A little lesson for sqrt(144) year olds.
Dirk Van de moortel
difficulties with the concept of square roots:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/PythagoRescue.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SqrtRev.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/NegSqrt.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SqrtAnswers.html
http://www.androc1es.pwp.blueyonder.co.uk/Dinkyfumblezero.htm .
So in order to prevent this little inconvenience from
spreading among others, I decided to publish a little
lesson that my sons got at school at the age of 12.
Remark - this is targeted at an audience aged between
12 and 14.
Definition of the square root of the real numbers:
sqrt:
R+ ==> R+
x |--> y = sqrt(x) if y^2 = x
where x and y belong to R+, being the set of real
positive (or zero) numbers.
In other words: sqrt takes a positive (or zero) number
and produces a positive (or zero) number.
We have:
x^2 = 4
<==> x^2-4 = 0
<==> (x-2)*(x+2) = 0
<==> x-2 = 0 or x+2 = 0
<==> x = 2 or x = -2
and we say that the equation
x^2 = 4
has two solutions. The first solution is
x = 2.
The second solution is
x = -2
Warning. Be careful for the following pitfall:
http://users.pandora.be/vdmoortel/dirk/Stuff/AndrolesAtSchool.gif
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Mutual.html
As we have just seen, this can be written as
x = 2 or x = -2.
Sometimes it is also written as
{ x_1 = 2
{ x_2 = -2
By the way, in a few years you'll learn about polynomials
and roots. Roots are things that make polynomials zero.
For example, we say that the polynomial
x^2 - 4
has two roots.
We say that the first root is 2 and the second root is -2.
But that is for later. Just remember now not to call these
things "square roots" or you will get utterly confused
later...
So what more do we have with square roots?
We have:
sqrt(4) = 2 because 2^2 = 4 and 2 and 4 belong to R+.
So we have:
x^2 = 4 <==> x = sqrt(4) or x = -sqrt(4) .
We have:
x^2 = 5 <==> x = sqrt(5) or x = -sqrt(5)
We have:
sqrt(anything) is positive or zero - by definition.
We have:
-sqrt(anything) is negative or zero
because "minus positive" is negative
We also have the "absolute value function"
|x| = x for all x >= 0
|x| = -x for all x <= 0
And of course we have:
sqrt(x^2) = |x| for all real x (positive and negative)
So we have:
sqrt(x^2) = x for all x >= 0
sqrt(x^2) = -x for all x <= 0
And we have:
IF x >= 0 THEN sqrt(x^2) = x
IF x <= 0 THEN sqrt(x^2) = -x
I hope this helps. If only just a little bit ;-)
Dirk Vdm
phil
>
>Remark - this is targeted at an audience aged between
>12 and 14.
>
>Definition of the square root of the real numbers:
> sqrt:
> R+ ==> R+
> x |--> y = sqrt(x) if y^2 = x
>where x and y belong to R+, being the set of real
>positive (or zero) numbers.
>
Are you kidding? You lost the 12 year olds right about here
phil
Dirk Van de moortel
"phil" <pfreed...@hotmail.com> wrote in message news:200406231252...@proapp.mathforum.org...
:-)
My sons are still alive and kicking though.
They belong to the generation that had sets and
relations and functions before the age of 12 :-P
Dirk Vdm
Paulo Jorge O. C. Matos
And I can bet that you lost it exactly here: R+ ==> R+
I'd like to help but unfortunately I'm very very very bad
teaching children. Sorry!
I don't have the slightest idea on how to explain them the sqrt
Cheers,
Paulo Matos
pfreed...@hotmail.com (phil) writes:
--
Paulo J. Matos : pocm [_at_] mega . ist . utl . pt
Instituto Superior Tecnico - Lisbon
Computer and Software Eng. - A.I.
- > http://mega.ist.utl.pt/~pocm
---
-> God had a deadline...
So, he wrote it all in Lisp!
phil
>difficulties with the concept of square roots:
> <a href="http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SqrtRev.html">http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SqrtRev.html</a>
> <a href="http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/NegSqrt.html">http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/NegSqrt.html</a>
> <a href="http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SqrtAnswers.html">http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SqrtAnswers.html</a>
> <a href="http://www.androc1es.pwp.blueyonder.co.uk/Dinkyfumblezero.htm">http://www.androc1es.pwp.blueyonder.co.uk/Dinkyfumblezero.htm</a> .
>
>So in order to prevent this little inconvenience from
>spreading among others, I decided to publish a little
>lesson that my sons got at school at the age of 12.
>
>12 and 14.
>
>Definition of the square root of the real numbers:
> sqrt:
> R+ ==> R+
> x |--> y = sqrt(x) if y^2 = x
>where x and y belong to R+, being the set of real
>positive (or zero) numbers.
>
>and produces a positive (or zero) number.
>
>We have:
> x^2 = 4
> <==> x^2-4 = 0
> <==> (x-2)*(x+2) = 0
> <==> x-2 = 0 or x+2 = 0
> <==> x = 2 or x = -2
>and we say that the equation
> x^2 = 4
>has two solutions. The first solution is
> x = 2.
>The second solution is
> x = -2
>
> Warning. Be careful for the following pitfall:
> <a href="http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Mutual.html">http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Mutual.html</a>
I suggest that the vast majority of 12 year olds have no idea
what a real number is, or rather, they do not know of any other kind.
So the way to teach square roots is by simple examples. Start with
3^2 = 9 and explain that this is equivalent to sqrt(9) = 3 . From
there move on to other rational numbers, like sqrt(10) is a little
larger than sqrt(9). Then maybe draw a graph y = sqrt(x) and point
out that the graph exists only in the first quadrant, because (for
12 year olds) negative numbers DON'T HAVE square roots.
Your boys are ready for graduate school at 12.
phil
phil
Mike
[snip]
>
> And we have:
> IF x >= 0 THEN sqrt(x^2) = x
> IF x <= 0 THEN sqrt(x^2) = -x
>
> I hope this helps. If only just a little bit ;-)
>
> Dirk Vdm
Having real troubles Dirk with basic algebra? No wonder. You are a
major crank.
The square root function of x, sqrt(x) is the inverse function of f(x)
= x^2, for all x greater or equal to 0.
Cranky Dirk, if x <= 0 then X^2 >= 0, let it be equal to X, and the
"principle" square root of X is a positive number x such that x^2 =
X, while there is another square root of X equal to -x.
Many crackpots like you confuse the definition of absolute value with
that of the square root simply because they do not understand the
meaning of an inverse function.
Mike
Dirk Van de moortel
"Mike" <ele...@yahoo.gr> wrote in message news:9c1b39be.04062...@posting.google.com...
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
news:<KDdCc.4526$Yf1....@news.cpqcorp.net>...
>
> [snip]
>
> >
> > And we have:
> > IF x >= 0 THEN sqrt(x^2) = x
> > IF x <= 0 THEN sqrt(x^2) = -x
> >
> > I hope this helps. If only just a little bit ;-)
> >
> > Dirk Vdm
>
> Having real troubles Dirk with basic algebra? No wonder. You are a
> major crank.
>
> The square root function of x, sqrt(x) is the inverse function of f(x)
> = x^2, for all x greater or equal to 0.
>
> Cranky Dirk, if x <= 0 then X^2 >= 0,
Brilliant!
"if x <= 0 then X^2 >= 0"
> let it be equal to X, and the
> "principle" square root of X is a positive number x such that x^2 =
> X, while there is another square root of X equal to -x.
HAHAHAHAHAHA!
Sucker!
>
> Many crackpots like you confuse the definition of absolute value with
> that of the square root simply because they do not understand the
> meaning of an inverse function.
>
> Mike
Dear "Mike" aka "Bill Smith" aka "Undeniable", I explicitly said
that my lesson was aimed at an audience aged between 12 and
14. With a mental age of 6 you don't qualify yet. Sorry.
Dirk Vdm
Dirk Van de moortel
"phil" <pfreed...@hotmail.com> wrote in message news:200406231524...@proapp.mathforum.org...
> On 23 Jun 2004, Dirk Van de moortel wrote:
> >A few posters recently showed having some 'minor'
> >difficulties with the concept of square roots:
[snip]
>
> I suggest that the vast majority of 12 year olds have no idea
> what a real number is, or rather, they do not know of any other kind.
> So the way to teach square roots is by simple examples. Start with
> 3^2 = 9 and explain that this is equivalent to sqrt(9) = 3 . From
> there move on to other rational numbers, like sqrt(10) is a little
> larger than sqrt(9). Then maybe draw a graph y = sqrt(x) and point
> out that the graph exists only in the first quadrant, because (for
> 12 year olds) negative numbers DON'T HAVE square roots.
No argument there :-)
As long as we don't tell them that sqrt(9) can be -3.
>
> Your boys are ready for graduate school at 12.
No, they have a reasonably good grasp of sets, elements, union,
intersection, product sets, relations and functions - mainly visually
and graphically... with Venn diagrams and arrows... nothing
spectacular really.
But as far as I see, the educational system has stopped teaching
elementary maths this way - they fear it's too abstract - for some.
Maybe they're right...
Dirk Vdm
Anthony Natoli
(snip)
>I hope this helps. If only just a little bit ;-)
>
>Dirk Vdm
>
Dirk,
While your exposition is correct and lucid, some
people out there, such as Androcles, don't
understand the motive for focusing on the
non-negative domain and range for the square root
function sqrt( ).
Specifically, in the Lorentz Transformation in
Special Relativity, the ubiquitous
gamma = sqrt(1 - (v^2)/(c^2)) is always positive,
but Androcles and others don't appreciate that
allowing sqrt to be negative will screw up the
coordinate transformations. And when it comes
to relativity, you have to know which twin is
going in which direction, right?
And relativistic length and mass don't
automatically become impossibly negative just
because some Androclesean thinker doesn't understand
why gamma is always positive.
Same with the quadratic equation in mathematics;
that is, basic algebra, in which the solution
is (1/2a)*(-b +- sqrt(b^2 - 4ac))
where one specifically includes +- before
sqrt(b^2 - 4ac), to insure that both roots/solutions
of the quadratic equation are obtained. In
Androcles' world, the "+-" is unnecessary and
could be replaced by just "+". But the unwary
may forget the "other root" in Androcles' world,
and get incomplete or wrong answers.
While one _could_ define sqrt to include the
negative value in its range, it will cause more
problems than solutions, which Androcles and his
brethren do not see or appreciate. So, by
convention in mathematics and mathematical physics,
sqrt has only the non-negative domain and range.
Side comment: Androcles and other cranks move
symbols around on paper (or in newsgroup posts),
and _think_ they known mathematics or physics
(such as relativity). See a recent thread in
sci.math where YET ANOTHER poster _thinks_ he
solved Fermat's Last Theorem by elementary methods,
while other James Harris wannabes focus on
the Goldbach Conjecture or the Riemann Hypothesis.
Move a few symbols around, and voila, stupendous
(but stupendously wrong) mathematics and/or physics.
Good luck teaching the masses!
Anthony Natoli
Dirk Van de moortel
"Anthony Natoli" <antho...@aol.com> wrote in message news:200406231731...@proapp.mathforum.org...
> On 23 Jun 2004, Dirk Van de moortel wrote:
>
> (snip)
>
> >I hope this helps. If only just a little bit ;-)
> >
> >Dirk Vdm
> >
>
> Dirk,
>
> While your exposition is correct and lucid, some
> people out there, such as Androcles, don't
> understand the motive for focusing on the
> non-negative domain and range for the square root
> function sqrt( ).
>
> Specifically, in the Lorentz Transformation in
> Special Relativity, the ubiquitous
> gamma = sqrt(1 - (v^2)/(c^2)) is always positive,
> but Androcles and others don't appreciate that
> allowing sqrt to be negative will screw up the
> coordinate transformations. And when it comes
> to relativity, you have to know which twin is
> going in which direction, right?
And end up with two gamma's - one for each "answer
of sqrt":
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SqrtAnswers.html
;-)
>
> And relativistic length and mass don't
> automatically become impossibly negative just
> because some Androclesean thinker doesn't understand
> why gamma is always positive.
>
> Same with the quadratic equation in mathematics;
> that is, basic algebra, in which the solution
> is (1/2a)*(-b +- sqrt(b^2 - 4ac))
> where one specifically includes +- before
> sqrt(b^2 - 4ac), to insure that both roots/solutions
> of the quadratic equation are obtained. In
> Androcles' world, the "+-" is unnecessary and
> could be replaced by just "+". But the unwary
> may forget the "other root" in Androcles' world,
> and get incomplete or wrong answers.
Can you imagine having to explain to these guys that +- or +/-
is nothin more than shorthand, where for example
a = +/- b
is merely an abbreviation of
a = b or a = -b
?
If you *can* imagine that, have a look at
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Gibberish.html
Sorry, couldn't resist ;-)
>
> While one _could_ define sqrt to include the
> negative value in its range,
Yes, one could define it as a "multivalued function" [Yuck!]
and say that Sqrt(4) = { -2, 2 } [Yuck!]
> it will cause more
> problems than solutions, which Androcles and his
> brethren do not see or appreciate. So, by
> convention in mathematics and mathematical physics,
> sqrt has only the non-negative domain and range.
... and in engineering, and everywhere else where mathematics
is used as a tool.
>
> Side comment: Androcles and other cranks move
> symols around on paper (or in newsgroup posts),
> and _think_ they known mathematics or physics
> (such as relativity). See a recent thread in
> sci.math where YET ANOTHER poster _thinks_ he
> solved Fermat's Last Theorem by elementary methods,
> while other James Harris wannabes focus on
> the Goldbach Conjecture or the Riemann Hypothesis.
I have seen them :-)
>
> Move a few symbols around, and voila, stupendous
> (but stupendously wrong) mathematics and/or physics.
>
> Good luck teaching the masses!
Some masses somehow seem to be so utterly immune to
being taught... but they do provide excellent entertainment :-)
>
> Anthony Natoli
Cheers,
Dirk Vdm
Bill Hobba
in message news:<KDdCc.4526$Yf1....@news.cpqcorp.net>...
>
> [snip]
>
> >
> > And we have:
> > IF x >= 0 THEN sqrt(x^2) = x
> > IF x <= 0 THEN sqrt(x^2) = -x
> >
> > I hope this helps. If only just a little bit ;-)
> >
> > Dirk Vdm
>
> Having real troubles Dirk with basic algebra? No wonder. You are a
> major crank.
