Standard crank division by zero thread - Page 24

xilman · 2011-04-21, 12:13

Quote:

Originally Posted by Don Blazys

chose to stand under it, and got "knocked down"!

Whoosh!

Paul

science_man_88 · 2011-04-21, 13:08

Quote:

Originally Posted by Don Blazys

Quoting xilman:

You are absolutely and unequivocally wrong xilman.

In fact, I can assure you that you don't even know
what almost all middle-schoolers know because
I happen to work at a high school / middle school,
so I see what the kids are being taught every day.

I found this article

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

just today, in several of our classrooms. Read it!
Especially the part under "1/0" where it says...

Quoting the "Math Path" article "What Does 0/0 mean?"

Don.

that fact that you're at a high school/middle school kinda scares me personally.

flouran · 2011-04-21, 14:16

Quote:

Originally Posted by Don Blazys

as they dance to
their "Justin Beaver" music.

It's actually "Justin Bieber"

Quote:

Originally Posted by Don Blazys

I found this article

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

just today, in several of our classrooms. Read it!
Especially the part under "1/0" where it says...

Quoting the "Math Path" article "What Does 0/0 mean?"

Don.

Read more about l'Hôpital's rule and also get a better understanding of indeterminate forms. 0/0 is one of these such forms. As xilman already pointed out, the article you reference is essentially saying 0/0 is indeterminate (the indeterminacy of 0/0 is a universal concept and exceptions do not exist for particular cases as you have misunderstood), but in much simpler terms.

As a side note, I suggest you use more mathematically professional articles to state your "case" (I think I'm being rather generous here to call what you have a "case"), else you won't be taken seriously (I'm also being generous here to say you're being taken seriously at this point).

Quote:

Originally Posted by science_man_88

that fact that you're at a high school/middle school kinda scares me personally.

Very true.

akruppa · 2011-04-21, 14:20

Quote:

Originally Posted by Don Blazys

Most importantly however, my use of colors brings joy and laughter
to many people

The whole thread brings laughter to many people!

rogue · 2011-04-21, 14:29

Division by 0 is indeterminate. It doesn't mean that 0/0 = n for some fixed value of n. It doesn't mean that n has a value. It just means that n is indeterminate.

You can't just say "Let T = c" later in your proof, especially when c can have the value 1 and T cannot. If T = c is then a condition of your proof, then your proof goes "poof" due to division by 0 because

${\frac{\frac{{z}*{\ln(c)}}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}}$

becomes:

${\frac{z-1}{0}$

which you imply is defined only when z = 1. In other words, that if z != 1, the fact that ${\frac{z-1}{0}$ is undefined means that z must be equal to 1.

Condor · 2011-04-21, 15:30

Quote:

Originally Posted by rogue

Division by 0 is indeterminate.

Division by zero is undefined. When an expression has zero divided by zero, it's a special kind of undefined we call indeterminate because it can have an alternate definition under certain assumptions.

But 1^(1/0) is similarly indeterminate, so it is just as "allowed" to have a zero in that place as in Don's form.

Quote:

It doesn't mean that 0/0 = n for some fixed value of n. It doesn't mean that n has a value. It just means that n is indeterminate.

Right. Don is misusing an over-simplified explanation aimed at high school students who are not mathematically inclined as though it were truth.

Quote:

You can't just say "Let T = c" later in your proof, especially when c can have the value 1 and T cannot. If T = c is then a condition of your proof, then your proof goes "poof" due to division by 0 because

${\frac{\frac{{z}*{\ln(c)}}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}}$

becomes:

${\frac{z-1}{0}$

Or, one could also expand $C^z$ the following way (and btw, I corrected an error in the derivation that I had cut-and-pasted before).

For any integer M:
$ \frac{T}{T}C^z =$\frac{T}{T}C^{\frac{z}{M}}$^M =$T\(\frac{C}{T}$^{\frac{ln(\frac{C^{\frac{z}{M}}}{T})}{ln(\frac{C}{T})}}\)^M =$T\(\frac{C}{T}$^{\frac{\frac{ln(\frac{C^{\frac{z}{M}}}{T})}{ln(T)}}{\frac{ln(\frac{C}{T})}{ln(T)}}}\)^M =$T\(\frac{C}{T}$^{\frac{\frac{ln(C^{\frac{z}{M}})-ln({T})}{ln(T)}}{\frac{ln(C)-ln(T)}{ln(T)}}}\)^M =$T\(\frac{C}{T}$^{\frac{\frac{{\frac{z}{M}}*{ln(C)}}{ln(T)}-1}{\frac{ln(C)}{ln(T)}-1}}\)^M $

Let's apply Don's logic - or is that illogic? - to this. We can substitute z=M and reduce this to $$T\(\frac{C}{T}$\)^z$ . Now substituting c=T is "allowed." But for any other value of z, including 1 if M is not 1, substituting c=T is "disallowed." But since we can do this for any integer M, we have "proved" that every integer z is both allowed and disallowed. This seems like a "classic contradiction" to me - I wonder what assumption we need to look for, that has been proven incorrect? Could it be Don's whole line of reasoning? Here's a fun new game: point out a new flaw in Don's proof, and see what abuse Don will do to math in order to discount that flaw.

Algebraic manipulation is a tool only, not a means of directly validating or invalidating a concept. If you find that what should be a defined quantity becomes undefined or "disallowed to evaluate," it says nothing about that quantity. It means you used the tool incorrectly.

Don Blazys · 2011-04-22, 13:31

Quoting "Condor":

Quote:

1^(1/0) is similarly indeterminate.

