
Re: FLT Discussion: Simplifying
Posted:
Jan 18, 2001 8:05 PM


Isn't the outcome worth it? If you can convince us that the proof is valid, WHY not?
Peter
<jstevh@mydeja.com> wrote in message news://947tfo$3lu$1@nnrp1.deja.com... > In article <3a64de9b.260659399@news.newsguy.com>, > randyp@visionplace.com (Randy Poe) wrote: > > On Tue, 16 Jan 2001 23:04:15 GMT, jstevh@mydeja.com wrote: > > > > >> > (x+sqrt(1)y)(xsqrt(1)y) = x^2 + y^2 = 0, so > > >> > > > >> > x = sqrt(1)y *or* x = sqrt(1)y. > > >> > > >> James, you have to *prove* that last step. You cannot rely on the > > >standard > > >> proof because it uses: x^2 + y^2 <> 0 whenever x != 0 or y != 0. > > > > > >I think some people may be surprised that you are essentially calling > > >this a gap. > > > > Quick poll: Anyone who is surprised that this is being called a "gap" > > (remove the word "essentially"), chime in. > > > > > > > >My comment to them is that it's the same thing being done with the > > >proof of Fermat's Last Theorem. > > > > Yes. This proof was introduced into this forum because the logical > > fault is similar. > > > > Ok folks, here you are hearing that strange statement again. > > I want to emphasize that this guy is arguing that what I have above is > not a proof. > > > > > > >As for your other statement, it doesn't make any sense to me. > > > > > >What's going on is simple though. I have that x^2 + y^2 = 0, and I > > >know that x^2 + y^2 = (x+sqrt(1)y)(xsqrt(1)y), > > > > You don't know this unless you define the context in which you are > > operating. > > > > See folks? I just don't get why I have to repeat over, and over, and > over again that x and y are integers. > > Do you? > > Given that x and y are integers, how many of you out there have any > doubt that x^2 + y^2 = (x+sqrt(1)y)(xsqrt(1)y)? > > Well, this guy certainly seems to doubt it, and not only that he's > doubts it so much that he decided he'd post that doubt with emphasis. > > What gives? > > Remember, he's taking this position as part of saying that my proof of > Fermat's Last Theorem is wrong. > > That's the big picture, and you should keep it in mind. > > > > so > > > > > > (x+sqrt(1)y)(xsqrt(1)y) = 0, > > > > > so > > > > > > x+sqrt(1)y = 0 or xsqrt(1)y=0, > > > > And you don't know this automatically. > > > > > >and that goes to the question of how long a proof has to be. > > > > No, it goes to the question of whether a statement preceded by the > > word "so" actually follows logically from the statement that precedes > > it. > > > > > > > >As you can see, some people will force you to outline simple steps > that > > >others would find unnecessary. > > > > And the person providing the proof should be able to outline those > > steps, even if he/she finds it unnecessary. > > > > So humor us. Even though it's unnecessary, prove that > > > > ab = 0, for complex a, b (they're COMPLEX, not integers, OK?) > > > > implies > > > > a = 0 or b = 0. > > > > It's not an axiom, it's a theorem. So it's provable. Just for grins, > > offer up a proof. Even though it's unnecessary. Waste a few electrons > > on us. It can't hurt. > > > It doesn't matter whether or not a and b are complex or not. > > The only possible issue might be made by using something like 3(2)=0 > (mod 6), but we're not talking about a finite ring. > > As for me proving that if ab = 0, a or b equals 0, I'm not interested > in doing that. > > If you need that proof to understand what I'm talking about, oh well. > > I'll just worry about the people who don't have a problem with that. > > > > > > >That bit of arbitrariness is used by some people to claim that a > proof > > >isn't. > > > > So isn't the obvious answer to add a few lines to prove the assertion? > > If you did, everyone would shut up. Saying "I don't have to" doesn't > > do anything to answer the objections. What's the harm in explaining > > what mathematical principle justifies the conclusion? > > > > I did that. You folks just keep asking for more and more detail, like > you asking for me to prove that if ab = 0, a or b = 0. > > > > > > >> > > >> > First off, it's worth noting that I'm treating the sqrt(1) as > an > > >> > *operation*. I think some of you live under the false notion > that > > >> > something like sqrt(2) is the actual number. > > >> > > >> I must be dense, but what the heck do you mean? If it is not a > number > > >> you have to define how to do multiplication of operations with > > >numbers. > > >> > > > > > >The sqrt(2) is a representation of the number that multiplies times > > >itself to give 2 (notice how circular that is). > > > > That makes it a number, not an operation. It stands for the NUMBER. > > That is not at all circular. > > Isn't it? Like I've said, we don't put down 1+1 to represent 2. > The '+' is an operator. The sqr() is an operator. We use the > operation to represent the square root of 2, or we write 1.414..., or > something similar. > > Why you would argue such a simple point is beyond me. > > > > > >I think I should mention that there's also a question of trueness. > > > > Don't know what you mean by this, except that perhaps you are looking > > for a vote on the truth value of an assertion. > > > > In mathematics, normally a sequence of logical deductions serves to > > demonstrate truth. Got that? > > > > SEQUENCE of LOGICAL deductions. > > > > You don't have to be condescending *and* obnoxious. > > I think most folks out there can conceive of a statement as being > either true or false. 1=2 is a statement that is false. 2 = 2 is true. > > Again, why you would argue over something this simple is beyond me. > > And I think it pertinent that you deleted out the statement that was > being referred to. > > > > For instance, please offer a sequence of logical steps showing that > > for two complex numbers a and b, ab=0 implies a=0 or b=0. > > > > And now we have that again... > > > Even a oneliner would be better than saying "it's obvious" over and > > over. Here, I'll start you out: "We can conclude that either a=0 or > > b=0 because... " > > > > If you want to argue about whether or not ab = 0 means that a = 0, or b > = 0, then go ahead, but you'll be doing it without me. > > > Sent via Deja.com > http://www.deja.com/

