Standard crank division by zero thread - Page 48

science_man_88 · 2012-10-16, 20:32

Quote:

Originally Posted by gd_barnes

It is now becoming increasingly obvious to everyone that Don Blazys is wrong. If you google "Don Blazys is wrong", this thread comes up as #1 and my "Don Blazys is wrong" web page comes up as #10. Looking good but we can do even better!

Come on everyone, keep clicking on this thread and my "Don Blazys is wrong" web pages at:
http://www.noprimeleftbehind.net/crus/
-and-
http://www.noprimeleftbehind.net/cru...zysiswrong.htm

"Don Blazys attempt" get your site as the only result.

gd_barnes · 2012-10-16, 21:23

Quote:

Originally Posted by science_man_88

"Don Blazys attempt" get your site as the only result.

I get a lot more than that. I get the same as "Don Blazys is wrong". This thread comes out as #1 and my page comes out as #10.

Flatlander · 2012-10-19, 18:02

He says they division by zero gives un undefined result except for 0/0.

What are the consequences of accepting that 0/0 has a value that would prove his assumption wrong?

rogue · 2012-10-19, 19:58

See next post

rogue · 2012-10-19, 20:03

Quote:

Originally Posted by Flatlander

He says they division by zero gives un undefined result except for 0/0.

What are the consequences of accepting that 0/0 has a value that would prove his assumption wrong?

If 0/0 has a value (or 1^(0/0) for that matter), then you must define that value. Don is trying to say that 1^(0/0) can have any value, i.e. it can be equal to 0, 1, 2, 123, 3234.345344, etc. Does "any value" mean 0, 1, 2, 123, OR 3234.345344, etc. or does it mean 0, 1, 2, 123, AND 3234.345344, etc.? If the former, then one can say that 1 = 2. The latter implies that it can have all values concurrently, which makes no mathematical sense. (The smarter math folks on this forum can correct that statement and Don is clearly not one of them.)

My point with Don was that I believe that 1^(0/0) is an indeterminate form just like 1^inf and 0/0. Indeterminate does not mean the same thing as "any value". Note that I haven't proven that it is indeterminate, although someone else on this forum probably could.

Imagine the equation x = y. Now say that x can have "any value" and y can have "any value". While that may appear to be true, you can't say "let x be 1 and let y be 2" (because both x and y can have "any value") and assume the equation to be true. x and y can have "any value", but they must have the "same value".

His use of the term "any value" is vague in the least.

BTW, 0/0 is both indeterminate and undefined whereas x/0 (where x != 0) is only undefined.

Dubslow · 2012-10-19, 20:55

Quote:

Originally Posted by rogue

BTW, 0/0 is both indeterminate and undefined whereas x/0 (where x != 0) is only undefined.

If you're a physicist, x/0 = sgn(x)*∞, where sgn(0) is undefined.

Don Blazys · 2012-10-20, 13:03

Those of you who think that the indeterminate form $\frac{0}{0}$ is somehow an
"issue" in my proof are simply wrong.

For one thing, the indeterminate form $\frac{0}{0}$ doesn't even exist in my proof!

Now, the reason why the indeterminate form $\frac{0}{0}$ doesn't even exist
in my proof is because I was clever enough to simplify the expressions
involving logarithms at $Z=1$ and $z=2$ before substituting $\left(\frac{c}{c}\right)$ for $\left(\frac{T}{T}\right)$ .

Those of you who don't know how to simplify or want to make the silly
argument that "simplification is impossible" should read this:

http://www.helpalgebra.com/onlinebook/simplifyingrationalexpressions.htm

You see, in my proof, at $Z=1$ and $z=2$ , the logarithms simply "cancel out".

Then, and only then do we substitute $\left(\frac{c}{c}\right)$ for $\left(\frac{T}{T}\right)$ .

Thus, those indeterminate forms that you youngsters are making such
a "big deal" about are really nothing more than "removable singularities"
which are so utterly trivial that thay can be removed simply by avioding
them in the first place!