Looks like "Condor" laid yet another "egg"!

0 cannot divide any number exept itself.

Dividing any number other than 0, by 0
is strictly dissalowed. That's the truth.

Quoting "roque":

Quote:

It doesn't mean that 0/0 = n for some fixed value of n.
It doesn't mean that n has a value.

Division "undoes" multiplication and vice versa.

Thus, if $n$ has a value in the multiplication $n*0=0$ ,
then $n$ must also have that same value in the division $$\frac{0}{0}$=n$ .

That's just common sense!

Thus, $n*0=0$ is true for any number $$\frac{0}{0}$=n$ .

Quoting "akruppa":

Quote:

The whole thread brings laughter to many people!

That's because humour must have an
element of truth in order to be funny.

Quoting

Quote:

Read more about l'Hôpital's rule...
0/0 is one of these such forms.

l'Hôpital's rule is used only if we need to determine
a particular value of $n$ in $$\frac{0}{0}$=n$ .

However, in my proof, we don't need to determine
any such particular value of $n$ .

In my proof, all we need to determine is that
$$\frac{0}{0}$=n$ is true for any number $n$ .

Quoting "flouran"
[QUOTE]
As a side note, I suggest you use more mathematically
professional articles to state your "case".

The article I cited told the truth. Namely, that
"0 cannot divide any number exept itself".

Moreover, I find the "education level" of that article
to be perfectly suitable for the denizens of this forum.

Quoting "flouran":

Quote:

... else you won't be taken seriously

Oh, I'm being taken seriously alright.

This thread is extrordinarily popular and some posters such as
"CRGreathouse", "Condor" and "science man 88", among others, are so
absolutely obsessed with my work, that they can't help but continue
posting on this topic, because deep inside, they know that all their
attempts at finding some "fatal flaw" in my proof failed miserably.

It just "sticks in their craw" that I'm right, and they're wrong.
Their egos simply can't handle it, and now they are desperate
because it is becoming painfully clear to everyone here that
the easily avoidable "indeterminate form" $\frac{0}{0}$ is a "non-issue".

Moreover, this thread is being followed by Don Blasius from U.C.L.A.
and probably has many other very distinguished readers as well.

If I was wrong, then everyone (including myself) would have lost
interest in this topic a long time ago. Instead, just the opposite
has happened! My readers are now drinking beer, eating popcorn,
and in general, having a great time watching my "opponents" make
complete and utter fools of themselves.

Quoting "flouran":

Quote:

It's actually "Justin Bieber"

You must be a fan.

Don.

xilman · 2011-04-22, 13:55

Quote:

Originally Posted by Don Blazys

That's just common sense!

What does common sense have to do with the validity of mathematical theorems?

Very little, as far as I can tell. For example, the Banach-Tarski "paradox" appears to violate common sense.

http://en.wikipedia.org/wiki/Banach%...Tarski_paradox

Paul

rogue · 2011-04-22, 14:17

Quote:

Originally Posted by Don Blazys

Thus, $n*0=0$ is true for any number $$\frac{0}{0}$=n$ .

No it doesn't. n*0=0 does not imply that 0/0 = n. You can't simply divide both sides by 0.

You are basically saying that n*0 = 0 becomes (n*0)/0 = 0/0 becomes n = 0/0. In reality, you could also do this: (n*0)/0 = (n/0)*(0/0) = 0/0. If 0/0 = n (which you state to be true), then (n/0)*n = n. Divide both sides by n, now you get n*0 = 1. That isn't right because n*0 = 0.

We are trying to get you to understand that you cannot use division by 0 (even 0/0) as the basis of any proof. Show your proof (and this thread) to any real mathematician (I have a degree in math BTW) and they tell you the same that we are trying to tell you.

Spherical Cow · 2011-04-22, 14:35

Quote:

Originally Posted by Don Blazys

because it is becoming painfully clear to everyone here that
the easily avoidable "indeterminate form" $\frac{0}{0}$ is a "non-issue".

You must be using the indeterminate form of "everyone"; I'm a regular reader, followed this thread, and consider myself part of "everyone". The only painful clarity is that your proof is flawed, and your presentation of your argument is worse. You should try to minimize the insults to other's intelligence; people might take you more seriously then, and you'd have a better chance of getting your point across.

By following this thread, I've learned a few things from CRGreathouse, Condor, and others, and its been entertaining sometimes. I didn't read it because I thought you were right; I read it because I was able to learn.

A poll of regular readers would be interesting- I'd be surprised if anyone thinks you're right, much less everyone.

Norm

CRGreathouse · 2011-04-22, 16:49

Quote:

Originally Posted by Spherical Cow

A poll of regular readers would be interesting- I'd be surprised if anyone thinks you're right, much less everyone.

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.

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2011-04-21, 14:29	#258
rogue "Mark" Apr 2003 Between here and the 1100011010000₂ Posts	Division by 0 is indeterminate. It doesn't mean that 0/0 = n for some fixed value of n. It doesn't mean that n has a value. It just means that n is indeterminate. You can't just say "Let T = c" later in your proof, especially when c can have the value 1 and T cannot. If T = c is then a condition of your proof, then your proof goes "poof" due to division by 0 because ${\frac{\frac{{z}*{\ln(c)}}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}}$ becomes: ${\frac{z-1}{0}$ which you imply is defined only when z = 1. In other words, that if z != 1, the fact that ${\frac{z-1}{0}$ is undefined means that z must be equal to 1.