Think back to grade nine I think it was when they gave you the general
formula for solving a quadratic - it s x = -b +- sqrt (b2 -4ac)/2a. Now
what do you think they put the +- in front it? If as you say it can be +
or - why bother? The answer is in grade 8 it was defined as the positive
number when multiplied by itself gives the original number - see
http://www.mathpath.org/concepts/principal.square.root.htm
Bill
Christophe Major
There is both the principal square root and the other square root as
you wrote. But Dirk's argument is on the notation, and he's not
pointing it out so that the fun lasts longer. The principal square
root is sqrt(x), and the other square root is -sqrt(x). So sqrt(x) is
a single valued positive function, but the natural language "square
root(s)" is not single valued as you pointed out.
http://mathworld.wolfram.com/SquareRoot.html
-- chris
Randy Poe
> = x^2, for all x greater or equal to 0.
Do you remember what "inverse function" means? Do you
remember the concept of "domain" and "range"? You just stated
that the square root function has a range consisting of the
non-negative reals. sqrt(x) is a non-negative real.
> Cranky Dirk, if x <= 0 then X^2 >= 0, let it be equal to X, and the
> "principle" square root of X is a positive number x such that x^2 =
> X, while there is another square root of X equal to -x.
You are confusing the square root function with the concept
of roots of a polynomial equation.
The "square root function" is defined to be what you are calling
the "principle square root". A function is single-valued. The
mapping x^2 -> +-x is a mapping to two numbers. This is not a
function.
> Many crackpots like you confuse the definition of absolute value with
> that of the square root simply because they do not understand the
> meaning of an inverse function.
An inverse function, being a function, is single valued. That is
why it is necessary to restrict the definition of sqrt(x), in
order to make something which qualifies as a function.
- Randy
Mike
Save your lessons for your crank club. Because someone is 12 years old
that does not mean he or she deserves your crap. I know some 12 year
olds who have more brain than you do and know the definition of sqrt
well.
At age 7 Gauss said to his teacher: Sum(n) = n(n+1)/2
He was lucky you were not around to teach him.
Mike
By the way: sum(36) = 666. Never go to a casino again. Devil is
watching you there. HAHAHAHAHAHA
Simplicio
> ... if x <= 0 then X^2 >= 0, let it be equal to X, and the
> "principle" square root of X is a positive number x such that x^2 =
> X, while there is another square root of X equal to -x.
>
This is a bit confusing. You start out with the symbol 'x' denoting a
non-positive number. You then change the meaning of 'x' so that it denotes
a positive number. So, by the time you reach the end of your sentence, it's
not clear which meaning of 'x' you are referring to when you write -x.
(Also, I believe that when you wrote X^2 you meant x^2.)
Simplicio
Franz Heymann
wrote in message news:<KDdCc.4526$Yf1....@news.cpqcorp.net>...
>
> [snip]
>
> >
> > And we have:
> > IF x >= 0 THEN sqrt(x^2) = x
> > IF x <= 0 THEN sqrt(x^2) = -x
> >
> > I hope this helps. If only just a little bit ;-)
> >
> > Dirk Vdm
>
> Having real troubles Dirk with basic algebra? No wonder. You are a
> major crank.
>
> The square root function of x, sqrt(x) is the inverse function of
f(x)
> = x^2, for all x greater or equal to 0.
>
> Cranky Dirk, if x <= 0 then X^2 >= 0,
Why only if x <= 0 ?
Where did this X crawl out of the woodwork?
> let it be equal to X,
What is it that you are letting be equal to X?
> and the
> "principle"
principal
> square root of X is a positive number x such that x^2 =
> X,
while there is another square root of X equal to -x.
> Many crackpots like you confuse the definition of absolute value
with
> that of the square root simply because they do not understand the
> meaning of an inverse function.
That was quite hilarious.
If you have more of the same up your sleeve, you should join a circus.
Franz
Mike
Sorry, that was a typo. I meant x^2. the correct statement is:
if x <= 0 then x^2 >= 0, let it be equal to X, and the
"principle" square root of X is a positive number x such that x^2 =
X, while there is another square root of X equal to -x.
Mike
Mike
> "Mike" <ele...@yahoo.gr> wrote in message
> news:9c1b39be.04062...@posting.google.com...
> > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
> wrote in message news:<KDdCc.4526$Yf1....@news.cpqcorp.net>...
> >
> > [snip]
> >
> > >
> > > And we have:
> > > IF x >= 0 THEN sqrt(x^2) = x
> > > IF x <= 0 THEN sqrt(x^2) = -x
> > >
> > > I hope this helps. If only just a little bit ;-)
> > >
> > > Dirk Vdm
> >
> > Having real troubles Dirk with basic algebra? No wonder. You are a
> > major crank.
> >
> > The square root function of x, sqrt(x) is the inverse function of
> f(x)
> > = x^2, for all x greater or equal to 0.
> >
> > Cranky Dirk, if x <= 0 then X^2 >= 0,
>
> Why only if x <= 0 ?
> Where did this X crawl out of the woodwork?
>
that was an obvious typo; must be x^2
> > let it be equal to X,
>
> What is it that you are letting be equal to X?
>
> > and the
> > "principle"
>
> principal
>
> > square root of X is a positive number x such that x^2 =
> > X,
>
> while there is another square root of X equal to -x.
>
> > Many crackpots like you confuse the definition of absolute value
> with
> > that of the square root simply because they do not understand the
> > meaning of an inverse function.
>
> That was quite hilarious.
> If you have more of the same up your sleeve, you should join a circus.
>
> Franz
being good at circus is a rough job, especially when you have to deal
with many untrained monkeys.
Mike
Dirk Van de moortel
"Mike" <ele...@yahoo.gr> wrote in message news:9c1b39be.04062...@posting.google.com...
> "Simplicio" <a...@nospam.com> wrote in message news:<JUqCc.81692$Hg2.8606@attbi_s04>...
> > "Mike" <ele...@yahoo.gr> wrote in message
> > news:9c1b39be.04062...@posting.google.com...
> >
> > > ... if x <= 0 then X^2 >= 0, let it be equal to X, and the
> > > "principle" square root of X is a positive number x such that x^2 =
> > > X, while there is another square root of X equal to -x.
> > >
> >
> > This is a bit confusing. You start out with the symbol 'x' denoting a
> > non-positive number. You then change the meaning of 'x' so that it denotes
> > a positive number. So, by the time you reach the end of your sentence, it's
> > not clear which meaning of 'x' you are referring to when you write -x.
> > (Also, I believe that when you wrote X^2 you meant x^2.)
> >
> > Simplicio
>
> Sorry, that was a typo. I meant x^2. the correct statement is:
>
>
> if x <= 0 then x^2 >= 0
x^2 is always positive.
The condition x <= 0 is not needed.
> , let it be equal to X, and the
> "principle" square root of X is a positive number x such that x^2 =
> X, while there is another square root of X equal to -x.
Yes, you can call it that way. But all this is about the
fact that the notation sqrt( ) is by definition reserved
to the *positive* value:
sqrt(X) = sqrt(x^2) = x for x >= 0
but
sqrt(X) = sqrt(x^2) = -x for x <= 0
This way sqrt(anything) is always positive.
There is no need to introduce a new variable X for x^2.
Dirk Vdm
Michael Varney
"Mike" <ele...@yahoo.gr> wrote in message
news:9c1b39be.04062...@posting.google.com...
> "Simplicio" <a...@nospam.com> wrote in message
news:<JUqCc.81692$Hg2.8606@attbi_s04>...
> > "Mike" <ele...@yahoo.gr> wrote in message
> > news:9c1b39be.04062...@posting.google.com...
> >
> > > ... if x <= 0 then X^2 >= 0, let it be equal to X, and the
> > > "principle" square root of X is a positive number x such that x^2 =
> > > X, while there is another square root of X equal to -x.
> > >
> >
> > This is a bit confusing. You start out with the symbol 'x' denoting a
> > non-positive number. You then change the meaning of 'x' so that it
denotes
> > a positive number. So, by the time you reach the end of your sentence,
it's
> > not clear which meaning of 'x' you are referring to when you write -x.
> > (Also, I believe that when you wrote X^2 you meant x^2.)
> >
> > Simplicio
>
> Sorry, that was a typo.
Oh bullshit. You were flaming Dirk, who is practically perfect in every
way... at least compared to crackpots like yourself.
Franz Heymann
[snip]
> if x <= 0 then x^2 >= 0, let it be equal to X, and the
> "principle"
principal
> square root of X is a positive number x such that x^2 =
> X, while there is another square root of X equal to -x.
Franz
Tom Potter
> watching you there. HAHAHAHAHAHA
This reminds me,
what happened to Harry Hanson?
Is he sick or on vacation?
--
Tom Potter http://home.earthlink.net/~tdp
Mike
> "Mike" <ele...@yahoo.gr> wrote in message
> news:9c1b39be.04062...@posting.google.com...
> > "Simplicio" <a...@nospam.com> wrote in message
> news:<JUqCc.81692$Hg2.8606@attbi_s04>...
> > > "Mike" <ele...@yahoo.gr> wrote in message
> > > news:9c1b39be.04062...@posting.google.com...
> > >
> > > > ... if x <= 0 then X^2 >= 0, let it be equal to X, and the
> > > > "principle" square root of X is a positive number x such that x^2 =
> > > > X, while there is another square root of X equal to -x.
> > > >
> > >
> > > This is a bit confusing. You start out with the symbol 'x' denoting a
> > > non-positive number. You then change the meaning of 'x' so that it
> denotes
> > > a positive number. So, by the time you reach the end of your sentence,
> it's
> > > not clear which meaning of 'x' you are referring to when you write -x.
> > > (Also, I believe that when you wrote X^2 you meant x^2.)
> > >
> > > Simplicio
> >
> > Sorry, that was a typo.
>
> Oh bullshit. You were flaming Dirk, who is practically perfect in every
> way... at least compared to crackpots like yourself.
You and Dirk are crackpots. It's an empirical fact. The domain of the
square root function is positive Reals + {0}. Confusing polynomial
roots with the sqrt function is only a sign of crackpot.
Hello crank
Dirk Van de moortel
"Mike" <ele...@yahoo.gr> wrote in message news:9c1b39be.04062...@posting.google.com...
> "Michael Varney" <varney@colorado_no_spam.edu> wrote in message news:<j_ACc.3$xr3....@news.uswest.net>...
> > "Mike" <ele...@yahoo.gr> wrote in message
> > news:9c1b39be.04062...@posting.google.com...
> > > "Simplicio" <a...@nospam.com> wrote in message
> > news:<JUqCc.81692$Hg2.8606@attbi_s04>...
> > > > "Mike" <ele...@yahoo.gr> wrote in message
> > > > news:9c1b39be.04062...@posting.google.com...
> > > >
> > > > > ... if x <= 0 then X^2 >= 0, let it be equal to X, and the
> > > > > "principle" square root of X is a positive number x such that x^2 =
> > > > > X, while there is another square root of X equal to -x.
> > > > >
> > > >
> > > > This is a bit confusing. You start out with the symbol 'x' denoting a
> > > > non-positive number. You then change the meaning of 'x' so that it
> > denotes
> > > > a positive number. So, by the time you reach the end of your sentence,
> > it's
> > > > not clear which meaning of 'x' you are referring to when you write -x.
> > > > (Also, I believe that when you wrote X^2 you meant x^2.)
> > > >
> > > > Simplicio
> > >
> > > Sorry, that was a typo.
> >
> > Oh bullshit. You were flaming Dirk, who is practically perfect in every
> > way... at least compared to crackpots like yourself.
>
> You and Dirk are crackpots. It's an empirical fact. The domain of the
> square root function is positive Reals + {0}.
From the original post of this same thread:
| Definition of the square root of the real numbers:
| sqrt:
| R+ ==> R+
| x |--> y = sqrt(x) if y^2 = x
| where x and y belong to R+, being the set of real
| positive (or zero) numbers.
> Confusing polynomial
> roots with the sqrt function is only a sign of crackpot.
From the original post of this same thread:
| By the way, in a few years you'll learn about polynomials
| and roots. Roots are things that make polynomials zero.
| For example, we say that the polynomial
| x^2 - 4
| has two roots.
| We say that the first root is 2 and the second root is -2.
| But that is for later. Just remember now not to call these
| things "square roots" or you will get utterly confused
| later...
From
http://groups.google.com/groups?&threadm=DjRCc.4715$Xp3....@news.cpqcorp.net
| The thing sqrt(anything) is BY DEFINITION a positive
| number.
| The *equation* x^2 = 9 has two *solutions* for x.
| The *polynomial* x^2 - 9 has two *roots* (things that
| make the polynomial zero).
| The roots of the polynomial x^2 - 9 are 3 and -3
| The roots of the polynomial x^2 - 5 are sqrt(5) and -sqrt(5).
| This way sqrt is a positive number - by fucking world-wide
| definition.
|
| In sloppy, confusing language people sometimes say
| that 9 has two "square roots", namely 3 and -3.
| You see where this bad habit gets them:
| http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SqrtAnswers.html
| http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/NegSqrt.html
| It makes you think that there must be two gammas.
| If you think like that, you can't possibly understand
| *any* equation involving a square root.
| Being so arrogant about it, is super-stupid and makes
| it qualify as an immortal fumble.
Pretending not to be able to read is the sign of a troll.
Doing it in such an obvious way like you do it, is the sigh of the
stupid troll.
Doing it with different names like "Mike", "Bill Smith" and
"Undeniable" is the sign of the stupid coward troll.
Dirk Vdm
The Ghost In The Machine
<ele...@yahoo.gr>
wrote
on 26 Jun 2004 03:52:27 -0700
<9c1b39be.04062...@posting.google.com>:
Depends on the desired range. The domain is actually C,
but the range is a subset of C as well (I don't know the
convention offhand as to which root to pick). If one
restricts to R, domain is {0} union R+, range is {0} union
R+. If one restricts to Q, domain is either {0} union Q+
and range is a subset of {0} union R+, or domain is a
proper subset of {0} union Q+ and range is {0} union Q+.
Pick one. :-)
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.