In fact, they are such a non-issue that we mathematicians refer to them
as being "cosmetic", and even those of you who are too inexperienced to
avoid them can still remove them in much the same way that the
singularity at $z=0$ in the function $f(z)=\frac{\sin z}{z}$ can be removed
simply by defining $f(0)=1$ . You can read more about that here:

http://en.wikipedia.org/wiki/Removable_singularity

Then, there is this to consider.

According to "rogue's" argument, we can never ever substitute
$\left(\frac{c}{c}\right)$ for $\left(\frac{T}{T}\right)$ in my proof, not even at $Z=1$ and $z=2$ .

Thus "rogue" is clearly implying that what I "really" discovered is
a way to absolutely and in all cases prevent the substitution of $1$ for $1$ .

Heck, that would really make me a mathematical miracle worker!

Don.

rogue · 2012-10-20, 13:29

Where is the precondition in the proof that c != T? It is when c = T that the problem of 1^(0/0) occurs.

What does "prevents the substitution" supposed to mean?

Regardless of anything we say, you will argue. If you want to convince us that you are correct, you must submit your proof to a math journal and get it published. Posting it online and saying "it's correct" doesn't make your proof correct. Why do you refuse to do that? Your purported "proof" is so simple that it would take little time for independent referees to determine if it has any flaws. IMO, we are asking you to do what is most fair and reasonable to everyone here, including yourself. It seems to me that your abject refusal to publish your proof is cowardly and unreasonable. I believe that you already know your proof is wrong and are being a troll just because nobody has found the flaws in your proof that you are already aware of.

Don Blazys · 2012-10-20, 14:57

Quoting "rogue":

Quote:

Where is the precondition in the proof that c != T?

Division by zero is strictly disallowed.
Thus, the condition c != T is obvious when Z>2 and z>2.

Quoting "rogue":

Quote:

What does "prevents the substitution" supposed to mean?

It means that the substitution of (c/c) for (T/T) cannot occur
when Z>2 and z>2 because division by zero is strictly disallowed.

Quoting "rogue":

Quote:

It is when c = T that the problem of 1^(0/0) occurs.

In my proof, 1^(0/0) does not occur because unlike you,
I simplified the logarithmic exponents at Z=1 and z=2
before letting T = c.

Don't you know how to simplify?

Look... Here again is my proof:

httр://unsolvedрroblems.org/S14b.рdf

Anyone can see that the trivial expression (0/0) never occurs in it.

Quoting "rogue":

Quote:

If you want to convince us that you are correct, you must submit your proof to a math journal and get it published.

I don't want to convince you of anything.
I honestly couldn't care less what you think.

Also, my proof is published in an online journal for amateurs,
where anyone can referee it. That's good enough for me.

Don.

rogue · 2012-10-20, 16:17

Quote:

Originally Posted by Don Blazys

I honestly couldn't care less what you think.

That's obvious, you only care about what you think.

Quote:

Also, my proof is published in an online journal for amateurs,
where anyone can referee it. That's good enough for me.

What journal is that? Unsolved Problems? Who determines what is accepted to that journal (or whatever journal you have "published" it on)? I wouldn't be surprised if there are so many garbage proofs on that online journal that nobody wastes their time looking at them. As has been stated before, just because something is online doesn't make it true.

If your proof is correct, then why haven't I seen one Masters or Doctorate level mathematician defend it? If your proof is correct, why hasn't it received any accolades in the math community?

rogue · 2012-10-20, 21:04

Let's take another look. Assume that c != T. Although you don't state it explicitly in your definition of T, you should do that up front rather than stating "division by zero prevents". It is rather important to explain how you choose T. Why can't T be any real number > 1? I see no reason for the restriction that it must be an integer > 2.

Where in your proof do you state that the final formulas in steps 1 and 2 are not equal to some value i^j for i and j > 2?

In other words, it appears to me that you haven't proven that $T*({\frac{c}{T}})^{\frac{{\frac{(Z)*ln(c)}{ln(T)}-1}}<br /> {{\frac{ln(c)}{ln(T)}-1}}$ != i^j where i and j are integers > 2.

