email@example.com wrote: > > In article <940akc$d9p$1@nntp.Stanford.EDU>, > Michael Hochster <michael@rgmiller.Stanford.EDU> wrote: > > > > : Given x^2 + y^2 = 0, x and y nonzero integers, show that no solution > > : exists. > > > > : Proof by contradiction: > > > > : (x+sqrt(-1)y)(x-sqrt(-1)y) = x^2 + y^2 = 0, so > > > > : x = sqrt(-1)y *or* x = -sqrt(-1)y. > > > > Still waiting for an explanation of this step. > > Here's a case where I've left out what I think are obvious steps, and > this person disagrees. Some may think it unnecessary for me to add > them, others may not. > > Here are the missing steps: > > Starting from (x+sqrt(-1)y)(x-sqrt(-1)y) = x^2 + y^2 = 0, >
While this is certainly a correct factorization in C[x,y] (C being the field of complex numbers), or Z[i][x,y] (Z[i] being the ring of Gaussian integers), without an understanding of what rules apply to products sqrt(-1)*a where a is either an element of the coefficient ring or a polynomial with suitable coefficients, and what rules apply to summation, the above expression is simply undefined.
One may certainly *assume* that the ring wherein these manipulations take place is one of the aforementioned rings, but then you're back in the situation where the result has been assumed for you already. As has been pointed out on multiple occasions, it is the *absence* of nonzero solutions to x^2 + y^2 = 0 that leads more or less directly to the fact that C (or Z[i]) is a domain. Thus, whenever you appeal to this machinery, you're in effect saying, "I'll assume that no nonzero reals (x,y) have x^2 + y^2 =0." When you then proclaim the result "therefore no nonzero integers (x, y) have x^2 + y^2 = 0, QED!", you have engaged in circular logic.
> (x+sqrt(-1)y)(x-sqrt(-1)= 0, so > > x + sqrt(-1)y = 0 or x -sqrt(-1)y = 0, so > > x = -sqrt(1)y or x = sqrt(-1)y. > > Some, for reasons I'd like them to explain, have complained that I > don't know that x + sqrt(-1)y = 0 or x -sqrt(-1)y = 0, if > > (x+sqrt(-1)y)(x-sqrt(-1)= 0. > > (Sort of like if AB = 0, A or B = 0. These people are saying that must > be proven, and that it is a "gap" in my proof that I don't do so.) > > If so, I'd like them to say that is their position here and we can see > if we can't work that one out. >
Congratulations. After a mere 3 weeks, you've managed to focus on an actual question that has been put to you directly.
Let me be precise: We are given a ring(which I'll assume are polynomials in x,y (commuting indeterminates) and sqrt(-1), with coefficients in Z, and where we assume the usual properties of sqrt(-1) relative to the integers: the ring is commutative, and sqrt(-1)^2 = -1. The issue is to prove that, for such objects, the condition AB=0 allows one to conclude that either A = 0 or B = 0 (or both).
The property must be proven without appeal that same property for complex numbers or Gaussian integers, and especially without appeal to the property that for reals x,y, the polynomial x^2+y^2 takes only positive values.
> > > > : There doesn't exist an integers that multiplies times itself to > give a > > : negative number, and an integer can't be the product of an integer > and > > : a non integer, so there's a contradiction. > >
When you employ this fact (which you haven't proven, but for which the proof is utterly trivial), together with trichotomy for the order relation among integers or rationals or reals, and the fact that the sum of a non-negative number with a positive number is strictly positive, you have already assumed the final result: no sum of nonzero squares can ever be zero. The factorization argument is totally superfluous. That's why the argument has been called circular.
> > > > Ok, let's say you're right, and it did need work, and you may think it > still does. I don't have a problem with that. > > What I want to emphasize is that there is a process that can lead to > resolution and it is clear that there are those of you willing to > engage in it based on the fact that you made those comments here. > > So, a reasonable person may now ask, why hasn't that process played out > this way with my claims of a simple proof of Fermat's Last Theorem? >
One might venture to ask whether you have access to any sort of mirror.