Anthony Natoli
>
(snip)
>Pretending not to be able to read is the sign of a troll.
>Doing it in such an obvious way like you do it, is the sigh of the
>stupid troll.
>Doing it with different names like "Mike", "Bill Smith" and
>"Undeniable" is the sign of the stupid coward troll.
>
>Dirk Vdm
>
Some trolls "say" they don't read others' posts (a la
James Harris) while some trolls "say" they don't
read entire posts, and then talk idiotic.
But not in this case. "Mike" wasn't pretending - he may
have read it, but he was just stooopid.
Anthony Natoli
Dirk Van de moortel
"Anthony Natoli" <antho...@aol.com> wrote in message news:200406261421...@proapp.mathforum.org...
Ah yes indeed, the stoopidity of obviously pretending has
but a double 'o'. Good point :-)
Dirk Vdm
>
> Anthony Natoli
>
Dirk Van de moortel
"The Ghost In The Machine" <ew...@aurigae.athghost7038suus.net> wrote in message news:t1e0r1-...@lexi2.athghost7038suus.net...
[snip]
> Depends on the desired range. The domain is actually C,
> but the range is a subset of C as well (I don't know the
> convention offhand as to which root to pick). If one
> restricts to R, domain is {0} union R+, range is {0} union
> R+. If one restricts to Q, domain is either {0} union Q+
> and range is a subset of {0} union R+, or domain is a
> proper subset of {0} union Q+ and range is {0} union Q+.
>
> Pick one. :-)
Allow me to go off topic <grin> here with an unimportant
question on notation...
I notice you write
{0} union R+,
where in some books and texts I have seen (as well as in the
books I (and my sons) had at school), this set is usually simply
written as
R+,
and the set without the 0, namely
R+ \ {0} ( or in other notation R+ - {0} )
is written with a superscript '+' and subscript '0' as
R+_0
What's the most common notation in your experience?
Dirk Vdm
Androcles
"Christophe Major" <cma...@yahoo.com> wrote in message
news:fd2ca785.0406...@posting.google.com...
[snip]
| There is both the principal square root and the other square root as
| you wrote. But Dirk's argument is on the notation,
Nope.
Dinky is jabbing at me because I said "sqrt(1) has two answers, -1 and +1."
which he wrote up as a fumble three years ago.
Although x= -1 doesn't need a minus sign (and should not have one) before
the x, Dinky thinks
x = sqrt(1) does, so x = 1 and never -1, because "this way, sqrt(1) is
ALWAYS positive", according to Dinky. By that logic and notation, x is
always positive too and must be written as -x = 1, never as x = -1.
Saying sqrt(1) has two answers is apparently very wrong. It MUST be written
in the form
" x = +/- sqrt(20) / 2
which is shorthand notation for
{ x = sqrt(20) / 2 = 2 sqrt(5) / 2 = sqrt(5)
{ or
{ x = -sqrt(20) / 2 = -2 sqrt(5) / 2 = - sqrt(5)" - Immoortel
wisdom by Dinky the Deranged,
or poor feeble-minded Dinky cannot understand it. I think the rest of the
world is capable of understanding "sqrt(1) has two answers, -1 and +1",
don't you?
Androcles.
Dik T. Winter
> I notice you write
> {0} union R+,
> where in some books and texts I have seen (as well as in the
> books I (and my sons) had at school), this set is usually simply
> written as
> R+,
> and the set without the 0, namely
> R+ \ {0} ( or in other notation R+ - {0} )
> is written with a superscript '+' and subscript '0' as
> R+_0
>
> What's the most common notation in your experience?
Depends on whether you are a Bourbakionist or an Anglo-Saxon. For the first
{0} is in R+, for the latter it is not.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Dirk Van de moortel
"Dik T. Winter" <Dik.W...@cwi.nl> wrote in message news:Hzy34...@cwi.nl...
> In article <zelDc.167043$Jk5.8...@phobos.telenet-ops.be> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
writes:
> > I notice you write
> > {0} union R+,
> > where in some books and texts I have seen (as well as in the
> > books I (and my sons) had at school), this set is usually simply
> > written as
> > R+,
> > and the set without the 0, namely
> > R+ \ {0} ( or in other notation R+ - {0} )
> > is written with a superscript '+' and subscript '0' as
> > R+_0
> >
> > What's the most common notation in your experience?
>
> Depends on whether you are a Bourbakionist or an Anglo-Saxon. For the first
> {0} is in R+, for the latter it is not.
I remember our geometry prof Bilo (who started his career as a
high school teacher, later as an inspector and finally as Univ. prof)
in Gent to mention Bourbaki in just about every lecture he gave us...
So I guess we were firmly moulded into Bourbakionists :-)
Nice to know - thanks.
Dirk Vdm
Dirk Van de moortel
"Androcles" <andr...@nospamblueyonder.co.uk> wrote in message news:JrnDc.3764$vQ4.40...@news-text.cableinet.net...
>
> "Christophe Major" <cma...@yahoo.com> wrote in message
> news:fd2ca785.0406...@posting.google.com...
> [snip]
>
> | There is both the principal square root and the other square root as
> | you wrote. But Dirk's argument is on the notation,
>
> Nope.
> Dinky is jabbing at me because I said "sqrt(1) has two answers, -1 and +1."
> which he wrote up as a fumble three years ago.
And then you decided
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SqrtAnswers.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/NegSqrt.html
Doesn't work Fartocles - no way out - dead in the corner ;-)
> Although x= -1 doesn't need a minus sign (and should not have one) before
> the x, Dinky thinks
> x = sqrt(1) does, so x = 1 and never -1, because "this way, sqrt(1) is
> ALWAYS positive", according to Dinky. By that logic and notation, x is
> always positive too and must be written as -x = 1, never as x = -1.
> Saying sqrt(1) has two answers is apparently very wrong. It MUST be written
> in the form
> " x = +/- sqrt(20) / 2
> which is shorthand notation for
> { x = sqrt(20) / 2 = 2 sqrt(5) / 2 = sqrt(5)
> { or
> { x = -sqrt(20) / 2 = -2 sqrt(5) / 2 = - sqrt(5)" - Immoortel
> wisdom by Dinky the Deranged,
>
> or poor feeble-minded Dinky cannot understand it. I think the rest of the
> world is capable of understanding "sqrt(1) has two answers, -1 and +1",
> don't you?
I think you are a very sick individual, don't you?
Dirk Vdm
Dirk Van de moortel
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message news:KDdCc.4526$Yf1....@news.cpqcorp.net...
> A few posters recently showed having some 'minor'
> difficulties with the concept of square roots:
[snip]
> So we have:
> sqrt(x^2) = x for all x >= 0
> sqrt(x^2) = -x for all x <= 0
>
> And we have:
> IF x >= 0 THEN sqrt(x^2) = x
> IF x <= 0 THEN sqrt(x^2) = -x
>
For fun, I asked a friend to check how packages like Maple
and Mathematica react to an expression like sqrt(x^2).
Result:
http://users.pandora.be/vdmoortel/dirk/Stuff/SqrtMapMath.gif
Dirk Vdm
The Ghost In The Machine
<Dik.W...@cwi.nl>
wrote
on Sun, 27 Jun 2004 02:05:19 GMT
<Hzy34...@cwi.nl>:
> In article <zelDc.167043$Jk5.8...@phobos.telenet-ops.be> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> writes:
> > I notice you write
> > {0} union R+,
> > where in some books and texts I have seen (as well as in the
> > books I (and my sons) had at school), this set is usually simply
> > written as
> > R+,
> > and the set without the 0, namely
> > R+ \ {0} ( or in other notation R+ - {0} )
> > is written with a superscript '+' and subscript '0' as
> > R+_0
> >
> > What's the most common notation in your experience?
>
> Depends on whether you are a Bourbakionist or an Anglo-Saxon. For the first
> {0} is in R+, for the latter it is not.
Color me Anglo-Saxon, then. :-) Unfortunately, the
notation one might use for R+ union {0} ('R+0'?) in that
case is somewhat ambiguous from an ASCII viewpoint, though
I suppose one could subscript it ('R_{+0}' in TeX).
As long as it's clear from the context it might not matter
too much.
Dirk Van de moortel
"The Ghost In The Machine" <ew...@aurigae.athghost7038suus.net> wrote in message news:6l33r1-...@lexi2.athghost7038suus.net...
> In sci.physics, Dik T. Winter
> <Dik.W...@cwi.nl>
> wrote
> on Sun, 27 Jun 2004 02:05:19 GMT
> <Hzy34...@cwi.nl>:
> > In article <zelDc.167043$Jk5.8...@phobos.telenet-ops.be> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
writes:
> > > I notice you write
> > > {0} union R+,
> > > where in some books and texts I have seen (as well as in the
> > > books I (and my sons) had at school), this set is usually simply
> > > written as
> > > R+,
> > > and the set without the 0, namely
> > > R+ \ {0} ( or in other notation R+ - {0} )
> > > is written with a superscript '+' and subscript '0' as
> > > R+_0
> > >
> > > What's the most common notation in your experience?
> >
> > Depends on whether you are a Bourbakionist or an Anglo-Saxon. For the first
> > {0} is in R+, for the latter it is not.
>
> Color me Anglo-Saxon, then. :-) Unfortunately, the
> notation one might use for R+ union {0} ('R+0'?) in that
> case is somewhat ambiguous from an ASCII viewpoint, though
> I suppose one could subscript it ('R_{+0}' in TeX).
The '+' being a superscript and the zero a subscript (like we
had it in our Bourbaki books), wouldn't that simply be
R_0^+
or
R^+_0
in TeX?
>
> As long as it's clear from the context it might not matter
> too much.
Indeed it wouldn't :-)
Dirk Vdm
The Ghost In The Machine
<dirkvand...@ThankS-NO-SperM.hotmail.com>
wrote
on Sun, 27 Jun 2004 15:21:23 GMT
<TBBDc.167751$385.8...@phobos.telenet-ops.be>:
Actually, that it might be -- and that's probably how
It Should Be Done(tm). :-) I've not given the matter
that much thought, admittedly.
>
>>
>> As long as it's clear from the context it might not matter
>> too much.
>
> Indeed it wouldn't :-)
>
> Dirk Vdm
>
Dik T. Winter
> In sci.physics, Dik T. Winter
> <Dik.W...@cwi.nl>
> wrote
> > Depends on whether you are a Bourbakionist or an Anglo-Saxon. For the
> > first {0} is in R+, for the latter it is not.
>
> Color me Anglo-Saxon, then. :-) Unfortunately, the
> notation one might use for R+ union {0} ('R+0'?) in that
> case is somewhat ambiguous from an ASCII viewpoint, though
> I suppose one could subscript it ('R_{+0}' in TeX).
Oh, it is only one of the differences between French and Anglo-Saxon
mathematics. In Bourbaki mathematics 0 is both positive and negative,
in Anglo-Saxon mathematics it is neither. Bourbaki distinguishes
positive numbers and strictly positive numbers.
There are quite a few other notable differences: what the Anglo-Saxon
calls a 'skew field', is simply a 'field' for Bourbaki, there the term
'commutative field' is used.
It does not matter what definition you use, as long as it is clear.
News articles have a tendency not to be clear about it. It is in
English, so many assume Anglo-Saxon mathematics. On the other hand
you will find the Bourbaki definitions in translations of the books
to English.
I was made aware of all this at university, where some of the
professors were Bourbakionists and others were Anglo-Saxons.
It depended on the course what was what.
Androcles
Oh my goodness, there IS another one. Tell that to Dinky the Deranged, would
you? He claims
"3 = sqrt(9)
-3 = -sqrt(9)" - Dinky the Deranged.
"This way sqrt is ALWAYS positive."-Dinky the Deranged.
Androcles
Androcles
[snip]
"3 = sqrt(9)
-3 = -sqrt(9)" -Dinky the Deranged.
"This way sqrt is ALWAYS positive."-Dinky the Deranged.
Androcles
Androcles
-3 = -sqrt(9)" -Dinky the Deranged.
"This way sqrt is ALWAYS positive."-Dinky the Deranged.
3 = x
-3 = -x
This way x is ALWAYS positive, right, Dinky?
ROFLMAO!
Androcles
Androcles
"Dik T. Winter" <Dik.W...@cwi.nl> wrote in message
news:Hzzvo...@cwi.nl...
"sqrt(1) has two answers, -1 and +1" - Androcles.
This is NOT clear to Dinky the Deranged.
"3 = sqrt(9)
-3 = -sqrt(9)" -Dinky the Deranged.
"This way sqrt is ALWAYS positive."-Dinky the Deranged.
Androcles
"This way sqrt is ALWAYS positive." Dinky the Deranged.
Androcles
-3 = -sqrt(9)" -Dinky the Deranged.
"This way sqrt is ALWAYS positive."-Dinky the Deranged.
3 = x
-3 = -x
This way x is ALWAYS positive, right, Dinky?
Androcles
Androcles
Let's see if it works then.
x = sqrt(1).
What is x?
Well well...
If x = 1 then x = 1.
If x = -1 then x = -1.
Let x = Dinky.
If Dinky is a moron then Dinky is a moron.
Does the term "tautology" come to mind?
This way, x = moron(Dinky) is ALWAYS positive.
Androcles :-)
Randy Poe
> Let's see now.
> 3 = x
> -3 = -x
> This way x is ALWAYS positive, right, Dinky?
It is if x is 3, as you asserted on line 1.
Do you have an example where 3 is negative?
Do you think the second statement can be true for negative x?
Randy
Dirk Van de moortel
"Androcles" <andr...@nospamblueyonder.co.uk> wrote in message news:NKPDc.4770$uy4.50...@news-text.cableinet.net...
Title: "Does the term "tautology" come to mind?"
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Tautology.html
It must be hell to be so clueless - I'm on the brink
of starting to feel a wee bit sorry for you :-)
Dirk Vdm
Dirk Van de moortel
"Randy Poe" <poespa...@yahoo.com> wrote in message news:df76407e.04062...@posting.google.com...
Ha Randy, you have been away for a while...
nice to see you're back :-)
You obviously missed some of the action.
