Standard crank division by zero thread - Page 49

rogue · 2012-11-02, 18:22

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

firejuggler · 2012-11-04, 08:04

ahhh .. Jinx!

Don Blazys · 2012-11-06, 06:45

Quoting "rogue":

Quote:

Let's take another look.

Before we do that,
do we now agree that my proof,
which can be found here:

httр://unsolvedрroblems.org/S14b.рdf

does not contain the expression (0/0)?

Don.

axn · 2012-11-06, 07:19

Quote:

Originally Posted by firejuggler

ahhh .. Jinx!

Indeed!

rogue · 2012-11-06, 13:36

Quote:

Before we do that,
do we now agree that my proof
does not contain the expression (0/0)?

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.

Don Blazys · 2012-11-08, 04:51

Quoting "rogue":

Quote:

You need to modify the preconditions of your proof to state that c != T.

If c != T, then clearly, (T/T) != (c/c).

Dubslow · 2012-11-08, 05:08

Quote:

Originally Posted by Don Blazys

If c != T, then clearly, (T/T) != (c/c).

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.)

Batalov · 2012-11-08, 05:12

Quote:

Originally Posted by Don Blazys

If c != T, then clearly, (T/T) != (c/c).

If 2 $\ne$ 3, then clearly (3/3) $\ne$ (2/2).
Did I get this right?

Don Blazys · 2012-11-08, 06:37

The foundations of mathematics are its axioms,

which are defined as "self evident truths".

Consider the "symmetric axiom of equality"

which states that if $c=T$ , then $T=c$

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

Well, if

$\left(\frac{c}{c}\right)*c^3= \left(\frac{T}{T}\right)*c^3$ where $T=c$

and the properties of logarithms allow the identity:

$<br /> \left(\frac{T}{T}\right)*c^{3}=T*\left(\frac{c}{T}\right)^{\frac{\frac{3*\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}}$ where $T\not=c$

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 $\left(\frac{c}{c}\right)$ for $\left(\frac{T}{T}\right)$ ,
then we must conclude that $\left(\frac{c}{c}\right)=\left(\frac{T}{T}\right)$ is not always true,
and that there do exist some identities in which $\left(\frac{c}{c}\right)\not=\left(\frac{T}{T}\right)$
which of course shakes the very foundations of mathematics.

LaurV · 2012-11-08, 08:00

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

gd_barnes · 2012-11-08, 08:58

Quote:

Originally Posted by LaurV

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

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

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2012-11-02, 18:22	#529
rogue "Mark" Apr 2003 Between here and the 18D2₁₆ Posts	What? No response? Have we finally put this to bed?

2012-11-04, 08:04	#530
firejuggler Apr 2010 Over the rainbow 2·1,303 Posts	ahhh .. Jinx!

2012-11-08, 06:37	#537
Don Blazys Feb 2011 163 Posts	The foundations of mathematics are its axioms, which are defined as "self evident truths". Consider the "symmetric axiom of equality" which states that if $c=T$ , then $T=c$ and the "substitution axiom of equality" which states that we can always substitute $\left(\frac{c}{c}\right)$ for $\left(\frac{T}{T}\right)$ . Well, if $\left(\frac{c}{c}\right)c^3= \left(\frac{T}{T}\right)c^3$ where $T=c$ and the properties of logarithms allow the identity: $<br /> \left(\frac{T}{T}\right)c^{3}=T\left(\frac{c}{T}\right)^{\frac{\frac{3*\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}}$ where $T\not=c$ 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 $\left(\frac{c}{c}\right)$ for $\left(\frac{T}{T}\right)$ , then we must conclude that $\left(\frac{c}{c}\right)=\left(\frac{T}{T}\right)$ is not always true, and that there do exist some identities in which $\left(\frac{c}{c}\right)\not=\left(\frac{T}{T}\right)$ which of course shakes the very foundations of mathematics.

2012-11-08, 08:00	#538
LaurV Romulan Interpreter Jun 2011 Thailand 5·1,931 Posts	Don, something is really wrong with you: you forgot to say "don't you feel happy about it?" or something like that...