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Topic: FLT Discussion: Simplifying
Replies: 65   Last Post: Mar 17, 2001 11:59 PM

 Messages: [ Previous | Next ]
 Franz Fritsche Posts: 4 Registered: 12/13/04
Re: FLT Discussion: Algebra...
Posted: Jan 26, 2001 2:05 AM

Hi James!

> Here's where there's the issue between what I've recently called
> patterns and regular rings.
>

You can call it how you want to... but it is a matter of fact that even
patterns *can* have a algebraic "behavior" [that YOU attribute to them]
that let them be isomorphic to a special ring...

> My example has been
>
> (x+sqrt(-1)y)(x-sqrt(-1)y) = x^2 + y^2,
>
> which as a pattern is simply true for valid x's and y's.
>

???

Yes, If the pattern follows some special rules...
But HOW to PROOF that?!

You can define a lott of things but shis does NOT ensure yout that ONE
sinble object of the defined things exist!!!

Eyample:

I define: If a number x < 0 then i call it a negativ number.

Seems to be perfectly clear, no?

But what if I deak with the non negativ integers ["natÃÂÃÂ¼rliche Zahlen"] :
no SINGLE "negative" number exists!

> What I've shown is a problem with the coupling so many of you have
been
> heatedly defending, since the statement is valid with x and y members
> of an infinite number of rings, where the ring of integers is *one*
> such, which happens to not have the result of sqrt(-1) as a member.
>

True...

> Note the distinction between calling sqrt(-1) a result, since we're
> looking at a number with an operator, versus the casual error of
> concluding that it is the number itself.
>

Nice try, James - but WHAT IS the result of the operation, an apple?!

Maybe it SHOULD be some sort of "number" - at least something one could
calculate with??? No?!

> After all, in once case that result is i...
>

Wrong side... James... Not the "result" ob some operation [which on?] is
i, but wie van show that the proof holds for defining the pattern
"sqrt(-1)" := i e C.

> The only remaining issue then has been an insistence that it must be
> proven that (x+sqrt(-1)y)(x-sqrt(-1)y) = 0 means that x+sqrt(-1)y = 0
> or x - sqrt(-1)y = 0,
>

That's true James... :-)

> as some have insisted I must move to complex
> numbers to get this result.
>

No not necessarily... - Some special matrices would do too... :-)

> But, I've repeatedly brought up the fact that it's true for other
> rings, so why if you guys are acting like using rings is so

fundamental
> and important...
>

Becouse a SET of "object" is CALLED a ring if ther are some special
operations like + and * and some rules [Axioms] are fulfilled...

If you use a ring everything is clear... maybe not for *your* proof
so...

> do you wish to make a result that's a hack depending on
> what ring you're using?
>

I think so...

>
> Don't understand what I mean?
>
> Let's say that I'm in some other ring besides integers. All the
> results I've given are still true. That is,
>
> (x+sqrt(-1)y)(x-sqrt(-1)y) = x^2 + y^2, and if x^2 + y^2 = 0, then
>
> (x+sqrt(-1)y)(x-sqrt(-1)y) = 0.
>
> You've all been insisting that I have to know what ring x and y are in

> to prove whether or not this means that
>
> (x+sqrt(-1)y) = 0 *or* (x-sqrt(-1)y) = 0.
>
> When my simple question to you is, when would it not be?
>

When the ring has zero divisors [there ARE such rings..]

> So, if all of you are right, then it's not possible to have the
general
> result over a large number of rings that (x+sqrt(-1)y)(x-sqrt(-1)y) =
0
> means that (x+sqrt(-1)y) = 0 *or* (x-sqrt(-1)y) = 0.
>

Exactly!!!

> Then, basically all of you are arguing that it's impossible to prove
> this result generally, and that you have to go ring, by ring.
>

Not really : you yust have to show that in their special ring
(x+sqrt(-1)y)(x-sqrt(-1)y) = 0 => (x+sqrt(-1)y) = 0 *or* (x-sqrt(-1)y) =
0 and
sqrt(-1) * sqrt (-1) = -1 holds!