In this case Androcles is using Village Idiot Tactic #7:
Use a few lines of someone and put them together
as if they are a quote. It goes like this:
"I am a fool" - Androcles
"like you have" - Androcles
"never seen" - Androcles
"a fool before" - Androcles
Dirk Vdm
Androcles
He said
Quote:
"3 = sqrt(9)
-3 = -sqrt(9)" -Dinky the Deranged.
"This way sqrt is ALWAYS positive."-Dinky the Deranged.
I merely substituted x for sqrt(9).
Make sure you never write x = -3 because x is always positive, Dinky says
so. You MUST write -x = -3. He'll get really angry and abusive if you
disagree with him. :-)
Androcles
Mike
Dinky <---> moron
Sooner or later the necesity condition is added to demonstrated
sufficiency to result in a glorious equivalence:
Dinky <---> Moron
Very soon I suspect you will see it as a synonym in Word Thesarus.
Mike
Androcles
"3 = sqrt(9)
-3 = -sqrt(9)" -Dinky the Deranged.
"This way sqrt is ALWAYS positive."-Dinky the Deranged.
"IF x <= 0 THEN sqrt(x^2) = -x" - Dinky the Deranged
is not written in the form -3 = -sqrt(9). There is a missing minus sign
after the word "THEN".
Perhaps -x = -sqrt(x^2) is equivalent to sqrt(x^2) = -x
because x is on the other side of the '='?
Maybe we should ask Dinky.
"A few posters recently showed having some 'minor'
difficulties with the concept of square roots:" - Dinky the Deranged.
Androcles
Jesper Pedersen
<snip>
> Yes, very probably, but remember that
> "3 = sqrt(9)
> -3 = -sqrt(9)" -Dinky the Deranged.
> "This way sqrt is ALWAYS positive."-Dinky the Deranged.
> and
> "IF x <= 0 THEN sqrt(x^2) = -x" - Dinky the Deranged
> is not written in the form -3 = -sqrt(9). There is a missing minus sign
> after the word "THEN".
> Perhaps -x = -sqrt(x^2) is equivalent to sqrt(x^2) = -x
> because x is on the other side of the '='?
> Maybe we should ask Dinky.
You need to understand the fundamental difference between solving x^2=2 for
x and applying the function sqrt(2). the sqrt is DEFINED as being positive.
Otherwise how would equalities such as sin(Pi/4) = sqrt(2)/2 make any sense
at all? Surely you wouldn't claim that sin is dual-valued?
/ Jesper P
Dirk Van de moortel
"Jesper Pedersen" <jes...@befunk.com> wrote in message news:cbu7sn$a4i$1...@news.net.uni-c.dk...
You cannot even *begin* to imagine what he would claim ;-)
Sin will probably depend on the GIVEN:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/GIVEN.html
Dirk Vdm
Androcles
| news:CmnEc.6449$MV.69...@news-text.cableinet.net...
| <snip>
| > Yes, very probably, but remember that
| > "3 = sqrt(9)
| > -3 = -sqrt(9)" -Dinky the Deranged.
| > "This way sqrt is ALWAYS positive."-Dinky the Deranged.
| > and
| > "IF x <= 0 THEN sqrt(x^2) = -x" - Dinky the Deranged
| > is not written in the form -3 = -sqrt(9). There is a missing minus sign
| > after the word "THEN".
| > Perhaps -x = -sqrt(x^2) is equivalent to sqrt(x^2) = -x
| > because x is on the other side of the '='?
| > Maybe we should ask Dinky.
| <snip more commented quotes>
|
| You need to understand the fundamental difference between solving x^2=2
for
| x
I think the solution is -1.4142135623730950488016887242097.
Let's see.
-1.4142135623730950488016887242097 *
-1.4142135623730950488016887242097 = 2.
Yep, I'm right.
and applying the function sqrt(2). the sqrt is DEFINED as being positive.
Is it?
if x = [-b +/- sqrt(b^2 - 4ac]/2a, is b always positive?
| Otherwise how would equalities such as sin(Pi/4) = sqrt(2)/2 make any
sense
| at all? Surely you wouldn't claim that sin is dual-valued?
That rather depends on whether you use (x, iy) or (x,y), doesn't it?
Androcles.
|
| / Jesper P
|
|
Randy Poe
> "Randy Poe" <poespa...@yahoo.com> wrote in message
> news:df76407e.04062...@posting.google.com...
> | "Androcles" <andr...@nospamblueyonder.co.uk> wrote in message
> news:<knPDc.4759$yy4.50...@news-text.cableinet.net>...
> |
> | > Let's see now.
> | > 3 = x
> | > -3 = -x
> | > This way x is ALWAYS positive, right, Dinky?
> |
> | It is if x is 3, as you asserted on line 1.
> |
> | Do you have an example where 3 is negative?
> |
> | Do you think the second statement can be true for negative x?
> |
> | Randy
> Those were Dinky the Deranged's ideas, you had better ask him.
I agree that the first statement and the second one are
equivalent. I'm asking why YOU have a problem with the second
one, or the assertion that if x = 3, then x is always positive.
> He said
> Quote:
> "3 = sqrt(9)
> -3 = -sqrt(9)" -Dinky the Deranged.
> "This way sqrt is ALWAYS positive."-Dinky the Deranged.
> Unquote.
Yes. That's correct. The symbol "sqrt(9)", like your x,
stands for a positive quantity. If you want to refer to a
negative quantity with the same absolute value, you can
write a negative sign on both sides. Thus, -3 = -sqrt(9)
certainly follows from the definition of sqrt(9) as the
positive value whose square is 3.
Just as -3 = -x follows from the statement that 3 = x.
> I merely substituted x for sqrt(9).
OK. And why is it that you think -3 = -x is wrong when
you write 3 = x?
> Make sure you never write x = -3 because x is always positive,
No. Make sure that if you claim x = 3, you don't claim
x is simultaneously equal to -3.
"x" is not a symbol that has a sign convention in it. Nowhere
does Dirk or anyone else claim that the symbol "x" can only
be used for positive quantities.
On the other hand, the symbol "sqrt(x)" IS defined, by convention,
to stand for the non-negative root of y^2 - x. Since sqrt, unlike
x, has a sign-convention associated with it, then it is incorrect
to write "sqrt(9) = -3".
The thing on the left is >=0, the thing on the right is <0. They
can't be equal. There certainly is a negative root to y^2 - 9.
The symbol we use for that root is "-sqrt(9)".
> Dinky says so. You MUST write -x = -3.
You seem to have trouble with a really elementary concept here.
If we define sqrt(9) to be 3, then -sqrt(9) is -3.
If we define x to be 3, then -x is -3.
- Randy
Androcles
Aha! At last someone has hit the nail on the head. I was waiting for it.
By CONVENTION.
Yes, totally agree that there is a convenience to the concept. However, the
convention is an arbitrary one, just as the shape of the letters of the
alphabet are arbitrary. The is an unknown number of type fonts available,
and if we don't happen to like any particular one of them, we are free to
create another. it is a convention that we right from left to right, but we
need not do so to be dootsrednu, or even
u
n
d
e
r
s
t
o
o
d.
This doesn't mean -3 isn't a sqrt() of 9, though, and why was Dinky so
anxious to display my "stupidity" for saying sqrt(1) has two answers, -1 and
1? It is easy enough to understand, but because he is a slave to convention
he finds it unreasonable.
| to stand for the non-negative root of y^2 - x. Since sqrt, unlike
| x, has a sign-convention associated with it, then it is incorrect
| to write "sqrt(9) = -3".
|
| The thing on the left is >=0, the thing on the right is <0. They
| can't be equal. There certainly is a negative root to y^2 - 9.
| The symbol we use for that root is "-sqrt(9)".
|
| > Dinky says so. You MUST write -x = -3.
|
| You seem to have trouble with a really elementary concept here.
Not at all, I'm simply not a slave to convention. That doesn't make it wrong
to say sqrt(1) has two answers, -1 and 1. The cube root of 1 has three
answers. The 4th root has 4 and the nth root has n.
|
| If we define sqrt(9) to be 3, then -sqrt(9) is -3.
Sure, and if we define sqrt(9) to be -3, then -sqrt(9) is 3.|
| If we define x to be 3, then -x is -3.
Exactly. The "If" is all important.
If we place x' = x-vt, then xi = (x-vt) *gamma,
but if we place x' = x+vt, then xi = (x+vt) * gamma.
So relativity isn't anything that Nature dictates, it is what convention
dictates, as you have just shown.
Androcles.
Jesper Pedersen
<snip>
> | You need to understand the fundamental difference between solving x^2=2
> for
> | x
>
> I think the solution is -1.4142135623730950488016887242097.
> Let's see.
> -1.4142135623730950488016887242097 *
> -1.4142135623730950488016887242097 = 2.
> Yep, I'm right.
Almost, but not quite right. There are two solutions to the equation, yours
is one of them. I suspect that you know this, or at least I sincerely hope
so.
> and applying the function sqrt(2). the sqrt is DEFINED as being positive.
>
> Is it?
> if x = [-b +/- sqrt(b^2 - 4ac]/2a, is b always positive?
What does that have to do with sqrt(x) being positive by definition?
sqrt(b^2-4ac) is always positive, but you may choose b to be whatever suits
your fancy. Why do you even choose to write the equation the way you do? If
you believe sqrt(x) to be both positive and negative, why the need for the
+/- in front of it?
> | Otherwise how would equalities such as sin(Pi/4) = sqrt(2)/2 make any
> sense
> | at all? Surely you wouldn't claim that sin is dual-valued?
> That rather depends on whether you use (x, iy) or (x,y), doesn't it?
> Androcles.
I fail to see how sin(Pi/4) can ever be anything other than sqrt(2)/2 > 0.
And let's just stick to x belonging to R shall we? Would you at least agree
then, that sin(x) is not dual-valued? Or am I wasting time on a troll?
/ Jesper P
Dirk Van de moortel
"Androcles" <andr...@nospamblueyonder.co.uk> wrote in message news:OBGEc.416$rE3.4...@news-text.cableinet.net...
[snip]
> | On the other hand, the symbol "sqrt(x)" IS defined, by convention,
>
> Aha! At last someone has hit the nail on the head. I was waiting for it.
> By CONVENTION.
> Yes, totally agree that there is a convenience to the concept. However, the
> convention is an arbitrary one, just as the shape of the letters of the
> alphabet are arbitrary. The is an unknown number of type fonts available,
> and if we don't happen to like any particular one of them, we are free to
> create another. it is a convention that we right from left to right, but we
> need not do so to be dootsrednu, or even
> u
> n
> d
> e
> r
> s
> t
> o
> o
> d.
> This doesn't mean -3 isn't a sqrt() of 9, though, and why was Dinky so
> anxious to display my "stupidity" for saying sqrt(1) has two answers, -1 and
> 1?
Because it makes you draw the following conclusions from it:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SqrtAnswers.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/PythagoRescue.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SqrtRev.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/NegSqrt.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Tautology.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/GIVEN.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/GrowUp.html
So apparently we now have a "YES" to question (1)
of our little experiment
"When dealing, in a public arena, with a malicious,
stupid would-be terrorist, is it possible to
(1) make him really understand his condition?
(2) make him admit it?
(3) make him apologize for his behaviour?"
I'm sure we will continue to have a "NO" to questions
(2) and (3).
Dirk Vdm
Androcles
| "Androcles" <andr...@nospamblueyonder.co.uk> wrote in message
| news:kwCEc.88$rd1.1...@news-text.cableinet.net...
| <snip>
| > | You need to understand the fundamental difference between solving
x^2=2
| > for
| > | x
| >
| > I think the solution is -1.4142135623730950488016887242097.
| > Let's see.
| > -1.4142135623730950488016887242097 *
| > -1.4142135623730950488016887242097 = 2.
| > Yep, I'm right.
|
| Almost, but not quite right.
maybe?
There are two solutions to the equation, yours
| is one of them.
Well done. How very perceptive of you.
I suspect that you know this, or at least I sincerely hope
| so.
So we agree. sqrt(2) has two solutions.
|
| > and applying the function sqrt(2). the sqrt is DEFINED as being
positive.
I see. Who by?
| > Is it?
| > if x = [-b +/- sqrt(b^2 - 4ac]/2a, is b always positive?
|
| What does that have to do with sqrt(x) being positive by definition?
Well, you seem to be following some rule I'm not aware of.
I'm trying to figure out what it is.
If -2 is a sqrt of 4, then sqrt(4) * sqrt(4)= 4
and also -sqrt(4) * -sqr(4) = 4.
What's nearly wrong with that?
| sqrt(b^2-4ac) is always positive, but you may choose b to be whatever
suits
| your fancy. Why do you even choose to write the equation the way you do?
To emphasize and describe algebraically there are two roots, of course. Its
shorthand.
What's nearly wrong with that?
| If
| you believe sqrt(x) to be both positive and negative, why the need for the
| +/- in front of it?
I never said it was both, I said there were two solutions.
| > | Otherwise how would equalities such as sin(Pi/4) = sqrt(2)/2 make any
| > sense
| > | at all? Surely you wouldn't claim that sin is dual-valued?
| > That rather depends on whether you use (x, iy) or (x,y), doesn't it?
| > Androcles.
|
| I fail to see how sin(Pi/4) can ever be anything other than sqrt(2)/2 > 0.
| And let's just stick to x belonging to R shall we?
Why? What's nearly wrong with z = {x + iy} ?
| Would you at least agree
| then, that sin(x) is not dual-valued?
Would you at least agree tht bight green flying elephants lay eggs, Or am I
being irrelevant?
| Or am I wasting time on a troll?
Ah, the first insinuation that someone you disagree with must have some evil
motive in mind, although there are two solutions to sqrt(), one of them
negative, and sqrt() is defined as positive. Next you'll begin snipping and
ignoring and name calling, just like the rest of the trolls that lurk around
here.
Androcles.
|
| / Jesper P
|
|
Wilbert Dijkhof
Androcles wrote:
>
> | Or am I wasting time on a troll?
>
> Ah, the first insinuation that someone you disagree with must have some evil
> motive in mind, although there are two solutions to sqrt(), one of them
> negative, and sqrt() is defined as positive. Next you'll begin snipping and
> ignoring and name calling, just like the rest of the trolls that lurk around
> here.
