Standard crank division by zero thread - Page 26

Don Blazys · 2011-04-22, 20:28

Quoting CRGreathouse:

Quote:

I can't think of anyone on this thread (apart from Don himself)
who think that he's right. CRGreathouse, Spherical Cow, rogue, xilman,
condor, akruppa, flouran, axn, tichy, NBtarheel_33, rajula, R.D. Silverman,
and yes science_man_88 have all made it clear that they think he's wrong.

Yes, it is quite amazing that all those strangely named denizens of this forum
think that the ridiculously trivial "indeterminate forms" that I so easily eliminated
are somehow an "issue" in my proof.

Funny thing is, if my opponents were right,
and indeterminate forms really were an "issue",
and we really couldn't allow T = c in the identity

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

then I would have discovered a way to prevent, in all cases,
the substitution of c for T (and hence from T/T = 1 ever "becoming"
c/c = 1) by using only the properties of logarithms, which would make me
the greatest mathematical miracle worker in the history of the entire universe!

Don.

Don Blazys · 2011-04-22, 21:59

Quoting "rogue":

Quote:

The website also says "Therefore, 0/0 does not mean any particular number - or even anything until we give it some new meaning".

Just before that, it says, very clearly,

Quoting http://www.mathpath.org/concepts/division.by.zero.htm

Quote:

"Yes! c can be any number"...

The website you linked to agrees with that and says

Quoting http://www.math.utah.edu/~pa/math/0by0.html.

Quote:

0/0 could be anything, we could argue that it's 1 or 2 ...

But nowhere does either article state that
we must disallow the indeterminate form 0/0
as we do divisions by zero such as 2/0.

In my proof, 1^(0/0) = 1, so (0/0) is both allowed and trivial,

but 1^(2/0) is disallowed because (2/0) implies that n*0 = 2

for some number n, which is absolutely nonsensical!

Here's the bottom line.

All the dummies who implied that (0/0) is an "issue" in my proof are wrong
and I'm right. In my proof, (0/0) is easily avoided and absolutely trivial.

A total non-issue.

Don.

Don Blazys · 2011-04-22, 22:10

Quoting "jyb":

Quote:

How do you define division?

It's essentially the inverse of multiplication.

Don.

Uncwilly · 2011-04-22, 23:40

Quote:

Originally Posted by jyb

I would like your definition of division. Please don't send me a link to what somebody else thinks. How do you define division?

Quote:

Originally Posted by Don Blazys

It's essentially the inverse of multiplication.

Please elaborate. Those of us with IQ's above 115 would like to hear a more detailed defining.

By saying it is "essentially", you leave a lot of room to play around. And by referencing multiplication you are referring to a different thing that also needs to be defined.

CRGreathouse · 2011-04-22, 23:43

You really are delusional. Even the sources you cite to support your position disagree with you.

jyb · 2011-04-22, 23:43

Quote:

Originally Posted by Don Blazys

Quoting "jyb":

Quote:

Originally Posted by jyb

How do you define division?

It's essentially the inverse of multiplication.

Don.

Ooh, good. Now we're getting somewhere. Can you make that precise? What does "essentially" mean here? And what does it actually mean to be the inverse of multiplication? I.e. can you give a real definition of division, rather than just hand-waving?

rogue · 2011-04-23, 02:03

Going back to version 1 of your proof (httр://donblazys.com/02.рdf).

Clearly if T = c, then you get c^z = T, which can only happen if c = 1 or 0. But c > 1 due to the first condition of the proof.

That means that T != c. What I'm missing is how this statement
"division by zero prevents the substitution" is determined. If T != c, then where is the division by 0 that you are mentioning. Division by 0 can only happen if T = c or T = 0, neither of which are conditions of the proof.

Condor · 2011-04-23, 13:31

Quote:

Originally Posted by CRGreathouse

You really are delusional. Even the sources you cite to support your position disagree with you.

But Don can't see that because he has already formed his conclusion, and is basing his interpretation of the sources on how he thinks he can make them agree. Not on what they say.

