Re: JSH: A simple error

jstevh@xxxxxxxxx wrote:
On May 27, 10:47 am, s...@xxxxxxxxxxxxx wrote:
James Harris wrote:
Posters in attacking my proofs showing inconsistency with the ring of
algebraic integers routinely move outside the ring. Here is a post
meant to show you how they do it.
In the ring of integers, consider x^2 + 3x + 2 = 0, which of course
factors as
x^2 + 3x + 2 = (x+2)(x+1)
and now solve for it using the quadratic formula, but kind of weird by
NOT resolving the square root then
x = (-3 +/- sqrt(1))/2
and now make the substitution, x=2y, so you get
4y^2 + 6y + 2 = 0, so you can divide by 2 to get
2y^2 + 3y + 1 = 0,
and your solution now becomes
y = (-3+/- sqrt(1))/4
which is two solutions where one is not an integer.
So you moved outside the ring of integers.
But isn't this completely analogous to what you do in your paper? You
define 7*g(x) = 5*a_2(x), and then show that a_2(x) and therefore
5*a_2(x) can be chosen to be an algebraic integer. You then claim that
the fact that g(x) is not an algebraic integer means you are "pushed
outside the ring". How is this different to what you have done above,
where you show that x is an integer and then define x = 2y?


It is different because in the paper 5a_2(x) = 7g(x) meaning ONE
SOLUTION but if you cannot resolve the square roots you cannot see the
two solutions.

Think carefully about how with

x^2 - 3x + 2 = 0

you can use the quadratic formula with

x = (3 +/- sqrt(1))/2

as usually you zip past the step of taking the square root because
it's trivial, but notice that if you think about it ONLY ONE ROOT HAS
2 AS A FACTOR!

If you have instead

x^2 - 5x + 2 = 0

what if only one root has 2 as a factor then?

Standard answer is to say, in what ring?

And mathematicians go to the ring of algebraic integers by training,
when that ring is provably inconsistent, as in that ring, it is
impossible for a monic quadratic with integer coefficients to have
just one root with 5 as a factor if the roots are non-rational.

What I prove is that is an arbitrary condition like if you say, take
only evens, 2 has no factors in common with 6 because 3 is not even,
which is just another arbitrary condition.

Trouble is, if you do this wrong, you can "prove" all kinds of things
which makes it very appealing, so Andrew Wiles might not want to admit
this problem exists as it allows him a faux proof of Fermat's Last
Theorem.

You can "prove" just about anything with a deep mathematical error.

It makes it very appealing for people who have nothing if the truth is
known, as all their "proofs" go out the window!

The only hope in this situation is that there are real mathematicians
who care more about being correct than being believed to be correct.

Today Andrew Wiles is a hero to a lot of math people, if he knew of
this error, what motivation would he have to tell the truth?

I say, only his love of mathematics, and his sense of conscience and
need to be honest for the good of humanity.

But who in this world today is a real humanitarian?
[...]

Do you know any sentences that begin with
"If James Harris" ?

David Bernier

.

Relevant Pages

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  • Re: A simple error
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  • Re: Now a change
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