wpihughes@hotmail.com (William Hughes) wrote in message news:<4d5e4663.0312151116.734c0ab7@posting.google.com>...
"Me" <no@spam.com> wrote in message news:<brkemu$eme$1@grapevine.wam.umd.edu>...
William Hughes <wpihughes@hotmail.com> wrote:

: 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.

This is different. His claim seems to be that *if* two of the terms in
the product *are* divisible by 7 for a *specific* value of x, then they
are divisible by 7 for *all* values of x. In your example neither is ever
divisible by 3.

Best regards,
Justin

Consider any integer a such that 1+3a is a perfect square. Then
g1(a) and g2(a) are both integers. Since their product is divisible
by 3 one of them must be divisible by 3.

Note that posters have gone off on a tangent with a made up example,
and they're even getting *it* wrong! Fascinating behavior, but not
surprising for the sci.math newsgroup.


Examples

g1(0)=3, g2(0)=5

That is incorrect as the sqrt() operator is ambiguous

Quit saying people lie when they make honest mistakes, there is a BIG
difference. The sqrt of 4 is 2 OR -2;

Consider that the constant term of g1(x) is 3 *or* 5 because the
sqrt() operator is ambiguous. I've explained that before in replying
to Arturo Magidin, but mathematics is a difficult discipline for some,
so repetition is necessary, and still often not enough.

In article <3c65f87.0312151213.7bbfb1b7@posting.google.com> jstevh@msn.com (James Harris) writes:
"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message news:<HpxxrM.Iy4@cwi.nl>...
In article <4d5e4663.0312150510.86bcfd1@posting.google.com> wpihughes@hotmail.com (William Hughes) writes:
...
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.

Ah, but the constant term of g1(x) = 3, and the constant term of g2(x) = 5,
so g1(x) should be divisible by 3. A core error in mathematics .

It's actually not that simple. It's an interesting case to highlight
your ineptitude with basic mathematics though.

Consider that the constant term of g1(x) is 3 *or* 5 because the
sqrt() operator is ambiguous. I've explained that before in replying
to Arturo Magidin, but mathematics is a difficult discipline for some,
so repetition is necessary, and still often not enough.

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.

You see Dik Winter, actually *knowing* mathematics versus talking as
if you know it can be two different things.

It appears so, yes.

"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message news:<HpypJB.KA9@cwi.nl>...
In article <3c65f87.0312151213.7bbfb1b7@posting.google.com> jstevh@msn.com (James Harris) writes:
"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message news:<HpxxrM.Iy4@cwi.nl>...
In article <4d5e4663.0312150510.86bcfd1@posting.google.com> wpihughes@hotmail.com (William Hughes) writes:
...
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.

Ah, but the constant term of g1(x) = 3, and the constant term of g2(x) = 5,
so g1(x) should be divisible by 3. A core error in mathematics .

It's actually not that simple. It's an interesting case to highlight
your ineptitude with basic mathematics though.

Consider that the constant term of g1(x) is 3 *or* 5 because the
sqrt() operator is ambiguous. I've explained that before in replying
to Arturo Magidin, but mathematics is a difficult discipline for some,
so repetition is necessary, and still often not enough.

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.

That doesn't change the *inherent* ambiguity in the sqrt() operator.

That's easy to show as consider sqrt(4), and you wish to say it's
*defined* to be 2, but what about -2?

Does your definition take away -2 as a solution?

And besides, you lied, it's by *convention* that the positive is
taken, not by definition.

You see Dik Winter, actually *knowing* mathematics versus talking as
if you know it can be two different things.

It appears so, yes.

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

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

James Harris

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

"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message news:<HpxxrM.Iy4@cwi.nl>...
In article <4d5e4663.0312150510.86bcfd1@posting.google.com> wpihughes@hotmail.com (William Hughes) writes:
...
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.

Ah, but the constant term of g1(x) = 3, and the constant term of g2(x) = 5,
so g1(x) should be divisible by 3. A core error in mathematics .

It's actually not that simple. It's an interesting case to highlight
your ineptitude with basic mathematics though.

Consider that the constant term of g1(x) is 3 *or* 5 because the
sqrt() operator is ambiguous. I've explained that before in replying
to Arturo Magidin, but mathematics is a difficult discipline for some,
so repetition is necessary, and still often not enough.

You say "the sqrt() operator is ambiguous." Well, that makes YOUR OWN

You see Dik Winter, actually *knowing* mathematics versus talking as
if you know it can be two different things.

And you clearly by posts like yours here do not actually know
mathematics.

You do post a lot though living in a fantasy world.


James Harris

"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message
news:<HpypJB.KA9@cwi.nl>...
In article <3c65f87.0312151213.7bbfb1b7@posting.google.com
jstevh@msn.com (James Harris) writes:
"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message
news:<HpxxrM.Iy4@cwi.nl>...
In article <4d5e4663.0312150510.86bcfd1@posting.google.com
wpihughes@hotmail.com (William Hughes) writes:
...
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.

Ah, but the constant term of g1(x) = 3, and the constant term of
g2(x) = 5,
so g1(x) should be divisible by 3. A core error in mathematics
.

It's actually not that simple. It's an interesting case to highlight
your ineptitude with basic mathematics though.

