Friday, October 01, 2004
Latest fuss, my apologies
Unfortunately I was just wrong recently as I fiddled with
x^2 + (as_1 + bs_2)x + ab s_1 s_2 = (x + as_1)(x + bs_2)
and tried to generalize beyond the result with
x^2 + (2s_1 + 3s_2)x + 6 s_1 s_2 = (x + 2s_1)(x + 3s_2)
where s_1 s_2 = 1, and 2s_1 + 3s_2 is an integer,
where it's easy to show that s_1 and s_2 cannot be irrational algebraic integers, to show that the unit case is blocked in the ring of algebraic integers, for irrationals.
I hoped that if I could show that in general 'a' and b can't be algebraic integers without the same thing happening, then I could show that my other work is quite correct and there is a problem with the ring.
However, the result does not travel beyond 'a' and b being integers, and I was just wrong.
I did make a leap.
My apologies for jumping at that one and pushing people to toss easy examples on math professors like with that x^2 + x + 6, thingy.
I continually get frustrated in what I call conventional attempts to prove my advanced polynomial factorization techniques are correct.
The proper and final coverage of this result is shown with
x^2 + (2s_1 + 3s_2)x + 6 s_1 s_2 = (x + 2s_1)(x + 3s_2)
where s_1 s_2 = 1, and 2s_1 + 3s_2 is an integer,
as s_1 and s_2 cannot be irrational algebraic integers, showing that the unit case is blocked for irrationals, though s_1 = s_2 = 1 or s_1 = s_2 = -1, work just fine.
The traditional teaching is that the irrationals are just different, and that in fact the unit case IS blocked, while I can show, using advanced techniques, that the unit case is not blocked, as it's actually just not available in the ring of algebraic integers.
I apologize again for the screw-ups, and only hope that maybe, some people will bother to acknowledge the advanced techniques, as I can't find a conventional approach that does more than just hint at the problem, like with s_1 and s_2 not
being able to be algebraic integers if irrational.
x^2 + (as_1 + bs_2)x + ab s_1 s_2 = (x + as_1)(x + bs_2)
and tried to generalize beyond the result with
x^2 + (2s_1 + 3s_2)x + 6 s_1 s_2 = (x + 2s_1)(x + 3s_2)
where s_1 s_2 = 1, and 2s_1 + 3s_2 is an integer,
where it's easy to show that s_1 and s_2 cannot be irrational algebraic integers, to show that the unit case is blocked in the ring of algebraic integers, for irrationals.
I hoped that if I could show that in general 'a' and b can't be algebraic integers without the same thing happening, then I could show that my other work is quite correct and there is a problem with the ring.
However, the result does not travel beyond 'a' and b being integers, and I was just wrong.
I did make a leap.
My apologies for jumping at that one and pushing people to toss easy examples on math professors like with that x^2 + x + 6, thingy.
I continually get frustrated in what I call conventional attempts to prove my advanced polynomial factorization techniques are correct.
The proper and final coverage of this result is shown with
x^2 + (2s_1 + 3s_2)x + 6 s_1 s_2 = (x + 2s_1)(x + 3s_2)
where s_1 s_2 = 1, and 2s_1 + 3s_2 is an integer,
as s_1 and s_2 cannot be irrational algebraic integers, showing that the unit case is blocked for irrationals, though s_1 = s_2 = 1 or s_1 = s_2 = -1, work just fine.
The traditional teaching is that the irrationals are just different, and that in fact the unit case IS blocked, while I can show, using advanced techniques, that the unit case is not blocked, as it's actually just not available in the ring of algebraic integers.
I apologize again for the screw-ups, and only hope that maybe, some people will bother to acknowledge the advanced techniques, as I can't find a conventional approach that does more than just hint at the problem, like with s_1 and s_2 not
being able to be algebraic integers if irrational.
# posted by James Harris @ 18:35 
