condor, you explained it better than I was able to. That's exactly the flaw. It's like the [url=http://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html]classic proof that 1 = 2[/url].

[QUOTE=akruppa;259128]For those who haven't tried, quote Don and marvel at his text coloring prowess. :smile:[/QUOTE]
:missingteeth:

[QUOTE=akruppa;259128]For those who haven't tried, quote Don and marvel at his text coloring prowess. :smile:[/QUOTE]

quote the equality 10 times fast it took me like an hour to figure out in tex, and the worst part is I quoted it so I have no easy way to get it again.

Quoting "Condor":
[QUOTE]
So, Don can't set z=1 in

[TEX] \(\frac{T}{T}\)*c^{z}=
T*(\frac{c}{T}\)^{\frac{\frac{{z}*{\ln(c)}}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}} ,
[/TEX]

"divide out" the now identical terms,
and replace it with 1 as he wants to do.
[/QUOTE]

"Condor" is wrong again. Here's why.

If [TEX]z=1[/TEX] , then

[TEX] \(\frac{T}{T}\)*c^{z}=
T*(\frac{c}{T}\)^{\frac{\frac{{z}*{\ln(c)}}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}}
[/TEX]

becomes

[TEX] \(\frac{T}{T}\)*c^{1}=
T*(\frac{c}{T}\)^{\frac{\frac{{\ln(c)}}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}}=
T*(\frac{c}{T}\)^{1}
[/TEX]

where letting [TEX]T=c[/TEX] results in

[TEX] \(\frac{c}{c}\)*c^{1}=
c*(\frac{c}{c}\)^{\frac{0}{0}}=
c*(\frac{c}{c}\)^{1}
[/TEX]

which evaluates to

[TEX] (1)*c^{1}=
c*(1)^{\frac{0}{0}}=
c*(1)^{1}
[/TEX]

and since [TEX]\frac{0}{0}=n[/TEX] because [TEX]n*0=0[/TEX] ,
we can substitute [B][I]any[/I][/B] [B][U]number[/U][/B] [TEX]n[/TEX] for [TEX]\frac{0}{0}[/TEX] which gives us

[TEX] (1)*c^{1}=
c*(1)^{n}=
c*(1)^{1}
[/TEX]

where even though the above equation is [B][I][U]true[/U][/I][/B] for any number [TEX]n[/TEX],
consistency in logic would still require that we let [TEX]\frac{0}{0}=n=1[/TEX].

My proof works because [TEX]0 [/TEX] cannot divide any number [B][COLOR=red]exept itself[/COLOR][/B].

Don

Quoting akruppa:
[QUOTE] For those who haven't tried, quote Don and marvel at his text coloring prowess. :smile:
[/QUOTE]

[COLOR=red]I'm[/COLOR] [COLOR=blue]so[/COLOR] [COLOR=red]glad[/COLOR] [COLOR=blue]that[/COLOR] [COLOR=red]you[/COLOR] [COLOR=blue]do[/COLOR], [COLOR=red]in[/COLOR] [COLOR=blue]fact[/COLOR], [COLOR=red]appreciate[/COLOR] [COLOR=blue]my[/COLOR] [COLOR=red]text[/COLOR] [COLOR=blue]coloring[/COLOR] [COLOR=red]prowess[/COLOR]!

I developed that skill in order to help those who are not very
good at math see where they are making all their mistakes.

Also, I use colors because colors are [B][I]popular[/I][/B] with youngsters
such as yourself, "CRGreathouse", "Condor" and "science man 88".
I hear that kids like you even "wear their colors" as they dance to
their "Justin Beaver" music. Thus, I'm sure that the incredible popularity
of this thread is due, at least in part, to my use of color!

Most importantly however, my use of colors brings joy and laughter
to many people who would otherwise be quite miserable.

Quoting Uncwilly:
[QUOTE]
:missingteeth:
[/QUOTE]

See what I mean?

Don.

[QUOTE=Don Blazys;259163]My proof works because [TEX]0 [/TEX] cannot divide any number [B][COLOR=red]exept itself[/COLOR][/B].[/QUOTE]And that is precisely the point where your version of mathematics diverges from that used by virtually everyone else.

In non-Blazys mathematics, the operation of dividing by zero is undefined unconditionally, whether the dividend is zero or not.

Paul

[QUOTE=Don Blazys;259163]
consistency in logic would still require that we let [TEX]\frac{0}{0}=n=1[/TEX]. [/QUOTE]

Consistency. You keep using that word. I do not think it means what you think it means.
/Inigo Montoya

Quoting xilman:
[QUOTE]
And that is precisely the point where your version of mathematics
diverges from that used by virtually everyone else.

In non-Blazys mathematics, the operation of dividing by zero is
undefined unconditionally, whether the dividend is zero or not.
[/QUOTE]

You are absolutely and unequivocally [B][COLOR=red]wrong[/COLOR][/B] 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

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

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?"
[QUOTE]
It follows that 0 cannot divide any number exept itself.
[/QUOTE]

Don.

We have a new player: xilman.

Double :popcorn::popcorn:

[QUOTE=Don Blazys;259177]Quoting xilman:


You are absolutely and unequivocally [B][COLOR=red]wrong[/COLOR][/B] 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

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

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.[/QUOTE]I followed the link you gave, where I found this statement: "[FONT=Verdana][SIZE=2][COLOR=#000000][FONT=Verdana][SIZE=2][COLOR=#000000][FONT=Verdana][SIZE=2][COLOR=#000000]Therefore, 0/0 does not mean any particular number - or even anything until we give it some new meaning."[/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT][/COLOR][/SIZE][/FONT]

I used the term "undefined". The article quoted uses the more colloquial "does not mean any particular number - or even anything" which is a pretty good if somewhat verbose definition for "undefined".

You are, of course, perfectly entitled to assign any meaning you wish to any word or phrase of your choice. The downside is that hardly anyone else will agree with you, or even understand you, unless you agree to abide by the meaning assigned by almost everyone else.

Ah, the glory of a knock down argument!


Paul

It's not that hard to understand, and the [B]meaning[/B] of 0/0 is clear.

In my proof, [TEX]0/0 = n[/TEX] is [B][COLOR=green]allowed[/COLOR][/B] because [TEX]n*0 = 0[/TEX]

and [TEX]2/0 = n[/TEX] is [COLOR=red][B]disallowed[/B][/COLOR] because [TEX]n*0 \neq 2[/TEX].

Quoting xilman:
[QUOTE]
Ah, the glory of a knock down argument!
[/QUOTE]

The "Condor" was perched on a tree. "You",
along with "CRGreathouse", "science man 88",
"akruppa", "Uncwilly", "axn", and several "others"
chose to stand under it, and got "knocked down"!

[B][SIZE=3][COLOR=sienna][COLOR=black]"[/COLOR]It[/COLOR][COLOR=black]"[/COLOR][/SIZE][/B] happens!

Don.

[SIZE=1]PS. Clever use of color, don't you think?[/SIZE]