Consider that the constant term of g1(x) is 3 *or* 5 because the
sqrt() operator is ambiguous. I've explained that before in replying
to Arturo Magidin, but mathematics is a difficult discipline for
some,
so repetition is necessary, and still often not enough.

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.

That doesn't change the *inherent* ambiguity in the sqrt() operator.

That's easy to show as consider sqrt(4), and you wish to say it's
*defined* to be 2, but what about -2?

Does your definition take away -2 as a solution?

And besides, you lied, it's by *convention* that the positive is
taken, not by definition.

You see Dik Winter, actually *knowing* mathematics versus talking as
if you know it can be two different things.

It appears so, yes.

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

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


James Harris

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:

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

"James Harris" <jstevh@msn.com> wrote in message
news:3c65f87.0312160631.117923f5@posting.google.com...
"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message
news:<HpypJB.KA9@cwi.nl>...
In article <3c65f87.0312151213.7bbfb1b7@posting.google.com
jstevh@msn.com (James Harris) writes:
"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message
news:<HpxxrM.Iy4@cwi.nl>...
In article <4d5e4663.0312150510.86bcfd1@posting.google.com
wpihughes@hotmail.com (William Hughes) writes:
...
[...]

That's easy to show as consider sqrt(4), and you wish to say it's
*defined* to be 2, but what about -2?

Does your definition take away -2 as a solution?

And besides, you lied, it's by *convention* that the positive is
taken, not by definition.

Quit saying people lie when they make honest mistakes, there is a BIG
difference.

The sqrt of 4 is 2 OR -2; if you dispute this, go back to
algebra I in high school. You apparently have gaps in your algebra.

You see Dik Winter, actually *knowing* mathematics versus talking as
if you know it can be two different things.

It appears so, yes.

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

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


James Harris

In article <brn968$50ftg$1@ID-206850.news.uni-berlin.de>,
David Moran <ktulwxwatcher@hotmail.com> wrote:

[.snip.]

Quit saying people lie when they make honest mistakes, there is a BIG
difference. The sqrt of 4 is 2 OR -2;

Ehr, no. Both 2 and -2 are square roots of 4 (that is, both of them,
when squared, give 4); but "the sqrt of 4" is defined to be the
principal branch, and therefore it is unabiguously equal to 2. That's
why sqrt(a^2) = |a| for real numbers, for example.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@math.berkeley.edu

Arturo, thanks for the correction. I stand corrected.

"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message news:<HpypJB.KA9@cwi.nl>...
In article <3c65f87.0312151213.7bbfb1b7@posting.google.com> jstevh@msn.com (James Harris) writes:
"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message news:<HpxxrM.Iy4@cwi.nl>...
In article <4d5e4663.0312150510.86bcfd1@posting.google.com> wpihughes@hotmail.com (William Hughes) writes:
...
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.

Ah, but the constant term of g1(x) = 3, and the constant term of g2(x) = 5,
so g1(x) should be divisible by 3. A core error in mathematics .

It's actually not that simple. It's an interesting case to highlight
your ineptitude with basic mathematics though.

Consider that the constant term of g1(x) is 3 *or* 5 because the
sqrt() operator is ambiguous. I've explained that before in replying
to Arturo Magidin, but mathematics is a difficult discipline for some,
so repetition is necessary, and still often not enough.

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.

That doesn't change the *inherent* ambiguity in the sqrt() operator.

That's easy to show as consider sqrt(4), and you wish to say it's
*defined* to be 2, but what about -2?

Does your definition take away -2 as a solution?

And besides, you lied, it's by *convention* that the positive is
taken, not by definition.

You see Dik Winter, actually *knowing* mathematics versus talking as
if you know it can be two different things.

It appears so, yes.

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

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0312151213.7bbfb1b7@posting.google.com>...

Consider that the constant term of g1(x) is 3 *or* 5 because the
sqrt() operator is ambiguous.

You say "the sqrt() operator is ambiguous." Well, that makes YOUR OWN
claims ambiguous. You probably dont realize it, but the a_1, a_2 and
a_3 in your "core error" CONTAIN SQUARE ROOTS! Just use the
Cardano(sp?) formulas to find the a_n's and you will see the square
roots.

x=sqrt(4) is not an equation.

brianscsmith@yahoo.com (Brian Smith) wrote in message news:<12f59340.0312160653.7a6c126@posting.google.com>...
jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0312151213.7bbfb1b7@posting.google.com>...

Consider that the constant term of g1(x) is 3 *or* 5 because the
sqrt() operator is ambiguous.

You say "the sqrt() operator is ambiguous." Well, that makes YOUR OWN
claims ambiguous. You probably dont realize it, but the a_1, a_2 and
a_3 in your "core error" CONTAIN SQUARE ROOTS! Just use the
Cardano(sp?) formulas to find the a_n's and you will see the square
roots.

No, a_1(x), a_2(x) and a_3(x) do not "CONTAIN SQUARE ROOTS!". They are
functions defined in terms of the roots of a polynomial with
coefficients depending on x (actually, James leaves out a couple of
steps needed to completely specify the functions).
While one can certainly use the Cardano formulae to
find expressions for the a_i, and these formulae contain square roots,
there are ways to express the a_i that do not involve the
symbol sqrt. One should be careful to distinguish between the
existence of the underlying function and the notation used to
express the function.

- William Hughes