2012-10-19, 19:58	#521
rogue "Mark" Apr 2003 Between here and the 2²·7·227 Posts	See next post Last fiddled with by rogue on 2012-10-19 at 20:04 Reason: See next post

2012-10-20, 13:29	#525
rogue "Mark" Apr 2003 Between here and the 18D4₁₆ Posts	Where is the precondition in the proof that c != T? It is when c = T that the problem of 1^(0/0) occurs. What does "prevents the substitution" supposed to mean? Regardless of anything we say, you will argue. If you want to convince us that you are correct, you must submit your proof to a math journal and get it published. Posting it online and saying "it's correct" doesn't make your proof correct. Why do you refuse to do that? Your purported "proof" is so simple that it would take little time for independent referees to determine if it has any flaws. IMO, we are asking you to do what is most fair and reasonable to everyone here, including yourself. It seems to me that your abject refusal to publish your proof is cowardly and unreasonable. I believe that you already know your proof is wrong and are being a troll just because nobody has found the flaws in your proof that you are already aware of. Last fiddled with by rogue on 2012-10-20 at 14:12

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2012-10-19, 18:02	#520
Flatlander I quite division it "Chris" Feb 2005 England 31×67 Posts	He says they division by zero gives un undefined result except for 0/0. What are the consequences of accepting that 0/0 has a value that would prove his assumption wrong?

2012-10-20, 13:03	#524
Don Blazys Feb 2011 163 Posts	Those of you who think that the indeterminate form $\frac{0}{0}$ is somehow an "issue" in my proof are simply wrong. For one thing, the indeterminate form $\frac{0}{0}$ doesn't even exist in my proof! Now, the reason *why* the indeterminate form $\frac{0}{0}$ doesn't even exist in my proof is because I was clever enough to simplify the expressions involving logarithms at $Z=1$ and $z=2$ *before* substituting $\left(\frac{c}{c}\right)$ for $\left(\frac{T}{T}\right)$ . Those of you who don't know how to simplify or want to make the silly argument that "simplification is impossible" should read this: http://www.helpalgebra.com/onlinebook/simplifyingrationalexpressions.htm You see, in my proof, at $Z=1$ and $z=2$ , the logarithms simply "cancel out". Then, and only then do we substitute $\left(\frac{c}{c}\right)$ for $\left(\frac{T}{T}\right)$ . Thus, those indeterminate forms that you youngsters are making such a "big deal" about are really nothing more than "removable singularities" which are so utterly trivial that thay can be removed simply by avioding them in the first place! In fact, they are such a non-issue that we mathematicians refer to them as being "cosmetic", and even those of you who are too inexperienced to avoid them can *still* remove them in much the same way that the singularity at $z=0$ in the function $f(z)=\frac{\sin z}{z}$ can be removed simply by defining $f(0)=1$ . You can read more about that here: http://en.wikipedia.org/wiki/Removable_singularity Then, there is this to consider. According to "rogue's" argument, we can never ever substitute $\left(\frac{c}{c}\right)$ for $\left(\frac{T}{T}\right)$ in my proof, not even at $Z=1$ and $z=2$ . Thus "rogue" is clearly implying that what I "really" discovered is a way to absolutely and in all cases prevent the substitution of $1$ for $1$ . Heck, that would *really* make me a mathematical miracle worker! Don.

2012-10-20, 21:04	#528
rogue "Mark" Apr 2003 Between here and the 2²×7×227 Posts	Let's take another look. Assume that c != T. Although you don't state it explicitly in your definition of T, you should do that up front rather than stating "division by zero prevents". It is rather important to explain how you choose T. Why can't T be any real number > 1? I see no reason for the restriction that it must be an integer > 2. Where in your proof do you state that the final formulas in steps 1 and 2 are not equal to some value i^j for i and j > 2? In other words, it appears to me that you haven't proven that $T({\frac{c}{T}})^{\frac{{\frac{(Z)ln(c)}{ln(T)}-1}}<br /> {{\frac{ln(c)}{ln(T)}-1}}$ != i^j where i and j are integers > 2.