JSH; Simple Construct
JSH
7x^2 + 5x - 8 = 0
and
7y^2 + 5y - 1 = 0
so subtracting the second from the first I have
7(x^2 - y^2) + 5(x-y) - 7 = 0
so
7(x+y) + 5 = 7/(x-y).
Therefore 1/y has 7 as a factor, right?
James Harris
biggus
"JSH" <DO...@spamless.com> wrote in message
news:46ba4393$0$97266$892e...@authen.yellow.readfreenews.net...
this crap again? I thought you deleted it.
your IQ is insufficient to handle anything above very simple algebra.
you are demonstrating the Bell Curve theory with your posts.
yup, on the bottom 10%.
fishfry
"JSH" <DO...@spamless.com> wrote:
JSH, this one is beneath even you.
local host
"fishfry" <BLOCKSPA...@your-mailbox.com> wrote in message
news:BLOCKSPAMfishfry-4B...@comcast.dca.giganews.com...
if 1/y has a factor of 7 then y = 1/7 * K
therefore 7 *(1/7*K)^2+5*1/7*K - 8 =0
then 1/7 K^2+5/7*K - 8 = 0
or K^2 +5*K - 56 = 0
-5+-SQRT(4*1*(-56))/2 are the answers
x = -5 +- i*7.483314......
Thar ya be, all solv-ed now, you dont has to post her anymore creep.
Rick Decker
Stop wasting our time, moron.
R.
The Ghost In The Machine
<DO...@spamless.com>
wrote
on Wed, 8 Aug 2007 17:29:24 -0500
<46ba4393$0$97266$892e...@authen.yellow.readfreenews.net>:
OK, Harris, one more time....
There are way too many problems with this logic for it
to work at all. For starters, one can parameterize the
problem with ease, by taking the pair of equations
7x^2 + 5x - N = 0
and
7y^2 + 5y - (N - 7) = 0
where N is any complex number at all.
After the subtraction/division one has
7(x+y) + 5 = 7/(x-y)
as before (unless x = y for more or less obvious reasons).
Since neither x nor y is an algebraic integer in the
general case (even if N is an integer), one can conclude
little if anything about whether 1/y has a factor of 7,
and in fact a simple counterexample can now be devised;
setting N = 9, one gets
7y^2 + 5y - 2 = 0
which has roots y1 = 2/7 and y2 = -1. Neither 1/y1 = 7/2
nor 1/y2 = -1 is divisible by 7.
As a check, the equation
7x^2 + 5x - 9 = 0
has roots -5/14 + sqrt(277)/14 and -5/14 - sqrt(277)/14, and
therefore x != y -- unless somehow 277 mutates into a perfect
square in your world. (The nearest squares in the normal integer ring
are 256 = 16^2 and 289 = 17^2.)
Another, even more obvious counterexample sets N = 7;
one of the y roots is therefore 0 (the other is -7/5)
and therefore the term 1/y = 1/0 has no meaning.
This logic therefore fails completely.
A direct approach to the original equation
is probably preferable. Reprising:
7y^2 + 5y - 1 = 0
has roots
y1 = -5/14 + sqrt(25+28)/14 = (-5+sqrt(53))/14
y2 = (-5-sqrt(53))/14
by the application of the usual quadratic formula and/or
completing the square.
Since 7 * (y - y1) * (y - y2) = 7*y^2 - 7 * (y1+y2) * y + 7*y1*y2,
y1*y2 = 1/7 or 7 * y1 = 1/y2. However, neither y1 nor y2 are
algebraic integers (the equation cannot be reduced to a form where
the y^2 coordinate is 1) so the mere fact that there's a 7 involved
in 7 * y1 = 1/y2 means little.
And of course -7*(y1+y2) = 5, as one can see by inspection.
(That's worth even less, but it serves as a bit of a check.)
One can also invert the equation, by substituting y = 1/z,
leading to
-1 + 5/z + 7/z^2 = 0
and, multiplying through by z^2 (yes, we know that's nonzero),
-z^2 + 5z + 7 = 0
or
z^2 - 5z - 7 = 0
This actually looks extremely promising, and proves that
both 1/y1 and 1/y2 are algebraic integers -- a mildly
interesting result.
The two roots of this equation are
z1 = 5/2 + sqrt(53)/2
z2 = 5/2 - sqrt(53)/2
and neither of these are divisible by 7; were z1 or z2
divisible by 7 then w = z/7 (= 7/y) would be an integer,
but w's defining equation (which can be found by simple
substitution, then multiplying by 49) is
49*w^2 - 35*w - 7 = 0
or, dividing by 7,
7*w^2 - 5*w - 1 = 0
This polynomial is irreducible over Q (as is the original).
Therefore, neither z is divisible by 7, and neither 1/y1 nor 1/y2
is divisible by 7.
--
#191, ewi...@earthlink.net
Linux. Because Windows' Blue Screen Of Death is just
way too frightening to novice users.
--
Posted via a free Usenet account from http://www.teranews.com
Gib Bogle
> OK, Harris,
Is not Harris.
Richard Tobin
Are you incapable of looking at the headers?
-- Richard
--
"Consideration shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.
Dear Leader
"Richard Tobin" <ric...@cogsci.ed.ac.uk> wrote in message
news:f9est0$10ao$1...@pc-news.cogsci.ed.ac.uk...
> In article
> <BLOCKSPAMfishfry-4B...@comcast.dca.giganews.com>,
> fishfry <BLOCKSPA...@your-mailbox.com> wrote:
>>JSH, this one is beneath even you.
>
> Are you incapable of looking at the headers?
>
it is a repost of JSH from 4 days ago that JSH deleted from Google.
Richard Tobin
Dear Leader <span...@spamless.com> wrote:
>it is a repost of JSH from 4 days ago that JSH deleted from Google.
That's no excuse. It's bad enough having endless replies to James's
own posts, let alone copies of them.
Dear Leader
> In article <46bb0b9a$0$97229$892e...@authen.yellow.readfreenews.net>,
> Dear Leader <span...@spamless.com> wrote:
>>it is a repost of JSH from 4 days ago that JSH deleted from Google.
>
> That's no excuse. It's bad enough having endless replies to James's
> own posts, let alone copies of them.
>
> -- Richard
I think that is what he is after, just the # of replies to each post.
It is not the math for sure.
The Ghost In The Machine
<bo...@ihug.too.much.spam.co.nz>
wrote
on Thu, 09 Aug 2007 19:12:48 +1200
<f9eej8$3bf$1...@lust.ihug.co.nz>:
> The Ghost In The Machine wrote:
>
>> OK, Harris,
>
> Is not Harris.
Maybe not, but wotthehell...the math's just as wrong. :-)
--
#191, ewi...@earthlink.net
Insert random misquote here.