Forget it, Jerper is correct. First sqrt() doesn't have a solution,
because
it's an expression (regardless whether you see it as single valued
or multi valued). Like Jesper said, it is custom to take sqrt(.) as the
principal value. Which implies that sqrt(4) is taken as 2.
Some mathematicians use the expression a^(1/2) to denote the set of
solutions of x^2=a. In that case 4^(1/2) = {+/-2}, while sqrt(4)
being the positive one.
Wilbert
Jesper Pedersen
<snip>
> There are two solutions to the equation, yours
> | is one of them.
>
> Well done. How very perceptive of you.
>
> I suspect that you know this, or at least I sincerely hope
> | so.
>
> So we agree. sqrt(2) has two solutions.
Not at all. We agree that the equation x^2=2 has two solutions, sqrt(2)
and -sqrt(2). How you could read what I wrote and perceive it as me agreeing
that sqrt(2) has two solutions is disturbing. The very notion of "solutions"
to sqrt(2) makes no sense.
> | > and applying the function sqrt(2). the sqrt is DEFINED as being
> positive.
> I see. Who by?
> | > Is it?
> | > if x = [-b +/- sqrt(b^2 - 4ac]/2a, is b always positive?
> |
> | What does that have to do with sqrt(x) being positive by definition?
>
> Well, you seem to be following some rule I'm not aware of.
> I'm trying to figure out what it is.
It is really very simple. By definition, sqrt(x) denotes the positive
solution of the equation y^2 = x. Why is this so difficult to comprehend?
<snip>
> | If
> | you believe sqrt(x) to be both positive and negative, why the need for
the
> | +/- in front of it?
>
> I never said it was both, I said there were two solutions.
It does not have "solutions". It's notation.
> | > | Otherwise how would equalities such as sin(Pi/4) = sqrt(2)/2 make
any
> | > sense
> | > | at all? Surely you wouldn't claim that sin is dual-valued?
> | > That rather depends on whether you use (x, iy) or (x,y), doesn't it?
> | > Androcles.
> |
> | I fail to see how sin(Pi/4) can ever be anything other than sqrt(2)/2 >
0.
> | And let's just stick to x belonging to R shall we?
> Why? What's nearly wrong with z = {x + iy} ?
Nothing at all. I am just trying very hard to explain to you why sqrt(x)
must be positive by definition, and why there can be no ambiguity in whether
the positive or negative root of y^2=x is chosen as the value of sqrt(x). No
need to include complex numbers for that purpose.
> | Would you at least agree
> | then, that sin(x) is not dual-valued?
>
> Would you at least agree tht bight green flying elephants lay eggs, Or am
I
> being irrelevant?
What is irrelevant about that question? Can't you see that if the sign of
sqrt(x) may be chose arbitrarily, there is a major problem with equalities
such as the one I posed? I was trying to give you an example that should
make it obvious that a convention is necessary on the sign of sqrt(x).
/ Jesper P
Randy Poe
> "Jesper Pedersen" <jes...@befunk.com> wrote in message
> news:cc0hvk$uf8$1...@news.net.uni-c.dk...
>
> | is one of them.
>
> Well done. How very perceptive of you.
You seemed so close to comprehension here...
>
> So we agree. sqrt(2) has two solutions.
But then you lost it.
"sqrt(2)" does not have "two solutions". "sqrt(2)"
is not an equation. It is a number. It is a single number.
That number is, by definition of the "sqrt" notation, a
positive number.
The equation x^2 = 2 has two solutions. You can tell it is
an equation by the presence of a variable in it (x), and an
equality sign (=). The notation "sqrt(2)" has neither of those
attributes. Therefore it makes no sense to talk about it having
ANY solutions, let alone two of them.
What are the two solutions to x^2 = 2? There are, we agree,
two of them.
One of them is a positive real number. It is designated sqrt(2).
There is a unique meaning for that notation, and it stands for
this positive number, the POSITIVE solution to x^2 = 2.
There is another solution to the equation x^2 = 2. It is a
negative real number, with the same absolute value as sqrt(2). We designate
that solution with the notation -sqrt(2). It is the negative of
the positive value which we designate "sqrt(2)". The fact that
"-sqrt(2)" is considered to be a negative value tells us that
"sqrt(2)" is considered to be a positive value. At no time does
anybody, anywhere, anytime, ever use the notation "sqrt(2)"
to stand for a negative value.
- Randy
Dirk Van de moortel
"Randy Poe" <poespa...@yahoo.com> wrote in message news:df76407e.04070...@posting.google.com...
Randy,
have you seen the original post of this thread?
The bigot simply does not want to know.
Dirk Vdm
Randy Poe
> Randy,
> have you seen the original post of this thread?
Nope, lost in the depths of google's thread-breaking.
> The bigot simply does not want to know.
Well, yeah I know. This is just for entertainment.
- Randy
Dirk Van de moortel
news:<oB9Fc.5232$cU3....@news.cpqcorp.net>...
> > Randy,
> > have you seen the original post of this thread?
>
> Nope, lost in the depths of google's thread-breaking.
Breath-taking you mean?
Click
http://groups.google.com/groups?&threadm=KDdCc.4526$Yf1....@news.cpqcorp.net
>
> > The bigot simply does not want to know.
>
> Well, yeah I know. This is just for entertainment.
First class :-))
Dirk Vdm
Androcles
"Wilbert Dijkhof" <w.j.d...@tue.nl> wrote in message
news:40E40190...@tue.nl...
|
|
| Androcles wrote:
| >
|
| > | Or am I wasting time on a troll?
| >
| > Ah, the first insinuation that someone you disagree with must have some
evil
| > motive in mind, although there are two solutions to sqrt(), one of them
| > negative, and sqrt() is defined as positive. Next you'll begin snipping
and
| > ignoring and name calling, just like the rest of the trolls that lurk
around
| > here.
|
| Forget it, Jerper is correct. First sqrt() doesn't have a solution,
| because
| it's an expression (regardless whether you see it as single valued
| or multi valued). Like Jesper said, it is custom
Exactly. Custom, convention, call it what you will, sqrt(1) has two
answers, -1 and +1. Showing a preference for one over the other is merely
prejudice.
to take sqrt(.) as the
| principal value. Which implies that sqrt(4) is taken as 2.
What makes '+ve' the principal?
|
| Some mathematicians use the expression a^(1/2) to denote the set of
| solutions of x^2=a.
Do they really? Well, well. They wouldn't happen to call the (1/2) a
logarithm, by any slim chance, would they? Or exponent?
a^(1/2) * a^( -1/2) wouldn't happen to be 1, would it?
| In that case 4^(1/2) = {+/-2}, while sqrt(4)
| being the positive one.
|
| Wilbert
Who says so? You?
Androcles
Androcles
Who's definition? Yours? Why must Americans drive on the right side of the
road? Why must Brits drive on the left side of the road? I'll tell you in
one word. 'Convention'. Not definition as you claim, but convention.
| Why is this so difficult to comprehend?
Because you claim it is by definition, and you are wrong. I am in no way
obligated to accept your definition, or your convention. The sqrt() of
anything has two possible answers, and that goes for sqrt(1-v^2/c^2) as
well.
|
| <snip>
| > | If
| > | you believe sqrt(x) to be both positive and negative, why the need for
| the
| > | +/- in front of it?
| >
| > I never said it was both, I said there were two solutions.
|
| It does not have "solutions". It's notation.
That doesn't entitle anyone to ignore the negative root of sqrt(1-v^2/c^2).
|
| > | > | Otherwise how would equalities such as sin(Pi/4) = sqrt(2)/2 make
| any
| > | > sense
| > | > | at all? Surely you wouldn't claim that sin is dual-valued?
| > | > That rather depends on whether you use (x, iy) or (x,y), doesn't it?
| > | > Androcles.
| > |
| > | I fail to see how sin(Pi/4) can ever be anything other than sqrt(2)/2
>
| 0.
| > | And let's just stick to x belonging to R shall we?
| > Why? What's nearly wrong with z = {x + iy} ?
|
| Nothing at all. I am just trying very hard to explain to you why sqrt(x)
| must be positive by definition,
Which it isn't.
and why there can be no ambiguity in whether
| the positive or negative root of y^2=x is chosen as the value of sqrt(x).
Chosen? Pick the root you prefer, huh?
No
| need to include complex numbers for that purpose.
|
| > | Would you at least agree
| > | then, that sin(x) is not dual-valued?
| >
| > Would you at least agree tht bight green flying elephants lay eggs, Or
am
| I
| > being irrelevant?
|
| What is irrelevant about that question? Can't you see that if the sign of
| sqrt(x) may be chose arbitrarily,
Yes, I see that only too well.
| there is a major problem with equalities
| such as the one I posed? I was trying to give you an example that should
| make it obvious that a convention is necessary on the sign of sqrt(x).
Aha! It is a convention, not a definition. Agreed.
Let's move on. Work on this "Definition" for me, please.
"If we place x'=x-vt, it is clear that a point at rest in the system k must
have a system of values x', y, z, independent of time. We first define tau
as a function of x', y, z, and t. To do this we have to express in equations
that tau is nothing else than the summary of the data of clocks at rest in
system k, which have been synchronized according to the rule given in § 1.
From the origin of system k let a ray be emitted at the time tau0 along the
X-axis to x', and at the time tau1 be reflected thence to the origin of the
co-ordinates, arriving there at the time tau2; we then must have
½(tau0+tau2) =tau1, or, by inserting the arguments of the function tau and
applying the principle of the constancy of the velocity of light in the
stationary system:-
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))"-
Albert Einstein.
See that "IF" at the beginning?
Does that mean
xi really is equal to (x-vt)*gamma,
or does it depend on the "If"?
What IF we place x' = x+vt?
Androcles
| / Jesper P
|
|
Jesper Pedersen
<snip>
> | Why is this so difficult to comprehend?
>
> Because you claim it is by definition, and you are wrong. I am in no way
> obligated to accept your definition, or your convention. The sqrt() of
> anything has two possible answers, and that goes for sqrt(1-v^2/c^2) as
> well.
sqrt(x) has ONE value - not "answers", it is not an equation - which by
definition, convention, dicatorship, call it whatever you want, is the
positive root of y^2=x. You may choose the negative one in your own work,
but it'll make everything you write nonsense to anyone but you. You wouldn't
choose "+" to mean "-" would you? That is just a convention as well. Why
don't you have any problems with those definitions? Or hell, why not
interchange sin and cosine? Those are just conventions as well.
So no, you are not obligated to accept this definition. You'll write utter
nonsense if you don't though. So I'd advise you to stick with the standard
notation.
/ Jesper P
Dirk Van de moortel
"Androcles" <andr...@nospamblueyonder.co.uk> wrote in message news:UReFc.1932$xl6.16...@news-text.cableinet.net...
>
> "Jesper Pedersen" <jes...@befunk.com> wrote in message
> news:cc11n6$3jb$1...@news.net.uni-c.dk...
[snip]
> | It does not have "solutions". It's notation.
>
> That doesn't entitle anyone to ignore the negative root of sqrt(1-v^2/c^2).
I guess this turns the answer to question (1) back
into "NO". Sometime overnight it must have changed
its little mind.
Perhaps, in order to have at least one question that
can be answered with "YES", we should adapt the
Androcles Experiment as follows:
"When dealing, in a public arena, with a malicious,
stupid would-be terrorist, is it possible to
(0) make him realize that there is some sort
of difference between him and the rest
of the world?
(1) make him really understand his condition?
(2) make him admit it?
(3) make him apologize for his behaviour?"
Dirk Vdm
Jesper Pedersen
<snip>
> Let's move on. Work on this "Definition" for me, please.
>
> "If we place x'=x-vt, it is clear that a point at rest in the system k
must
> have a system of values x', y, z, independent of time. We first define tau
I am assuming that your system k is moving along the positive x-axis with
constant velocity v>0, since this will make the sentence make sense. The
fact that you don't bother telling me these vital details already hints at
the fact that you will end up drawing a false conclusion. So we have v > 0
and k moving with constant velocity v along the x-axis.
<snip intermediates>
> See that "IF" at the beginning?
> Does that mean
> xi really is equal to (x-vt)*gamma,
> or does it depend on the "If"?
> What IF we place x' = x+vt?
> Androcles
If you choose x'=x+vt, then a point at rest in system k no longer has values
x', y, z independent of time. For this to be true you would have to invert
the direction of movement of k along the x-axis, meaning that you new v'
= -v, in which case you end up with the same expression that you get
choosing x'=x-vt. Sounds like you have not understood the fundamental
reasoning behind the choice of x'=x-vt. It is not arbitrary, given the
initial movement of k.
/ Jesper P
Robin Chapman
> Who's definition? Yours? Why must Americans drive on the right side of
> the road? Why must Brits drive on the left side of the road? I'll tell you
> in one word. 'Convention'. Not definition as you claim, but convention.
It's the convention to make the definition of sqrt(x), for real x > 0,
as the positive solution of y^2 = x.
If you must bravely defy convention, please come over here and
flout the convention of driving on the left.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9"
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
luke
> [...]
> Exactly. Custom, convention, call it what you will, sqrt(1) has two
> answers, -1 and +1. Showing a preference for one over the other is merely
> prejudice.
>
You're tripping Andorcles. Did you or did you not memorize:
y = [ - b +- sqrt(b^2 - 4ac) ] / 2a
do you know why you memorized the "plus or minus"?
>
> |
> | Some mathematicians use the expression a^(1/2) to denote the set of
> | solutions of x^2=a.
>
> Do they really? Well, well. They wouldn't happen to call the (1/2) a
> logarithm, by any slim chance, would they? Or exponent?
> a^(1/2) * a^( -1/2) wouldn't happen to be 1, would it?
>
They could, yes. Not if a=0, no.
Androcles
"luke" <fun...@yahoo.com> wrote in message
news:e1b04639.04070...@posting.google.com...
| "Androcles" <andr...@nospamblueyonder.co.uk> wrote in message
news:<hyeFc.1912$rc6.16...@news-text.cableinet.net>...
| > "Wilbert Dijkhof" <w.j.d...@tue.nl> wrote in message
| > news:40E40190...@tue.nl...
|
| > [...]
| > Exactly. Custom, convention, call it what you will, sqrt(1) has two
| > answers, -1 and +1. Showing a preference for one over the other is
merely
| > prejudice.
| >
|
| You're tripping Andorcles. Did you or did you not memorize:
|
| y = [ - b +- sqrt(b^2 - 4ac) ] / 2a
|
| do you know why you memorized the "plus or minus"?