[This is the case for C or special matices rings... i.e...]

NOW... as you want to have x and y to be INTEGERS, it's enough to use a
ring in which the integers can be embeded...

Or in other words... an ring for which each integer number x corresponds
to a number in the ring...

x <--> [x], with x e Z and [x] e Ring...

That's enough? Why?

Because wie stated that

x^2 + y^2 = 0 (1)

this can then be re-stated as

[ x]^2 + [y]^2 = [0] (1)

With the corresponding values of the Ring...; (1) holds exactly when (2)
holds!

NOW your proof shows that these ONLY holds for [x] = [y] = [0], and it
follows that (1) only can hold for x=y=0.

Otherwise wie should have a z!=0 with z^2 + y^2 = 0 (1) and this wold
lead to [z]^2 + [y]^2 = [0] : [z] != [0], but this is impossible as

> Well folks, there are an infinite number of rings that x and y could
> belong to for which the above would be valid.
>

You see: ONE is enogh!

So long,
Franz

P.S.
But be carefull: you HAVE TO stick on x and y to be integers...!

Otherwise in C we would have:
1^2 + i^2 = 1 + (-1) = 0.

Date Subject Author
1/15/01 jstevh@my-deja.com
1/15/01 Dik T. Winter
1/16/01 Charles H. Giffen
1/16/01 jstevh@my-deja.com
1/16/01 Randy Poe
1/18/01 jstevh@my-deja.com
1/18/01 Michael Hochster
1/18/01 Peter Johnston
1/18/01 Randy Poe
1/18/01 Doug Norris
1/16/01 Doug Norris
1/16/01 Randy Poe
1/16/01 Dik T. Winter
1/18/01 jstevh@my-deja.com
1/19/01 Dik T. Winter
1/19/01 Randy Poe
1/20/01 jstevh@my-deja.com
1/20/01 oooF
1/21/01 hale@mailhost.tcs.tulane.edu
1/21/01 Peter Percival
1/21/01 Randy Poe
1/26/01 Franz Fritsche
1/19/01 gus gassmann
1/20/01 jstevh@my-deja.com
1/20/01 Doug Norris
1/26/01 Franz Fritsche
1/16/01 hale@mailhost.tcs.tulane.edu
1/16/01 Randy Poe
1/17/01 hale@mailhost.tcs.tulane.edu
1/18/01 jstevh@my-deja.com
1/19/01 hale@mailhost.tcs.tulane.edu
1/20/01 jstevh@my-deja.com
1/21/01 hale@mailhost.tcs.tulane.edu
1/18/01 Peter Percival
1/19/01 hale@mailhost.tcs.tulane.edu
3/17/01 Ross A. Finlayson
1/16/01 hale@mailhost.tcs.tulane.edu
1/18/01 jstevh@my-deja.com
1/19/01 hale@mailhost.tcs.tulane.edu
1/29/01 jstevh@my-deja.com
1/19/01 Dik T. Winter
1/21/01 Dennis Eriksson
1/15/01 Michael Hochster
1/16/01 jstevh@my-deja.com
1/16/01 Michael Hochster
1/18/01 jstevh@my-deja.com
1/18/01 Peter Percival
1/18/01 Randy Poe
1/19/01 oooF
1/21/01 Dik T. Winter
1/21/01 oooF
1/18/01 Edward Carter
1/19/01 W. Dale Hall
1/19/01 Michael Hochster
1/16/01 Randy Poe
1/16/01 Randy Poe
1/17/01 W. Dale Hall
1/17/01 W. Dale Hall
1/19/01 oooF
1/16/01 Charles H. Giffen
1/16/01 David Bernier
1/16/01 jstevh@my-deja.com
1/18/01 Arthur
1/30/01 plofap@my-deja.com
1/30/01 plofap@my-deja.com
1/30/01 plofap@my-deja.com