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Topic: Finishing up, explaining FLT Proof conclusion
Replies: 45   Last Post: Aug 1, 2001 2:53 AM

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 JAMES HARRIS Posts: 9,787 Registered: 12/4/04
Re: Finishing up, explaining FLT Proof conclusion
Posted: Jul 28, 2001 10:49 AM

Paul Sperry <plsperry@sc.rr.com> wrote in message news:<plsperry-C8BAC4.22472227072001@news1.southeast.rr.com>...
> jstevh@msn.com (James Harris) wrote:
>

> > Paul Sperry <plsperry@sc.rr.com> wrote in message
> > news:<plsperry-3C9F3E.00043026072001@news2.southeast.rr.com>...

> > > In article <3c65f87.0107251356.16778672@posting.google.com>,
> > > jstevh@msn.com (James Harris) wrote:
> > > [...]

> > > > So, let me re-emphasize my point, by asking you a simple question:
> > > >
> > > > Can you start with integers and end up in the field of rationals with
> > > > fractions...your regular old garden variety fraction like 1/2...using
> > > > only addition and multiplication?
> > > >

> > >
> > > Yes, I can. The construction is standard, routine and well known:
> > > On the subset S of the set Z x Z of ordered pairs of integers consisting
> > > of all pairs (a,b) with b != 0, define (a,b)==(a',b') iff
> > > a * b' = a' * b where "*" is integer multiplicaion.

> >
> > "standard", "well known" and CIRCULAR.
> >
> > Folks, he's saying that in integers, just say that 2b = 3, claim that
> > b "exists" (which he can't prove, so he must base it on some
> > principle, which basically just says it exists), and presto
> > magico!--you have a fraction.
> >

> [...]
> Am I the only one that finds James' response odd? Nowhere in my post did
> I allude to solutions to equations of the form a*x=b yet James comes up
> with 2b=3. He then accuses me of using the non existent "b" as a
> fraction.
>
> Of course that is exactly what one would like to do: adjoin to the
> integers all solutions to equations of the form a*x=b. If one tries
> that, there are immediate problems but why not use the polynomials
> themselves? Let Q be the set of all linear polynomials with integer
> coefficients so that 2x-b becomes the solution to itself - sort of.
> Since 2x-3 and 4x-6 have the same roots an equivalence relation must be
> introduced. The move to ordered pairs just simplifies notation.

Finally, at least you get to the heart of the problem.

Mathematicians want to be able to solve everything, and

2x = 1, looks like something that's always solvable.

After all, it's simple enough, but the trouble comes in because it
involves set operations.

So, properly written it's given a set 1, x equals half its members if
the total number of members of the set is even.

Basically, mathematicians are mixing in a lot of different things.

The 1 is a set, with members, whereas the x is a partial count of
members.

Let 1 represent a set with an odd number of members, then

2x = 1 is FALSE.

The problem is that mathematicians DECIDED that 2x = 1 is true in this
case as well.

The issue has to do with a problem with the concept that everything in
mathematics ultimately involves set operations. So people LOOK at 2x
= 1 and figure there should just be some general way where it's
always, always, ALWAYS true, which violates set theory.

But that kind of error doesn't show up in the real world because
you're forced to follow set theory (humanity has no choice), while for
"pure" math, you can just get it all wrong as long as other people
will tell you it's right.

But ultimately you run into an impasse, with something like Fermat's
Last Theorem, because doing the math wrong has a price.

That price extends beyond math that isn't considered practical because
the Universe doesn't have to do all of Her math with counting
numbers--like human beings have chosen to do up until this point.

So, if you can't get past the baby steps, what does that tell Her?

James Harris