Actually I derived it, luek.
Androcles
Androcles
Do you mean this kind of utter nonsense, which, smart as you are
(or think you are), you snip and ignore?
Ballisticus
sqrt(4)*sqrt(4) can also be -4
You simply multiply the negative root by the positive one.
Whoever proved that (-A)*(-B)=+AB anyway?
Is (-^2) = (+) just another convention?
>
>Androcles.
>
>
>
>|
>| / Jesper P
>|
>|
>
See why relativity is wrong:
www.users.bigpond.com/hewn/index.htm
Dirk Van de moortel
"Ballisticus" <B@..> wrote in message news:q8sce0521crlt0kt2...@4ax.com...
> On Thu, 01 Jul 2004 12:01:20 GMT, "Androcles"
> <andr...@nospamblueyonder.co.uk> wrote:
>
> >
> >"Jesper Pedersen" <jes...@befunk.com> wrote in message
> >news:cc0hvk$uf8$1...@news.net.uni-c.dk...
> >| "Androcles" <andr...@nospamblueyonder.co.uk> wrote in message
> >| news:kwCEc.88$rd1.1...@news-text.cableinet.net...
[snip]
> >| > and applying the function sqrt(2). the sqrt is DEFINED as being positive.
> >I see. Who by?
> >| > Is it?
> >| > if x = [-b +/- sqrt(b^2 - 4ac]/2a, is b always positive?
I hadn't seen this one :-)
"Androcles and the quadratic equation":
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Quadratic.html
Sorry I'm late.
> >|
> >| What does that have to do with sqrt(x) being positive by definition?
> >
> >Well, you seem to be following some rule I'm not aware of.
> >I'm trying to figure out what it is.
> >If -2 is a sqrt of 4, then sqrt(4) * sqrt(4)= 4
> >and also -sqrt(4) * -sqr(4) = 4.
> >What's nearly wrong with that?
>
> sqrt(4)*sqrt(4) can also be -4
>
> You simply multiply the negative root by the positive one.
The battle of the Village Idiots:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Nearly.html
Devastating argument though ;-)
Dirk Vdm
Androcles
| "Androcles" <andr...@nospamblueyonder.co.uk> wrote in message
| news:UReFc.1932$xl6.16...@news-text.cableinet.net...
| <snip>
| > Let's move on. Work on this "Definition" for me, please.
| >
| > "If we place x'=x-vt, it is clear that a point at rest in the system k
| must
| > have a system of values x', y, z, independent of time. We first define
tau
|
| I am assuming that your system k
| is moving along the positive x-axis with
| constant velocity v>0, since this will make the sentence make sense.
Assumptions are not wise. It is Einstein's system k. Otherwise your guess
was correct.
| The
| fact that you don't bother telling me these vital details already hints at
| the fact that you will end up drawing a false conclusion.
The fact that you have somehow poked your nose into sci.physics.relativity,
probably because somebody I was replying to added sci.math in the list of
newsgroups, is the most probable cause of your lack of understanding.
Newsgroups are not scientific papers where all references are indicated with
footnotes. However, since you are not familiar with the author's work, I
will cite: http://www.fourmilab.ch/etexts/einstein/specrel/www/
So we have v > 0
| and k moving with constant velocity v along the x-axis.
Very good.
|
| <snip intermediates>
| > See that "IF" at the beginning?
| > Does that mean
| > xi really is equal to (x-vt)*gamma,
| > or does it depend on the "If"?
| > What IF we place x' = x+vt?
| > Androcles
|
| If you choose x'=x+vt, then a point at rest in system k no longer has
values
| x', y, z independent of time.
What does it have values of, then?
| For this to be true you would have to invert
| the direction of movement of k along the x-axis, meaning that you new v'
| = -v, in which case you end up with the same expression that you get
| choosing x'=x-vt.
Are you telling me that the RHS of the equation which you carefuly snipped
to deliberately induce confusion doesn't change?
| Sounds like you have not understood the fundamental
| reasoning behind the choice of x'=x-vt. It is not arbitrary, given the
| initial movement of k.
Is that what is "sounds like"?
It sounds to me as if you are guesser and a snipper, ignoring what you have
no answer for, and certainly you have not understood the fundamental
reasoning behind (2/3 + 1/3)/2 = 1/3, or that cars may be driven forwards as
well as backwards, in which case (2/3 + 1/3)/2 = 2/3.
Or am I discussing special relativity with a troll?
Androcles
| / Jesper P
|
Jesper Pedersen
This makes no sense at all. I haven't poked my nose into
sci.physics.relativity. I only monitor psi.physics. The reason I responded
to your message was that I wanted to help you out in understanding the
definitions of sqrt(x). And though I certainly understand that newsgroups
are not scientific papers, I would expect that if someone wants to challenge
special relativity they should at least make clear all the specifics of
their derivations. And I am familiar with Einstein's work on special
relativity. Seems like you have a bit of learning to do though. "Einstein's
system k" for one makes no sense at all. In which direction is it moving, IS
it moving at all, is it accelerated, etc? These are quite vital specifics.
> So we have v > 0
> | and k moving with constant velocity v along the x-axis.
>
> Very good.
Thank you.
> | <snip intermediates>
> | > See that "IF" at the beginning?
> | > Does that mean
> | > xi really is equal to (x-vt)*gamma,
> | > or does it depend on the "If"?
> | > What IF we place x' = x+vt?
> | > Androcles
> |
> | If you choose x'=x+vt, then a point at rest in system k no longer has
> values
> | x', y, z independent of time.
>
> What does it have values of, then?
Well, x' is no longer independent of time, which should be painfully obvious
to anyone who has ever studied physics. Look at time t. The system has now
moved v*t along the positive x-axis. Thus a point at rest in k which was
originally at (x, y, z)=(x1, y1, z1) is now at (x, y, z)=(x1+v*t, y1, z1).
The coordinate system of your choice, (x', y, z), then places this same
point in (x', y, z)=(x1+2v*t, y1, z1) which is by no means independent of
time. This is the reasoning behind choosing x'=x-vt, since this effectively
eliminates the time dependence of the x' coordinate.
> | For this to be true you would have to invert
> | the direction of movement of k along the x-axis, meaning that you new v'
> | = -v, in which case you end up with the same expression that you get
> | choosing x'=x-vt.
>
> Are you telling me that the RHS of the equation which you carefuly snipped
> to deliberately induce confusion doesn't change?
I did not snip anything to induce confusion. If anything, I am trying very
hard to avoid confusing you. But yes, I am saying that the entire equation
stays the same.
> | Sounds like you have not understood the fundamental
> | reasoning behind the choice of x'=x-vt. It is not arbitrary, given the
> | initial movement of k.
>
> Is that what is "sounds like"?
> It sounds to me as if you are guesser and a snipper, ignoring what you
have
> no answer for, and certainly you have not understood the fundamental
> reasoning behind (2/3 + 1/3)/2 = 1/3, or that cars may be driven forwards
as
> well as backwards, in which case (2/3 + 1/3)/2 = 2/3.
I have throughout this "debate" tried to maintain a sober tone of writing.
You are the one who has ignored several of my questions, and who has now
lowered yourself to name-calling. I have clearly indicated where I snipped
your original text, and it has been done only to ensure that any casual
readers could more easily find their way around the posts. I even indicated
that I snipped intermediate steps in your derivation, to make it sure that
noone believed that you jumped even faster to your false conclusion.
As for the rest of this sentence...(2/3+1/3)=1/3? I am hoping that this is
either a joke or a typo. In either case, the entire thing makes no sense.
> Or am I discussing special relativity with a troll?
> Androcles
I have had just a single course on special relativity in my physics
education, and since then only met it briefly in various other courses.
Special relativity isn't really that vital in the branch of physics I have
chosen. However, even with that slight introduction to the field, I feel
quite capable of discussing the theory with you. I wonder if you have any
physics background at all. The notion of a non-time-dependent coordinate
system (x', y', z') = (x' - vt, y', z') is not something related solely to
special relativity. If you have not caught on to the fundamental reasoning
behind this choice of coordinates, then you would have had a hard time
grasping basic mechanics as well.
I think if you could swallow enough of your pride you might be able to
actually learn something from all of this.
/ Jesper P
Richard Henry
"Ballisticus" <B@..> wrote in message
news:q8sce0521crlt0kt2...@4ax.com...
>
> Is (-^2) = (+) just another convention?
No. It's algebra.
Ahmed Ouahi, Architect
.................... ...In recent years logicians and semanticists
have carried out a very thorough analysis of the symbols, in terms of which
men do their thinking. Linguistics has become a science, and one may even
study a subject to which the late Benjamin Whorf gave the name of
meta-linguistics. All this is greatly to the good; but it is not enough.
Logic and semantics, linguistics and meta-linguistics--these are purely
intellectual disciplines. They analyse the various ways, correct and
incorrect, meaningful and meaningless, in which words can be related to
things, processes and events. But they offer no guidance, in regard to the
much more fundamental problem of the relationship of man in his
psychophysical totality, on the one hand, and his two worlds, of data and of
symbols, on the other.
In every region and at every period of history, the problem has
been repeatedly solved by individual men and women. Even when they spoke or
wrote, these individuals created no systems--for they knew that every system
is a standing temptation to take symbols too seriously, to pay more
attention to words than to the realities for which the words are supposed to
stand. Their aim was never to offer ready-made explanations and panaceas; it
was to induce people to diagnose and cure their own ills, to get them to go
to the place where man's problem and its solution present themselves
directly to experience!!!!!!!!!!!!!!!!!!!!!!!!.......................
...
-- Aldous Huxley
--
Ahmed Ouahi, Architect
Best Regards!
"Jesper Pedersen" <jes...@befunk.com> wrote in message
news:cc66ej$ia7$1...@news.net.uni-c.dk...
Androcles
I only post my own material to sci.physics.relativity. When somebody I'm
responding to includes sci.physics and sci.math, I have no idea which the
originator of the message is monitoring. You have sent your missive to
sci.physics.relativity,sci.physics,sci.math, and that is where I'm replying.
I do not know (or particularly care) who changed the cross posting to three
newsgroups, but you have perpetuated it as much as I. If that makes no sense
to you, then try thinking about it.
| The reason I responded
| to your message was that I wanted to help you out in understanding the
| definitions of sqrt(x). And though I certainly understand that newsgroups
| are not scientific papers, I would expect that if someone wants to
challenge
| special relativity they should at least make clear all the specifics of
| their derivations.
That has been done frequently and often. If you respond to my posts, I
expect you to understand that I'm responding primarily to
sci.physics.relativity, whereas your claim to monitor only sci.physics has
lead you to post to sci.physics.relativity and sci.math as well. Why did you
do that? Same reason as I, I'm sure.
| And I am familiar with Einstein's work on special
| relativity.
If that had any ring of truth to it, you might have recognised the passage
I quoted.
| Seems like you have a bit of learning to do though.
Seems like you have a bit of lying to do.
| "Einstein's
| system k" for one makes no sense at all.
For someone that cannot figure out which newsgroup he is posting to, I doubt
very much anything makes sense to you.
| In which direction is it moving, IS
| it moving at all, is it accelerated, etc? These are quite vital specifics.
Read Einstein, since you claim you are "familiar with Einstein's work on
special relativity", but are clearly not.
|
| > So we have v > 0
| > | and k moving with constant velocity v along the x-axis.
| >
| > Very good.
|
| Thank you.
|
| > | <snip intermediates>
| > | > See that "IF" at the beginning?
| > | > Does that mean
| > | > xi really is equal to (x-vt)*gamma,
| > | > or does it depend on the "If"?
| > | > What IF we place x' = x+vt?
| > | > Androcles
| > |
| > | If you choose x'=x+vt, then a point at rest in system k no longer has
| > values
| > | x', y, z independent of time.
| >
| > What does it have values of, then?
|
| Well, x' is no longer independent of time, which should be painfully
obvious
| to anyone who has ever studied physics.
You really are arrogant, aren't you?
| Look at time t. The system has now
| moved v*t along the positive x-axis.
Well, well. So it has. How painfully obvious.
| Thus a point at rest in k which was
| originally at (x, y, z)=(x1, y1, z1) is now at (x, y, z)=(x1+v*t, y1, z1).
| The coordinate system of your choice,
Einstein's choice.
(x', y, z), then places this same
| point in (x', y, z)=(x1+2v*t, y1, z1) which is by no means independent of
| time.
Really?
| This is the reasoning behind choosing x'=x-vt, since this effectively
| eliminates the time dependence of the x' coordinate.
Explain "effectively". Here, I'll give you a diagram. One that should be
painfully obvious to anyone who has ever studied physics.
t0 x'------------x---x''-------
t1 ----x'--------x-------x''---
t2 ---------x'---x------------x''
> vt >
Which is also
t0 ---------x'------------x---x''
t1 ---------x'--------x-------x''
t2 ---------x'---x------------x''
> vt >
Now tell me how choosing x' = x-vt makes it "effectively" independent of
time and chosing x'' = x+vt does not.
|
| > | For this to be true you would have to invert
| > | the direction of movement of k along the x-axis, meaning that you new
v'
| > | = -v, in which case you end up with the same expression that you get
| > | choosing x'=x-vt.
| >
| > Are you telling me that the RHS of the equation which you carefuly
snipped
| > to deliberately induce confusion doesn't change?
|
| I did not snip anything to induce confusion. If anything, I am trying very
| hard to avoid confusing you. But yes, I am saying that the entire equation
| stays the same.
Really. Proof, please.
稼tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))"
is the same as
稼tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c+v))"
because
tau(x',0,0,t+x'/(c+v)) = tau(x',0,0,t+x'/(c-v))
Yes, I suppose it is if v = 0, which is trivial.
Conclusion: special relativity is valid for all values of v that are equal
to zero.
|
| > | Sounds like you have not understood the fundamental
| > | reasoning behind the choice of x'=x-vt. It is not arbitrary, given the
| > | initial movement of k.
| >
| > Is that what is "sounds like"?
| > It sounds to me as if you are guesser and a snipper, ignoring what you
| have
| > no answer for, and certainly you have not understood the fundamental
| > reasoning behind (2/3 + 1/3)/2 = 1/3, or that cars may be driven
forwards
| as
| > well as backwards, in which case (2/3 + 1/3)/2 = 2/3.
|
| I have throughout this "debate" tried to maintain a sober tone of writing.
