012-12-11, 11:04

Quote:

Originally Posted by Don Blazys

Which is it?

As requested: no comments. If you are unable to follow the logic in the answer, then so be it.

Don Blazys · 2012-12-11, 11:09

So, for the record, rajula's answer is "yes or no"!

We now have three different answers!

Doesn't that make you happy?

Don.

Don Blazys · 2012-12-11, 12:10

So here, yet again, is the question that should convince all of you
who are cringing behind your fake names that those so called
symmetric and substitution axioms of equality are 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 $\left(\frac{T}{T}\right)$ ?

"LaurV" says:

Quote:

Yes.

"xilman" says:

Quote:

No.

"rajula" says:

Quote:

Yes or no.

Clearly, those "axioms" which are supposedly "self evident truths"
are a joke, because even the computer aided "geniuses" in this
funky forum can't seem to agree on an answer to this ridiculously
simple question!

Now, can the rest of you answer the above yes or no question
without any commentary? We need a "consensus of idiots"!

rogue · 2012-12-11, 14:27

Quote:

Originally Posted by Don Blazys

Why should I respond to anything that you write when you can't
even answer the simple yes or no question in this post!

What you think of me as a person is not important to this thread, but if you choose to continue insulting me, then I'm sure that the moderators will be all too glad to ban you again.

Now back to your post. What do you mean by "this post"? Why should I have to answer when others have already done so? Presuming you mean this question:

Quote:

Originally Posted by Don Blazys

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

then the quick answer is no. The answer would be yes if you add the appropriate conditions for c and T.

I made two points in post 533. The first is that you need to fix the language of your proof, i.e. specify the conditions of T in the beginning of your proof and remove "division by zero prevents" as stuff like that makes real mathematicians cringe. The second presumes that your fomulae are correct. You have to prove that $T*({\frac{c}{T}})^{\frac{{\frac{(Z)*ln(c)}{ln(T)}-1}} {{\frac{ln(c)}{ln(T)}-1}}$ != i^j where i and j are integers > 2. Note that i could be an integer that is not equal to $T*({\frac{c}{T}})$ .

Don, step back for a second and look at steps 1 and 2 of your proof. Would you agree that $a^2 + b^2 = c^z$ has an infinite number of solutions for a, b, c > 1 and z > 2? If I changed the conditions of your proof to state that x and y >= 2, then your proof could be used to prove my statement to be false even though we both know it is true. Where in your proof do you prove that x and y must be greater than 2? In other words, why should your proof be true when x and y > 2, but not when x and y = 2? That is what you must prove and you have not done so.

The point I'm trying to make, a point that others have tried to make and one that you are clearly missing, is that the values of a, b, x, and y have absolutely NO BEARING on your proof. I could just as easily prove that $f = c^z$ is impossible for f, c and z > 2 using your logic.

LaurV · 2012-12-12, 05:48

Why not? (x/x) is 1 by definition, when x is fixed, regardless of the value of x. May x be integer, real, complex, even function, x/x is always 1 as long as the division works.

(c/c)*function1(c,T)=(T/T)*function2(c,T), anytime when function1(x,y) is equal with function2(x,y). What is the problem?
And this proves nothing. He still has to show when function1 is equal to function2.

He says nothing like replacing c with T or viceversa. (c/c) is 1 and (T/T) is 1, even when c is 0 or not, T is zero or not, the expression is 1 always, by definition. Take them complex numbers, functions, whatever you like, they always simplify to 1. Mind that replacing c/c with T/T DOES NOT mean replacing c with T.

So, yes, you can replace T/T with c/c in

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

and you get

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

Or, in a single row:

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

The right answer to the question is "YES". You can substitute c/c with whatever you like, even with pope/pope, as I heard all popes are gays.

And then what?

You can't replace c with T, or viceversa. They are different entities, and replacing them will change one side of the equality. If you understand this, the next question is: does your logic changes when you replace 3 with 2? Why?

