> 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!!!
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 ["natÃÂÃÂ¼rliche 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 =  (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] = , 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 =  : [z] != , but this is impossible as your proof shows!
> 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.