"Sounds like you have not understood..." is arrogance, not "tried to
maintain a sober tone". If you want to discuss physics, keep the personal
crap out of it, sonny, or you'll find I retaliate in like manner. Try a
little harder. "x' is no longer independent of time, which should be
painfully obvious to anyone who has ever studied physics". Who do you think
you are bamboozling?
| You are the one who has ignored several of my questions, and who has now
| lowered yourself to name-calling.
What "name" have I called you? You were the first to use the word "troll".
| I have clearly indicated where I snipped
| your original text, and it has been done only to ensure that any casual
| readers could more easily find their way around the posts. I even
indicated
| that I snipped intermediate steps in your derivation,
What do you mean, MY derivation?
| to make it sure that
| noone believed that you jumped even faster to your false conclusion.
Well done. You've made it very clear that you consider Einstein's false
conclusion was jumped to. Make sure you tell that to "noone" so they'll
believe you when you say are "familiar with Einstein's work on special
relativity".
|
| As for the rest of this sentence...(2/3+1/3)=1/3? I am hoping that this is
| either a joke or a typo. In either case, the entire thing makes no sense.
But
稼tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
does, I suppose, since you are "familiar with Einstein's work on special
relativity".
|
| > Or am I discussing special relativity with a troll?
| > Androcles
|
| I have had just a single course on special relativity in my physics
| education, and since then only met it briefly in various other courses.
| Special relativity isn't really that vital in the branch of physics I have
| chosen.
It isn't vital to anything, my friend, unless poppycock is vital.
| However, even with that slight introduction to the field, I feel
| quite capable of discussing the theory with you.
We'll see.
| I wonder if you have any
| physics background at all.
Back to the personal crap again. You are boring me with your arrogance.
I wonder if you have any background at all.
| The notion of a non-time-dependent coordinate
| system (x', y', z') = (x' - vt, y', z') is not something related solely to
| special relativity.
It isn't related to anything at all.
Since when did x' = x'-vt? When t = 0 or v = 0, I suppose. Heck, if you
can't even get your notation right, it's little wonder nothing makes sense
to you. Spelling mistakes I can tolerate, but that (X, y', z') = (X - vt,
y', z') is a monumental gaff.
| If you have not caught on to the fundamental reasoning
| behind this choice of coordinates, then you would have had a hard time
| grasping basic mechanics as well.
Personal crap again? Blustering gets you nowhere.
|
| I think if you could swallow enough of your pride you might be able to
| actually learn something from all of this.
Good advice. Take it yourself. Start with losing some of that arrogance, cut
out the personal crap and maintain a sober tone of writing.
Androcles.
| / Jesper P
Dirk Van de moortel
"Androcles" <andr...@nospamblueyonder.co.uk> wrote in message news:TJyFc.2789$WY.24...@news-text.cableinet.net...
>
> "Jesper Pedersen" <jes...@befunk.com> wrote in message
> news:cc66ej$ia7$1...@news.net.uni-c.dk...
[snip]
> | I think if you could swallow enough of your pride you might be able to
> | actually learn something from all of this.
>
> Good advice. Take it yourself. Start with losing some of that arrogance, cut
> out the personal crap and maintain a sober tone of writing.
Yes, a sober, less arrogant tone like
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Androrgasm.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/LoadCrap.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/CrapHuh.html
Dirk Vdm
Androcles
"Ballisticus" <B@..> wrote in message
news:q8sce0521crlt0kt2...@4ax.com...
Yes. "Well-defined" is the term to use.
Dirk Van de moortel
"Androcles" <andr...@nospamblueyonder.co.uk> wrote in message news:k0zFc.2804$fU.24...@news-text.cableinet.net...
>
> "Ballisticus" <B@..> wrote in message
> news:q8sce0521crlt0kt2...@4ax.com...
> | On Thu, 01 Jul 2004 12:01:20 GMT, "Androcles"
> | <andr...@nospamblueyonder.co.uk> wrote:
> |
> | >
> | >"Jesper Pedersen" <jes...@befunk.com> wrote in message
> | >news:cc0hvk$uf8$1...@news.net.uni-c.dk...
[snip]
> | >| What does that have to do with sqrt(x) being positive by definition?
> | >
> | >Well, you seem to be following some rule I'm not aware of.
> | >I'm trying to figure out what it is.
> | >If -2 is a sqrt of 4, then sqrt(4) * sqrt(4)= 4
> | >and also -sqrt(4) * -sqr(4) = 4.
> | >What's nearly wrong with that?
> |
> | sqrt(4)*sqrt(4) can also be -4
> |
> | You simply multiply the negative root by the positive one.
> Yep.
hm.... that argument turned out less devastating than I thought.
So it's not the Battle of the Village Idiots...
it's a Conference of Village Idiots:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Nearly.html
Dirk Vdm
Jesper Pedersen
>
<snip>
> | And I am familiar with Einstein's work on special
> | relativity.
>
> If that had any ring of truth to it, you might have recognised the passage
> I quoted.
>
>
> | Seems like you have a bit of learning to do though.
>
> Seems like you have a bit of lying to do.
>
>
> | "Einstein's
> | system k" for one makes no sense at all.
>
> For someone that cannot figure out which newsgroup he is posting to, I
doubt
> very much anything makes sense to you.
>
> | In which direction is it moving, IS
> | it moving at all, is it accelerated, etc? These are quite vital
specifics.
>
> Read Einstein, since you claim you are "familiar with Einstein's work on
> special relativity", but are clearly not.
For someone who continuously acuses me of arrogance and bad manners, you
sure do mount a lot of personal attacks.
No, Einstein's choice is x-v*t. You are the one going on about choosing
x+v*t instead.
> (x', y, z), then places this same
> | point in (x', y, z)=(x1+2v*t, y1, z1) which is by no means independent
of
> | time.
>
> Really?
Yes, really.
> | This is the reasoning behind choosing x'=x-vt, since this effectively
> | eliminates the time dependence of the x' coordinate.
>
> Explain "effectively". Here, I'll give you a diagram. One that should be
> painfully obvious to anyone who has ever studied physics.
> t0 x'------------x---x''-------
> t1 ----x'--------x-------x''---
> t2 ---------x'---x------------x''
> > vt >
How is it that you think that the translation of the x' and the x''
coordinate is in the same direction, even though they are given as x+/- v*t
respectively?
> Which is also
> t0 ---------x'------------x---x''
> t1 ---------x'--------x-------x''
> t2 ---------x'---x------------x''
> > vt >
> Now tell me how choosing x' = x-vt makes it "effectively" independent of
> time and chosing x'' = x+vt does not.
Because x' = x(t)-vt gives, since x(t)=x(0)+vt, that x' = x(0)+vt-vt = x(0),
which is time-independent. Obviously, choosing x'' = x+vt does not yield the
same result.
> | > | For this to be true you would have to invert
> | > | the direction of movement of k along the x-axis, meaning that you
new
> v'
> | > | = -v, in which case you end up with the same expression that you get
> | > | choosing x'=x-vt.
> | >
> | > Are you telling me that the RHS of the equation which you carefuly
> snipped
> | > to deliberately induce confusion doesn't change?
> |
> | I did not snip anything to induce confusion. If anything, I am trying
very
> | hard to avoid confusing you. But yes, I am saying that the entire
equation
> | stays the same.
>
> Really. Proof, please.
> 稼tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))"
> is the same as
> 稼tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c+v))"
> because
> tau(x',0,0,t+x'/(c+v)) = tau(x',0,0,t+x'/(c-v))
> Yes, I suppose it is if v = 0, which is trivial.
> Conclusion: special relativity is valid for all values of v that are equal
> to zero.
You misunderstood what I said. If you go through the deriviation, choosing
v<0 rather than v>0 and then x' = x+vt instead of x' = x-vt, then it is
quite obvious that you arrive at the same result.
> | > | Sounds like you have not understood the fundamental
> | > | reasoning behind the choice of x'=x-vt. It is not arbitrary, given
the
> | > | initial movement of k.
> | >
> | > Is that what is "sounds like"?
> | > It sounds to me as if you are guesser and a snipper, ignoring what you
> | have
> | > no answer for, and certainly you have not understood the fundamental
> | > reasoning behind (2/3 + 1/3)/2 = 1/3, or that cars may be driven
> forwards
> | as
> | > well as backwards, in which case (2/3 + 1/3)/2 = 2/3.
> |
> | I have throughout this "debate" tried to maintain a sober tone of
writing.
>
>
<snip personal stuff>
Let's keep this to a minimum.
> | to make it sure that
> | noone believed that you jumped even faster to your false conclusion.
>
> Well done. You've made it very clear that you consider Einstein's false
> conclusion was jumped to. Make sure you tell that to "noone" so they'll
> believe you when you say are "familiar with Einstein's work on special
> relativity".
No, I made it very clear that I believe that you jumped to a wrong
conclusion regarding the choice of x'.
<snip more personal stuff>
> | The notion of a non-time-dependent coordinate
> | system (x', y', z') = (x' - vt, y', z') is not something related solely
to
> | special relativity.
>
> It isn't related to anything at all.
> Since when did x' = x'-vt? When t = 0 or v = 0, I suppose. Heck, if you
> can't even get your notation right, it's little wonder nothing makes sense
> to you. Spelling mistakes I can tolerate, but that (X, y', z') = (X - vt,
> y', z') is a monumental gaff.
Yes. That was an obvious typo.
<snip remainding personal stuff>
I am not gonna take part in this discussion anymore, since it is clear that
it will get neither of us anywhere.
/ Jesper P
Androcles
| "Androcles" <andr...@nospamblueyonder.co.uk> wrote in message
| news:TJyFc.2789$WY.24...@news-text.cableinet.net...
| >
| <snip>
| > | And I am familiar with Einstein's work on special
| > | relativity.
| >
| > If that had any ring of truth to it, you might have recognised the
passage
| > I quoted.
| >
| >
| > | Seems like you have a bit of learning to do though.
| >
| > Seems like you have a bit of lying to do.
| >
| >
| > | "Einstein's
| > | system k" for one makes no sense at all.
| >
| > For someone that cannot figure out which newsgroup he is posting to, I
| doubt
| > very much anything makes sense to you.
| >
| > | In which direction is it moving, IS
| > | it moving at all, is it accelerated, etc? These are quite vital
| specifics.
| >
| > Read Einstein, since you claim you are "familiar with Einstein's work on
| > special relativity", but are clearly not.
|
| For someone who continuously acuses me of arrogance and bad manners, you
| sure do mount a lot of personal attacks.
I respond in kind. Note the operative word is "respond". In this case, I was
merely echoing your own words. If you find that to be "a lot of personal
attacks" look to yourself.
So the "If" isn't necessary then?
Simply dictate x' = x-vt and have done with it.
|
| > (x', y, z), then places this same
| > | point in (x', y, z)=(x1+2v*t, y1, z1) which is by no means independent
| of
| > | time.
| >
| > Really?
|
| Yes, really.
Assertion carries no weight, sorry. You'll have to prove it.
|
| > | This is the reasoning behind choosing x'=x-vt, since this effectively
| > | eliminates the time dependence of the x' coordinate.
| >
| > Explain "effectively". Here, I'll give you a diagram. One that should be
| > painfully obvious to anyone who has ever studied physics.
| > t0 x'------------x---x''-------
| > t1 ----x'--------x-------x''---
| > t2 ---------x'---x------------x''
| > > vt >
|
| How is it that you think that the translation of the x' and the x''
| coordinate is in the same direction, even though they are given as x+/-
v*t
| respectively?
Egads. Oh well, I'll try to answer. x' is approaching x and x'' is receding
from x. They remain the same distance from each other. Eventually, x' will
pass x, still going in the same "direction" by the definition of direction,
as the diagram shows. Maybe this will help. I'll simplify it for you, since
you seem to be confused.
t0 x'-------------x
t1 ----x'---------x
t2 ---------x'----x
t3 -------------x'x
t4 ---------------x-x'
t5 ---------------x-----x'
t6 ---------------x----------x'
Please notice that the distance between x and x' is given by vt.
In the line that begins with "t6", it is more than it is on the line marked
"t5".
It has increased by (t6-t5)*v.
This is dependent on time, and we write it as x' = x+vt.
Now, reversing (that means coming the other way),
t7 ---------------x----------x'
t8 ---------------x-----x'
t9 ---------------x-x'
The distance between x and x' has decreased.
So for this case (of reversal) we write x' = x-vt, which is now miraculously
independent of time.
| > Which is also
| > t0 ---------x'------------x---x''
| > t1 ---------x'--------x-------x''
| > t2 ---------x'---x------------x''
| > > vt >
| > Now tell me how choosing x' = x-vt makes it "effectively" independent
of
| > time and chosing x'' = x+vt does not.
|
| Because x' = x(t)-vt gives, since x(t)=x(0)+vt, that x' = x(0)+vt-vt =
x(0),
| which is time-independent. Obviously, choosing x'' = x+vt does not yield
the
| same result.
x' = x(0)?
I don't follow, unless v = 0, of course. Did you wish to discuss the trivial
case then?
Hmm... let's see if we can resolve it. vt is a distance, right? vt is the
distance between the points x and x'. Since these two points are in relative
motion, this distance can become zero, but that isn't independent of time.
Why does it matter that x' is the left of x to be time independent, but if
it is to the right it is time dependent?
I can agree that x is time independent, even arbitrarily set it to 1 so
that x(0) =1, x(1) = 1, x(2) = 1, but x' = 1 - vt or x' = 1+vt takes a
strong imagination to be independent of time. Then again, we COULD make x'
fixed and independent of time, but then makes x dependent, doesn't it?
Surely if x' = x-vt, then x = x'+vt?
Quite possibly.
| If you go through the deriviation, choosing
| v<0 rather than v>0 and then x' = x+vt instead of x' = x-vt, then it is
| quite obvious that you arrive at the same result.
I have to stubbornly disagree. I have been through the derivation, and it is
quite obvious that you arrive at a different result.