It ultimately boils down to what "n has any value in the multiplication n*0=0," means. Don wants to pick a number - 1, in this case - and use that in place of the 0/0. That is literally what he does when he divides out the offending term. It is really F(c,T)/F(c,T), where F(T,T)=0, which he deludes himself into believing isn't 0/0 since he sets c=T after the elimination. But it is 0/0 at that point if you make the substitution later. This is why you must put domain restrictions on variables in denominators.

What "n has any value in the multiplication n*0=0" actually means is "n could be undefinable, or it could have any value under a certain assumption. But only one value, and we can't tell what value that is without further analysis." Any student in High School calculus can tell you how.

The assumption is the assumption of continuity. The analysis is to use limits. You can show that 1 is the only value F(c,T)/F(c,T) can have in Don's equation when z=1. The trouble is, that same assumption, and analsyis, lets us evaluate his equation even when z is not 1.

Don doesn't seem to realize that the existence of a zero in a denominator is not what is "disallowed." It is the act of treating it as if it has a defined value. It does not, regardless of whether there is a zero in the numerator. Don is doing treating it like id does when he eliminates the term that could be zero.

And Don still hasn't figured out an argument against the last comment I made in post #260. If his approach were correct, it means every integer is both "allowed" and "disallowed." And no amount of bold, colored type can help his cause until he can come up with a supportable reason - and by supportable, I mean not based on - or meaningless words like "essentially,""trivial," "non-issue," and "obvious" - for why his methods apply only when M=1.

rogue · 2011-04-23, 13:32

I have a couple of additional points regarding the "proof".

First, $\sqrt{(c^z)^2}$ has two roots. One is $(c^z)$ . The other is $-(c^z)$ . Your proof only covers one, but not the other.

Second, take this example using your transformations:

$3^2$ = $(\frac{3}{3})3^2$ = $3(\frac{3}{3})^{\frac {\frac{{3}*{ln(3)}}{ln(3)}-1}{\frac{ln(3)}{ln(3)}-1}}$ = $3(1)^{\frac {\frac{{3}*{ln(3)}}{ln(3)}-1}{\frac{ln(3)}{ln(3)}-1}}$ = $3(1)$ = $3$

This means that one of the steps in this, i.e. your, transformation has a flaw. because clearly $3^2$ != $3$ .

There are certain transformations that if you go in one direction you get a single result, but if you in the opposite direction you get multiple results. Here is one (like my first point):

$2$ = $\sqrt{2^2}$ = $\sqrt{4}$ = $-2$

Another is the classic division by 0 error.

$0$ = $0$

becomes

$\frac{0}{0}$ = $\frac{0}{0}$

becomes

$0$ = $1$ .

This second example is where you and I diverge. You state that $\frac{0}{0}$ can have any value. If so, then the results are perfectly valid (albeit incorrect). I state that $\frac{0}{0}$ is indeterminate, thus it makes no sense to continue beyond any point after you get the term $\frac{0}{0}$ .

I stand by my claim that your proof has fatal flaws.

BTW, if anyone discovers typos, feel free to correct them.

Condor · 2011-04-23, 20:23

Just to be clear, we can make a more general expansion the following way:

$C^Z = $T\(\frac{C}{T}$^{\frac{\ \ \frac{{\frac{Z}{M}}{ln(C)}}{ln(T)}-1\ \ }{\frac{ln(C)}{ln(T)}-1}}\)^M$

Following Don's logic, because of a division by zero C=T is "disallowed" in this expansion unless Z=M. But in that case, you can reduce the internal exponent to 1, there is no zero in any denominator, and Z=M is "allowed." I call this an "incohesive term of order M."

The problem is, with this exansion we are free to choose M. Whatever integer we choose is "allowed" to be Z, and any other integer is "disallowed" to be Z. So every integer is both "allowed" and "disallowed."

