Re: JSH: Keep it simple


jstevh@xxxxxxx wrote:
> Rupert wrote:
> > On your website you state that if f is a non-zero non-unit rational
> > integer, co-prime to 3 and also co-prime to the non-zero rational
> > integer x, then the roots of the polynomial in a
> >
> > a^3 + 3(-1+x f^2)a^2 - f^2(x^3 f^4 - 3x^2 f^2 + 3x) = 0
> >
> > Let f=2, x=3. Then we get a^3+33a^2-1332=0. This is irreducible over Q.
> > So the Galois group of this polynomial over Q must act transitively on
> > the roots. So it's impossible that two of them could be divisible by 2
> > and the other coprime to 2.
>
> The base mathematical ideas you are relying on are wrong, but cannot
> easily be seen to be wrong because the solutions aren't rational, which
> is why I focus on rational solutions.
>

The argument I am making only applies in the case where the polynomial
is irreducible over Q, in which case no rational solutions exist.

There's nothing wrong with the argument. If there's an error, point it
out.

> My results--as they are mathematically correct--apply to rational and
> non-rational solutions allowing me to emphasize their perfect
> prediction with numbers that people can see.
>
> You had another post asking about the ring, and I say use x and f from
> the ring of algebraic integers, both non-zero, where f is not a unit
> and is coprime to x, but find cases where you have at least one
> rational root of the resulting cubic.
>

Is that another part of the hypothesis - that the cubic has to have a
rational root?

> I strongly suggest you do the search and see for yourself, hoping that
> mathematical evidence will have some meaning here for you.
>

In the case where the cubic is irreducible over Q, your claim is wrong.
If you want to add the extra hypothesis that the cubic has a rational
root, go ahead and then we can investigate the new claim.

> The numbers are where mathematical ideas show their real power because
> the numbers do not lie.
>
> One good example found by a poster claiming falsely that it overturned
> my result is f=2 and x=1, so people can see an example of what I mean.
>
> The Galois group of the polynomial has no mathematical meaning in terms
> of where factors go--that's the failure of the old theory.
>

Yes, it does.

> Sorry, but Galois Theory really just tells you how you can present a
> solution, and nothing about the solution itself.
>

No, it tells me that your claim is wrong.

> A simple rule of thumb to understand what Galois Theory does or does
> not tell you is to ask, does it say anything about rationals?
>
> Like, does the Galois group of
>
> x^2 + 3x + 2
>
> say anything about the solutions to that quadratic?
>

The Galois group is trivial. We can infer from that that the solutions
are rational.

>
> James Harris

.

Relevant Pages

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  • Re: Brutal result, actually unanswerable
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  • Re: Brutal result, actually unanswerable
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