IF x' = x-vt, THEN xi = (x-vt) / sqrt(1-v^2/c^2)
IF x' = x+vt, THEN xi = (x+vt) / sqrt(1-v^2/c^2)
It would seem that your definition of "quite obvious" differs from mine.
In support of my position, I'll add the words of Assistant Professor
Andersen, and quote:
"Through previous very long-winded posts,
That is, we can reverse the directions of the frames
which is the same as interchanging the frames,
which - as I have told you a LOT of times,
OBVIOUSLY will lead to the transform:
t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2)
x = (xi - v*tau)/sqrt(1-v^2/c^2)
or:
tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
xi = (x + vt)/sqrt(1-v^2/c^2)" -Paul B. Andersen
See, he likes to use "obviously" as well. He emphasizes it with capitals.
Now, why not be a good chap and go through the derivation CAREFULLY
before you claim the "quite obvious". Who knows, you might learn something.
Check this out:
"An analogous consideration--applied to the axes of Y and Z--it being borne
in mind that light is always propagated along these axes, when viewed from
the stationary system, with the velocity sqrt(c^2-v^2)"- Albert Einstein.
http://www.fourmilab.ch/etexts/einstein/specrel/www/
The velocity of light is ALWAYS positive, according to Dinky the Deranged
and your good self, because sqrt is defined as positive. Rotating
lighthouses don't happen.
|
| > | > | Sounds like you have not understood the fundamental
| > | > | reasoning behind the choice of x'=x-vt. It is not arbitrary, given
| the
| > | > | initial movement of k.
| > | >
| > | > Is that what is "sounds like"?
| > | > It sounds to me as if you are guesser and a snipper, ignoring what
you
| > | have
| > | > no answer for, and certainly you have not understood the fundamental
| > | > reasoning behind (2/3 + 1/3)/2 = 1/3, or that cars may be driven
| > forwards
| > | as
| > | > well as backwards, in which case (2/3 + 1/3)/2 = 2/3.
| > |
| > | I have throughout this "debate" tried to maintain a sober tone of
| writing.
| >
| >
| <snip personal stuff>
| Let's keep this to a minimum.
Good. I totally agree. Cut out the trite comments such as wondering about my
background and we'll get along fine.
|
| > | to make it sure that
| > | noone believed that you jumped even faster to your false conclusion.
| >
| > Well done. You've made it very clear that you consider Einstein's false
| > conclusion was jumped to. Make sure you tell that to "noone" so they'll
| > believe you when you say are "familiar with Einstein's work on special
| > relativity".
|
| No, I made it very clear that I believe that you jumped to a wrong
| conclusion regarding the choice of x'.
I ask for proof.
|
| <snip more personal stuff>
| > | The notion of a non-time-dependent coordinate
| > | system (x', y', z') = (x' - vt, y', z') is not something related
solely
| to
| > | special relativity.
| >
| > It isn't related to anything at all.
| > Since when did x' = x'-vt? When t = 0 or v = 0, I suppose. Heck, if you
| > can't even get your notation right, it's little wonder nothing makes
sense
| > to you. Spelling mistakes I can tolerate, but that (X, y', z') = (X -
vt,
| > y', z') is a monumental gaff.
|
| Yes. That was an obvious typo.
Okay, accepted.
|
| <snip remainding personal stuff>
|
| I am not gonna take part in this discussion anymore, since it is clear
that
| it will get neither of us anywhere.
Ok, accepted.
Because sqrt(c^2-v^2) is always positive, rotating lighthouses are
impossible.
Androcles
| / Jesper P
|
|
Androcles
"I am not gonna take part in this discussion anymore, since it is clear that
"I think if you could swallow enough of your pride you might be able to
actually learn something from all of this. - / Jesper P.
Androcles
Ballisticus
Can you prove logically, from basics, that (-A)*(-B)=+AB?
Please don't make a gigantic fumble...
Dik T. Winter
...
> Can you prove logically, from basics, that (-A)*(-B)=+AB?
Yup. Start at the ring axioms and it is easy.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Mitchell
Maybe jesper saw you for who you are Androcles.
Mitch Raemsch
-- Light's still falling --
Ballisticus
>In article <2deee0ldfhdfnc6s4...@4ax.com> B@the.. writes:
>...
> > Can you prove logically, from basics, that (-A)*(-B)=+AB?
>
>Yup. Start at the ring axioms and it is easy.
Oh gord! You're not related to dickhead Van de moortel, I hope!
Androcles
"Mitchell" <macro...@internetCDS.com> wrote in message
news:9c3da975.04070...@posting.google.com...
Maybe you should try swallowing a little of your own pride, Mitch. You might
learn something.
Androcles
Bilge
>> special relativity", but are clearly not.
>
>For someone who continuously acuses me of arrogance and bad manners, you
>sure do mount a lot of personal attacks.
concept of a coordinate transformation despite numerous attempts
from many people probably going back many years. He calls everyone
who can perform simple arithemetic operations in their head, ``arrogant''
and whatever else helps divert attention from the point he can't make.
It's not worth the effort to take him seriously. His logic is about
on par with the logic of someone who would set up a 1-900 number for
palm readings.
Androcles
"Ballisticus" <B@..> wrote in message
Ease up, H. You are getting out of your depth and that wasn't called for.
Nobody could come close to the stupidity of Dinky the Deranged with the age
of sqrt(100) and the IQ of sqrt(9).
There are any number of ways to prove a given statement, but all depend on
primitive axioms which are accepted without proof.
Consider a lever balance, the fulcrum of which is at 0, the left pan at -x
and the right pan at +x.
It is impossible to have two apples of equal mass. When you place them on
the balance, the turning moment of the negative apple cancels the turning
moment of the positive apple by
-x * -m = xm
Hence the apple on the left has negative mass and the apple on the right has
positive mass. :-)
Androcles.
Mitchell
He who humbles himself shall be made great.
Maybe you should swallow God. Then you would know you are nothing Androcles.
Mitch Raemsch -- Light Falls --
MorituriMax
> He who makes himself great shall be humbled.
> He who humbles himself shall be made great.
yada yada cliched sayings
> Maybe you should swallow God. Then you would know you are nothing Androcles.
Maybe you should do something useful with your time..
Ballisticus
Any SRian would believe you.
>
>Androcles.
Eric Gisse
> | <snip>
> | > Yes, very probably, but remember that
> | > "3 = sqrt(9)
> | > -3 = -sqrt(9)" -Dinky the Deranged.
> | > "This way sqrt is ALWAYS positive."-Dinky the Deranged.
> | > and
> | > "IF x <= 0 THEN sqrt(x^2) = -x" - Dinky the Deranged
> | > is not written in the form -3 = -sqrt(9). There is a missing minus sign
> | > after the word "THEN".
> | > Perhaps -x = -sqrt(x^2) is equivalent to sqrt(x^2) = -x
> | > because x is on the other side of the '='?
> | > Maybe we should ask Dinky.
> | <snip more commented quotes>
> |
> | You need to understand the fundamental difference between solving x^2=2
> for
> | x
>
> I think the solution is -1.4142135623730950488016887242097.
> Let's see.
> -1.4142135623730950488016887242097 *
> -1.4142135623730950488016887242097 = 2.
sqrt(2) is irrational. Carrying out to all those decimal places does
nothing to prove your point.
> Yep, I'm right.
>
> and applying the function sqrt(2). the sqrt is DEFINED as being positive.
>
> Is it?
> if x = [-b +/- sqrt(b^2 - 4ac]/2a, is b always positive?
Quadratic equation, moron. Learn to use it.
x^2 = 2
x^2 - 2 = 0
a = 1
b = ?
c = -2
What is b? It's blindingly obvious that you haven't had any math past
algebra in high school.
>
> | Otherwise how would equalities such as sin(Pi/4) = sqrt(2)/2 make any
> sense
> | at all? Surely you wouldn't claim that sin is dual-valued?
> That rather depends on whether you use (x, iy) or (x,y), doesn't it?
> Androcles.
You know nothing of complex numbers.
>
> |
> | / Jesper P
> |
> |
Dirk Van de moortel
"Eric Gisse" <fs...@uaf.edu> wrote in message news:fd0fc2fa.04070...@posting.google.com...
You'd be surprized what his point is going to be.
Watch out...
> > Yep, I'm right.
> >
> > and applying the function sqrt(2). the sqrt is DEFINED as being positive.
> >
> > Is it?
> > if x = [-b +/- sqrt(b^2 - 4ac]/2a, is b always positive?
>
> Quadratic equation, moron. Learn to use it.
>
> x^2 = 2
> x^2 - 2 = 0
>
> a = 1
> b = ?
> c = -2
>
> What is b? It's blindingly obvious that you haven't had any math past
> algebra in high school.
I tried this approach (message 59):
http://groups.google.com/groups?&threadm=lwfDc.166830$oQ6.8...@phobos.telenet-ops.be
Just have a look at his reply (message 60)
Nothing like the Smell of Andropalm in the Morning :-)
Dirk Vdm
Androcles
| > Yep, I'm right.
| >
| > and applying the function sqrt(2). the sqrt is DEFINED as being
positive.
| >
| > Is it?
| > if x = [-b +/- sqrt(b^2 - 4ac]/2a, is b always positive?
|
| Quadratic equation, moron.
*plonk* the name caller who couldn't answer the question.
Androcles
The Ghost In The Machine
<fs...@uaf.edu>
wrote
on 5 Jul 2004 05:23:32 -0700
<fd0fc2fa.04070...@posting.google.com>:
An easy proof:
Assume sqrt(2) = a/b, with a and b relatively prime integers.
Since a^2/b^2 = 2, a has to be even. Write a = 2*c.
c^2/b^2 = 1/2; b^2/c^2 = 2; b must therefore be even.
Contradiction.
Of course the OP should be aware that multiplication of the
numeric approximations (m * 2^e) is done approximately,
so it's entirely possible that the above approximate
multiplication result is in fact 2. A rather naive
implementation of this issue, with loss of precision
ramifications, leads to 3.11 - 3.10 = 0.00 in a rather
famous case involving a Windows calculator. Fortunately,
it has since been fixed -- but it took them a very long
time (probably because it wasn't all that high on their
priority list).
I got bitten by a variant of the approximation bug once,
at work. The results weren't horribly pretty.
It also turns out that, when represented in hex (standard IEEE754 stuff):
3.11: 3.11 = 4008e147ae147ae1
3.10: 3.1 = 4008cccccccccccd
3.11 - 3.10: 0.01 = 3f847ae147ae1400
(note that sprintf() is doing it right!)
but
1/100: 0.01 = 3f847ae147ae147b
Definite loss of precision there.
Also:
M_SQRT2: 1.41421 = 3ff6a09e667f3bcd
(M_SQRT2 is #defined in math.h)
and
sqrt(2): 1.41421 = 3ff6a09e667f3bcd
which means these two numbers are equal in all relevant respects,
but
sqrt(2)^2: 2 = 4000000000000001
which means the guard bit let its guard down. :-) (This on an Intel x86
micro.)
BTW, the OP's number is just a smidge off:
$ gp
GP/PARI CALCULATOR Version 2.1.5 (released)
...
? default("realprecision",200)
realprecision = 202 significant digits (200 digits displayed)
? default("format","g0.200")
format = g0.200
? sqrt(2)
%1 = 1.41421356237309504880168872420969807...
^
but then, any finite decimal representation of sqrt(2)
will be a smidge off, for some value of "smidge". :-)
I also note that
1.4142135623730950488016887242097^2 = 2.0000000000000000000
00000000000005434625658377656176426889574090000000000000000...
so the smidge has bitten again. (They're nasty little math bugs,
aren't they? :-) )
>
>
>> Yep, I'm right.
>>
>> and applying the function sqrt(2). the sqrt is DEFINED as being positive.
>>
>> Is it?
>> if x = [-b +/- sqrt(b^2 - 4ac]/2a, is b always positive?
>
> Quadratic equation, moron. Learn to use it.
>
> x^2 = 2
> x^2 - 2 = 0
>
> a = 1
> b = ?
> c = -2
>
> What is b? It's blindingly obvious that you haven't had any math past
> algebra in high school.
And maybe not even that; the quadratic solution, after all, can
be derived by "completing the square":
ax^2 + bx + c = 0
x^2 + bx/a + c/a = 0
x^2 + bx/a + b^2/(4a^2) + c/a = b^2/(4a^2)
x^2 + bx/a + b^2/(4a^2) = b^2/(4a^2) - c/a = (b^2 - 4ac)/(4a^2)
Recall that (c + d)^2 = c^2 + 2cd + d^2; therefore the left side
is rewritable:
(x + b/(2a))^2 = (b^2 - 4ac)/(4a^2)
Carefully taking square roots, one gets:
x + b/(2a) = +/- sqrt(b^2 - 4ac) / (2a)
and the formula follows.
An alternative, slightly more logical, derivation, might involve
x^2 + bx/a + c/a = 0
x^2 + bx/a = -c/a
and then asking the question "How do I complete the square here?".
Algebra, IIRC, is taught in 8th-9th grade level, so this wouldn't be
that much of a stretch.
>
>
>>
>> | Otherwise how would equalities such as sin(Pi/4) = sqrt(2)/2 make any
>> sense
>> | at all? Surely you wouldn't claim that sin is dual-valued?
>> That rather depends on whether you use (x, iy) or (x,y), doesn't it?
>> Androcles.
>
> You know nothing of complex numbers.
(x, iy)? Somebody's extremely confused. The usual representations
are (x,y) (or, if one prefers, (Re(z), Im(z))) or x + iy = Re(z) + i*Im(z).
Or perhaps Re(z) + Im(z); I don't remember offhand whether Im(z) returns
a multiple of sqrt(-1) or not. (I doubt it.)
Of course completing the square works even for equations with
b^2 - 4ac < 0; the result is, of course, a complex number.
x^2 + 1 = 0 in particular has a = 1, b = 0, c = 1; b^2 - 4ac = -4;
(-b + sqrt(b^2 - 4ac))/(2a) = +sqrt(-1) = i.
(The other root is of course -i.)
>
>>
>> |
>> | / Jesper P
>> |
>> |
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.
Eric Gisse
[snip]
I can see all this math misunderstanding coming full circle to his
great disproof of relativity using idiotic notions about the square
root expression.