[QUOTE=Don Blazys;261680]This has absolutely nothing to do with limits. We are dealing strictly with positive integers.[/quote]
Limits have everything to do with it, since that is how you can evaluate some indeterminate forms at integers. It is the only legitimate way to "remove" the logarithms when z=1 and z=2.
[QUOTE]My work is highly original, and I am indeed the first mathematician to unequivocally state that 1^(0/0) = 1^n = 1. :smile:[/QUOTE]
And also to be proven wrong about it, since we have a case where its continuous extension is e=2.71828.
[QUOTE]The [B][U]fact[/U][/B] is, 0 cannot divide any number [B][COLOR=blue]exept itself [/COLOR][/B]and when it [B][I]does[/I][/B] divide itself, its quotient is [B][COLOR=red]any number[/COLOR][/B] n.[/QUOTE]
No, the fact is that 0 can't divide any number, period; but the continuous extension of the expression could be any number, or it could be unbounded. You don't know, and can't tell, until you take the limits.

When it is unbounded, the continuous extension of 1^(0/0) could be any number, or be unbounded.
[quote]Besides, in my proof ... [B]there are no indeterminate forms[/B] because logarithms are not even involved when either z = 1 or z = 2.[/quote]
(1) You can't remove indeterminate forms by dividing them out, so yes there are there, even for z=1 and z=2.
(2) You still haven't addressed the fact that there are just as valid expressions with z=1 and z=2 where the logarithms cannot be "removed" by your method, but they can for other integers.
[QUOTE]Thus, what you all think is an "issue" is in fact, utterly trivial![/QUOTE]
If it is trivial, stop ignoring the points that prove it is wrong.

Quoting "Condor":
[QUOTE]
No, the fact is that 0 can't divide any number, period.
[/QUOTE]
No, the fact is that [COLOR=red][B][I]0 can't divide any number[/I][/B][/COLOR] [B][I][COLOR=blue]exept itself[/COLOR][/I][/B],

.........................................................................period. :smile:

Even middle-schoolers can understand that!
Here, you should read [I][B][U]this[/U][/B][/I] again.

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

Don.

Maybe Don should try reading it again:
[QUOTE][B]What does "0/0" mean?[/B]
0/0 means the the number c such that 0xc = 0. What value of c will make this equation true? How about 1? or 2, or -26/31? Yes! c can be any number and still satisfy 0xc = 0. Therefore, 0/0 does not mean any particular number - [COLOR="Red"][B]or even anything until we give it some new meaning.[/B][/COLOR][/QUOTE]
Don needs to understand the difference between the words "can be" used in this oversimplified explanation meant for middle schoolers, and the word "is" that he keeps using. The first expresses the impossibility of choosing any particular number for the expression since any other number "can be" used. But Don wants to pick a number he likes, which is explicitly disallowed by the part he ignores in this text. And what is left out, that the c could be unbounded, leading to the form this text omits: "What does [tex]1^\infty[/tex] mean?" I've given a case where it does not mean 1.

What the text says, is that you need to define a value for the expression some other way. That way requires the assumption of continuity and the use of limits (yes, even for integers).

Quoting "CRGreathouse":
[QUOTE]
Let's use / to denote division and "/" to denote
the inverse image of (real) multiplication.
The two are similar:

6 / 3 = 2
6 "/" 3 = {2}

21 / 7 = 3
21 "/" 7 = {3}

but they have differences:

5 / 0 is undefined, i.e., meaningless
5 "/" 0 = {}, the empty set

0 / 0 is undefined, i.e., meaningless
0 "/" 0 = R, the set of real numbers.

So now what happens when you attempt to take 1 ^ (0 "/" 0)?
You're raising a number to the power of a set,
not an allowed operation.
[/QUOTE]

That's pure gibberish!

In [B][U]mathematics[/U][/B], indeterminate forms are treated
exactly the same as unknowns or variables... [B][I]not[/I][/B] sets!

Thus, if n*0 = 0, then 0/0 = n where n E R (n is an element of R).

Don.

[QUOTE]
[B]What does "0/0" mean?[/B]
[B][COLOR=red]0/0 means the number c such that c*o = 0. [/COLOR][/B]
[/QUOTE]
That's final.

Quoting "Condor".
[QUOTE]
The first expresses the impossibility of choosing any particular number
for the expression since any other number "can be" used.
[/QUOTE]

"Condor" just keeps on laying more and more "eggs"!
His silly and retarted notion that there somehow exists...

"the impossibility of choosing any particular number... "
would render the concept of [B][I][U]variables[/U][/I][/B] impossible!

Indeed, in mathematics, indeterminate forms are treated
[B][I]exactly[/I][/B] the same as unknowns or variables.

Don.

Don, again I ask you these three questions:

1) Is 1^(0/0) determinate?
2) Is 1^n = 1 when n is undefined?
3) Is 1^n = 1 when n is indeterminate?

Or, on could look up a web site with an explanation intended for adults, not children: [url=http://mathworld.wolfram.com/DivisionbyZero.html]Wolfram's Mathworld[/url] says "The uniqueness of division breaks down when dividing by zero, since the product 0*y=0 is the same for any y, so [B][COLOR="Red"]y cannot be recovered[/COLOR][/B]."

Or [url=http://mathforum.org/library/drmath/view/57322.html]the Math Forum at Drexel[/url]: "'What is the value of 0/0 ? (is it really undefined or are there aninfinite number of values)?' There's a special word for stuff like this, where you could conceivably give it any number of values. [B][COLOR="Red"]That word is 'indeterminate.'[/COLOR][/B]"

[url=http://mathworld.wolfram.com/Indeterminate.html]Wolfram[/url] adds that "a limit of the form 0/0" - meaning the expression f(x)/g(x) when x approaches a value where f(x)=g(x)=0 - "is indeterminate." Theyt also say that a limit of the form [tex]1^\infty[/tex] - meaning f(x)^g(x) when f(x)=1 and g(x) is unbounded - is allso indeterminate. The same rules apply to either expression - they could mean any value, but you can only recover what that value is using the limits.

This one is quite old, but it seemed appropriate for this thread. It is the transcript of an exchange in a Court between an unrepresented suitor and a Judge. It seems the Judge was being asked to do something about the rules for multiplying and dividing by 0, nuclear physics and a good deal more besides.

[url]http://www.ratbags.com/rsoles/comment/routhighcourt.htm[/url]

[QUOTE=99.94;263259]This one is quite old, but it seemed appropriate for this thread. It is the transcript of an exchange in a Court between an unrepresented suitor and a Judge. It seems the Judge was being asked to do something about the rules for multiplying and dividing by 0, nuclear physics and a good deal more besides.

[url]http://www.ratbags.com/rsoles/comment/routhighcourt.htm[/url][/QUOTE]

His rantings sound like a typical day in sci.math.

Is there any value in sci.math anymore?

[QUOTE=Christenson;263273]Is there any value in sci.math anymore?[/QUOTE]

A little, but not much IMO. It is totally inundated by cranks, kooks,
trolls, and other unwelcome intruders. I find their psychology to be
a mystery --> which is one of my limitations: I don't understand their motives.

Back in the mid 80's it was a very pleasant and very productive venue for
discussion. Many world-class mathematicians (which I am not) would
ask questions, present problems, and give answers to questions asked
by others. Mathematicians with such stature as Noam Elkies, Andrew Odlyzko, etc. would post there.

Not any more......