In article <email@example.com> firstname.lastname@example.org writes: > In article <G7A90y.email@example.com>, > "Dik T. Winter" <Dik.Winter@cwi.nl> wrote: ... > > You assume that whenever AB = 0 for your funny > > things that either A = 0 or B = 0 or both. But you have to give a proof > > of that because there are many funny things in mathematics where that > > does *not* hold. And it has been *proven* for the complex numbers, but > > that proof actually uses that x^2 + y^2 can not be 0 unless both x and > > y are 0. Just the thing you want to prove. So because you rely on that > > fact for the complex numbers you are reasoning in a circular fashion. It > > is like (with either x or y nonzero or both): > > x^2 + y^2 != 0 because whenever AB = 0 either A = 0 or B = 0 and the > > latter is true because x^2 + y^2 != 0. > > > > Then I notice that (x+sqrt(-1)y)(x-sqrt(-1)y) = 0 because x^2 + y^2 = 0. > > And remember, at this point as far as I'm concerned x and y are still > integers!
Up to this point I had no complaint at all, so why pull it out of the box again?
> Why? > > Because if I know they aren't at this point then I already must have > reached the point of contradiction. And then, your argument must be > that the proof I've given is too long!!!
Up to this point I see *no* contradiction at all.
> What some of you appear to be arguing is that when I realize that this > thing, sqrt(-1) is not an integer, I must stop, and pull out a book on > complex number theory.
Nope. You conclude that either (x + sqrt(-1)y) = 0 or (x - sqrt(-1)y) = 0, without giving proof of that. In another article you said you declined to give a proof of this. Interesting, because a proof is *essential*. You can not pull such statements out of your hat and claim they are true.
> I'm looking for something that pushes me outside of integers because > that's what I want to prove must happen!
Tsk. Your initial use of sqrt(-1) pushes you out of the integers already.
> You say the textbooks do it a different way, and I assume that assaults > your sense of order for me to show it this way because if everybody has > done it one way, you seem to assume that any other way is wrong. > > Then prove it's wrong!!!
Textbooks prove that in this case AB = 0 implies either A = 0 or B = 0 or both. Something you are not willing to do for some reason. I presume you are not able to do it? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/