=======================

@rogue: man, the guy is here only for the flame war. Don't take him so serious. He has no idea what he is talking about, but he likes the flame and "popularity" derived from it. The only way to deal with it is to be sarcastic and set him intelligent traps. See my post 551, he practically admitted in post 552 that the equation a^2=b^2+c^2 has no solution.

Dubslow · 2012-12-12, 06:13

Quote:

Originally Posted by LaurV

(c/c) is 1 and (T/T) is 1, even when c is 0 or not, T is zero or not, the expression is 1 always, by definition.

Not true. If c or T = 0, then the expressions in question are meaningless.

Quote:

Originally Posted by LaurV

The only way to deal with it is to be sarcastic and set him intelligent traps.

Or just ignore him and let the thread die. (This should have been done a while ago.)

Batalov · 2012-12-12, 06:27

DonBlazys, LaurV - talk amonst yourselves. The rest are feeling verklemmt!

Second verse of Don's song will surely be the same as the first.

LaurV · 2012-12-12, 06:35

Quote:

Originally Posted by Dubslow

Not true. If c or T = 0, then the expressions in question are meaningless.

Now you are on his side?

Quote:

Or just ignore him and let the thread die. (This should have been done a while ago.)

And lose all the fun?

@Batalov: haha, nice one!

P.S. I never said I am not a crank!

2012-12-12, 05:48	#569
LaurV Romulan Interpreter Jun 2011 Thailand 9655₁₀ Posts	Why not? (x/x) is 1 by definition, when x is fixed, regardless of the value of x. May x be integer, real, complex, even function, x/x is always 1 as long as the division works. (c/c)function1(c,T)=(T/T)function2(c,T), anytime when function1(x,y) is equal with function2(x,y). What is the problem? And this proves nothing. He still has to show when function1 is equal to function2. He says nothing like replacing c with T or viceversa. (c/c) is 1 and (T/T) is 1, even when c is 0 or not, T is zero or not, the expression is 1 always, by definition. Take them complex numbers, functions, whatever you like, they always simplify to 1. Mind that replacing c/c with T/T DOES NOT mean replacing c with T. So, yes, you can replace T/T with c/c in $\left(\frac{T}{T}\right)c^{3}=T\left(\frac{c}{T}\right)^{\frac{\frac{3\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}}$ and you get $\left(\frac{c}{c}\right)c^{3}=T\left(\frac{c}{T}\right)^{\frac{\frac{3\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}}$ Or, in a single row: $\left(\frac{c}{c}\right)c^{3}=\left(\frac{T}{T}\right)c^{3}=T\left(\frac{c}{T}\right)^{\frac{\frac{3\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}}$ The right answer to the question is "YES". You can substitute c/c with whatever you like, even with pope/pope, as I heard all popes are gays. And then what? You can't replace c with T, or viceversa. They are different entities, and replacing them will change one side of the equality. If you understand this, the next question is: does your logic changes when you replace 3 with 2? Why? ======================= @rogue: man, the guy is here only for the flame war. Don't take him so serious. He has no idea what he is talking about, but he likes the flame and "popularity" derived from it. The only way to deal with it is to be sarcastic and set him intelligent traps. See my post 551, he practically admitted in post 552 that the equation a^2=b^2+c^2 has no solution. Last fiddled with by LaurV on 2012-12-12 at 06:19

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2012-12-11, 11:01	#564
Don Blazys Feb 2011 163 Posts	Which is it?

2012-12-11, 11:09	#566
Don Blazys Feb 2011 163₁₀ Posts	So, for the record, rajula's answer is "yes or no"! We now have three *different* answers! Doesn't that make you happy? Don.

2012-12-12, 06:27	#571
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 9494₁₀ Posts	DonBlazys, LaurV - talk amonst yourselves. The rest are feeling verklemmt! Second verse of Don's song will surely be the same as the first.