"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message news:<HpypJB.KA9@cwi.nl>...
....
Yup, it appears so. In mathematics the sqrt when applied to reals and
delivering a real is defined to give the positive result. This has
been said before, but it is apparently to difficult for you to understand.
....
And besides, you lied, it's by *convention* that the positive is
taken, not by definition.

That's necessary because, like sqrt(4) has *either* 2 or -2 as a
solution, as is easily proven:

sqrt(4) = -2, square both sides, 4 = 4. QED

Now then, can you *prove* that -2 is not a solution to sqrt(4)?

The whole discussion is silly in any case. There certainly
exits a single valued function, g(z) defined on the complex plane
such that for all complex z:

g(z)*g(z) = z,
either real(g(z)) > 0 or (real(g(z)) = 0 and imag(g(z))>=0)

If you have a deep psycho-sexual need not to call g(z) sqrt(z),
then call it sqrt_pb(z) or harris(z) or hughes(z) or whatever
else you want [1]. Define

g_1(x) = 4 - g(1+3x), g_2(x) = 4 + g(1+3x)

(note that both g_1 and g_2 are single valued)

Then g_1(x) g_2(x) = 3(5-x)

g_1(0) = 3, g_2(0) = 5

g_1(1) = 2 is not divisible by 3

wpihughes@hotmail.com (William Hughes) writes:

x=sqrt(4) is not an equation.

Surely, you didn't mean to say that.

"Me" <no@spam.com> wrote in message news:<brkrud$j66$1@grapevine.wam.umd.edu>...
Okay, so you're dividing both sides by 49. Point taken.

Now, would you please answer the question I posed when I wrote:

:> In other words, are you
:> claiming that (5 a_1(x) + 7) and (5 a_2(x) + 7) are divisible by 7 for
:> all x? If so, could you please explain why the fact that it's true for
:> x=0 implies it's true for other x also?

If you are not claiming this, then please explain what you are claiming.

Many thanks,
Justin

Given, where x is in the ring of algebraic integers, I've shown the
factorization

(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =

49(300125 x^3 - 18375 x^2 - 360 x + 22)

where b_3(x) = a_3(x) - 3 and the a's are roots of

a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)

so when x=0, a_1(0) = a_2(0) = b_3(0) = 0.

Now you can divide both sides by 49.

Some posters have claimed that when you divide by 49, what results
varies depending on what value x has.

Do you understand?

James Harris wrote:

Since people working with square roots *usually* want the positive
root, by convention the positive is taken, not by definition.

That's necessary because, like sqrt(4) has *either* 2 or -2 as a
solution, as is easily proven:

sqrt(4) = -2, square both sides, 4 = 4. QED

Nice logic! -1 is a solution of 1 (whatever that means), since
squaring both sides gives 1=1. QED

Wilbert

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0312141928.49c9b250@posting.google.com>...
Given, where x is in the ring of algebraic integers, I've shown the
factorization

(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =

49(300125 x^3 - 18375 x^2 - 360 x + 22)

where b_3(x) = a_3(x) - 3 and the a's are roots of

a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)

so when x=0, a_1(0) = a_2(0) = b_3(0) = 0.

I'm curious about the mental processes that allow *some* of you to
claim that 49 divides off as a *variable* dependent on x, so I'm
giving another opportunity for you to speak your minds.

To my knowledge, in the history of mathematics, no one has ever
presented such a proposition, so it is a unique one, and I must say
that I'm intrigued.

Speak your minds.


James Harris

Why does this not count as presenting "such a proposition"?

Let g1(x)=4-sqrt(1+3x) and g2(x)=4+sqrt(1+3x).

then
g1(x)*g2(x) = 3(5-x)

but neither g1 nor g2 is divisible by 3 for all x.

-"William Hughes"

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0312141928.49c9b250@posting.google.com>...
Given, where x is in the ring of algebraic integers, I've shown the
factorization

(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =

49(300125 x^3 - 18375 x^2 - 360 x + 22)

where b_3(x) = a_3(x) - 3 and the a's are roots of

a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)

so when x=0, a_1(0) = a_2(0) = b_3(0) = 0.

I'm curious about the mental processes that allow *some* of you to
claim that 49 divides off as a *variable* dependent on x, so I'm
giving another opportunity for you to speak your minds.

To my knowledge, in the history of mathematics, no one has ever
presented such a proposition, so it is a unique one, and I must say
that I'm intrigued.

Speak your minds.


James Harris

Why does this not count as presenting "such a proposition"?