Well, Don? If we use order 3 "incohesive terms," your logic says a^2+b^2=c^2 can't be true for co-prime (a,b,c), but a^3+b^3=c^3 is. Please explain, and assume nothing is "obvious" or "clear to everybody." In other words, treat it like a real proof.

R.D. Silverman · 2011-04-23, 23:10

Quote:

Originally Posted by Condor

But Don can't see that because he has already formed his conclusion, and is basing his interpretation of the sources on how he thinks he can make them agree. Not on what they say.

.

Don is unteachable. I suggest that he be ignored.
He is already on my ignore list.

2011-04-23, 20:23	#285
Condor Apr 2011 11111₂ Posts	Just to be clear, we can make a more general expansion the following way: $C^Z = \(T\(\frac{C}{T}\)^{\frac{\ \ \frac{{\frac{Z}{M}}{ln(C)}}{ln(T)}-1\ \ }{\frac{ln(C)}{ln(T)}-1}}\)^M$ Following Don's logic, because of a division by zero C=T is "disallowed" in this expansion unless Z=M. But in that case, you can reduce the internal exponent to 1, there is no zero in any denominator, and Z=M is "allowed." I call this an "incohesive term of order M." The problem is, with this exansion we are free to choose M. Whatever integer we choose is "allowed" to be Z, and any other integer is "disallowed" to be Z. So every integer is both "allowed" and "disallowed." Well, Don? If we use order 3 "incohesive terms," your logic says a^2+b^2=c^2 can't be true for co-prime (a,b,c), but a^3+b^3=c^3 is. Please explain, and assume nothing is "obvious" or "clear to everybody." In other words, treat it like a real proof. Last fiddled with by Condor on 2011-04-23 at 20:26

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2011-04-22, 23:43	#280
CRGreathouse Aug 2006 1011101011011₂ Posts	You really are delusional. Even the sources you cite to support your position disagree with you.

2011-04-23, 02:03	#282
rogue "Mark" Apr 2003 Between here and the 2⁴·397 Posts	Going back to version 1 of your proof (httр://donblazys.com/02.рdf). Clearly if T = c, then you get c^z = T, which can only happen if c = 1 or 0. But c > 1 due to the first condition of the proof. That means that T != c. What I'm missing is how this statement "division by zero prevents the substitution" is determined. If T != c, then where is the division by 0 that you are mentioning. Division by 0 can only happen if T = c or T = 0, neither of which are conditions of the proof.

2011-04-23, 13:32	#284
rogue "Mark" Apr 2003 Between here and the 2⁴×397 Posts	I have a couple of additional points regarding the "proof". First, $\sqrt{(c^z)^2}$ has two roots. One is $(c^z)$ . The other is $-(c^z)$ . Your proof only covers one, but not the other. Second, take this example using your transformations: $3^2$ = $(\frac{3}{3})3^2$ = $3(\frac{3}{3})^{\frac {\frac{{3}{ln(3)}}{ln(3)}-1}{\frac{ln(3)}{ln(3)}-1}}$ = $3(1)^{\frac {\frac{{3}{ln(3)}}{ln(3)}-1}{\frac{ln(3)}{ln(3)}-1}}$ = $3(1)$ = $3$ This means that one of the steps in this, i.e. your, transformation has a flaw. because clearly $3^2$ != $3$ . There are certain transformations that if you go in one direction you get a single result, but if you in the opposite direction you get multiple results. Here is one (like my first point): $2$ = $\sqrt{2^2}$ = $\sqrt{4}$ = $-2$ Another is the classic division by 0 error. $0$ = $0$ becomes $\frac{0}{0}$ = $\frac{0}{0}$ becomes $0$ = $1$ . This second example is where you and I diverge. You state that $\frac{0}{0}$ can have any value. If so, then the results are perfectly valid (albeit incorrect). I state that $\frac{0}{0}$ is indeterminate, thus it makes no sense to continue beyond any point after you get the term $\frac{0}{0}$ . I stand by my claim that your proof has fatal flaws. BTW, if anyone discovers typos, feel free to correct them.