Let g1(x)=4-sqrt(1+3x) and g2(x)=4+sqrt(1+3x).

then
g1(x)*g2(x) = 3(5-x)

but neither g1 nor g2 is divisible by 3 for all x.

-"William Hughes"

Wilbert Dijkhof <w.j.dijkhof@tue.nl> wrote in message news:<3FDF198B.1767B62A@tue.nl>...
James Harris wrote:

Since people working with square roots *usually* want the positive
root, by convention the positive is taken, not by definition.

That's necessary because, like sqrt(4) has *either* 2 or -2 as a
solution, as is easily proven:

sqrt(4) = -2, square both sides, 4 = 4. QED

Nice logic! -1 is a solution of 1 (whatever that means), since
squaring both sides gives 1=1. QED

Wilbert

-1 IS a solution for sqrt(1).

It's interesting that is a point of debate, but not surprising for the
sci.math newsgroup!

James Harris

Why does this not count as presenting "such a proposition"?

Let g1(x)=4-sqrt(1+3x) and g2(x)=4+sqrt(1+3x).

then
g1(x)*g2(x) = 3(5-x)

I noted there's a *sign* ambiguity in the sqrt() operator, which
sparked a lot of debate.

One thing I found interesting is that posters ignored that if you
divide both sides by 3, with the convention that you're taking the
positive of the sqrt() operator for *integer* results, you have

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

as a factor, which is an integer (remember x is an integer and
remember 1+3x is a square).

but neither g1 nor g2 is divisible by 3 for all x.

-"William Hughes"

Prove it. Readers should note that this poster presented a *later*
post claiming a result that covers integer results of the square root
operator but made a rather simple mistake. In my reply to that post I
noted the sign ambiguity in the square root operator.

However sqrt(x), where x is a square can be taken to be the positive
result since you can *give* the result, but the ambiguity remains if
you see sqrt(x) without a given value.

That is, for instance, sqrt(4) is 2 *or* -2, but by convention, it's
*usually* taken as 2, though if you do enough analysis you will run
into situations where you need the negative!!!

Mathematics is *logical* and consistent, which is something that many
people seem to have a problem with, as they try to twist it to their
own needs and interests.

Note lack of proper explication from other sci.math posters.


James Harris

-1 IS a solution for sqrt(1).

It's interesting that is a point of debate, but not surprising for the
sci.math newsgroup!

OOPS! I just replied to this post with my own mistake, as I had

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

being an integer if 1+3x is a square, when x=40 refutes that notion.

James Harris

James Harris wrote:

OOPS! I just replied to this post with my own mistake, as I had

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

being an integer if 1+3x is a square, when x=40 refutes that notion.

James Harris

Your "OOPS!" count just went up by one. Funny how *your* mistakes are
simple, easily corrected faults, but other
posters mistakes are signs of ignorance, incompetence or dishonesty.

Get a grip, James. You have been thoroughly and conclusively refuted. Your
*research* is a worthless pile of
pseudo-scientific, incomprehensible and error-ridden junk

--
There are two things you must never attempt to prove: the unprovable --
and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com


Heh, did you see that message a few days ago from someone who asked a

wpihughes@hotmail.com (William Hughes) wrote in message news:<4d5e4663.0312150510.86bcfd1@posting.google.com>...

snip

Why does this not count as presenting "such a proposition"?

Let g1(x)=4-sqrt(1+3x) and g2(x)=4+sqrt(1+3x).

then
g1(x)*g2(x) = 3(5-x)


I noted there's a *sign* ambiguity in the sqrt() operator, which
sparked a lot of debate.

One thing I found interesting is that posters ignored that if you
divide both sides by 3, with the convention that you're taking the
positive of the sqrt() operator for *integer* results, you have

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

as a factor, which is an integer (remember x is an integer and
remember 1+3x is a square).


but neither g1 nor g2 is divisible by 3 for all x.

-"William Hughes"


Prove it.

Wilbert Dijkhof <w.j.dijkhof@tue.nl> wrote in message
news:<3FDF198B.1767B62A@tue.nl>...
James Harris wrote:

Since people working with square roots *usually* want the positive
root, by convention the positive is taken, not by definition.

That's necessary because, like sqrt(4) has *either* 2 or -2 as a
solution, as is easily proven:

sqrt(4) = -2, square both sides, 4 = 4. QED

Nice logic! -1 is a solution of 1 (whatever that means), since
squaring both sides gives 1=1. QED

Wilbert

-1 IS a solution for sqrt(1).

It's interesting that is a point of debate, but not surprising for the
sci.math newsgroup!

I noted there's a *sign* ambiguity in the sqrt() operator, which
sparked a lot of debate.