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| Tim Smith |
 Posted: Sat Feb 19, 2005 8:01 pm |
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In article <37pqbeF5e9jtaU1@individual.net>, Schweinkolben wrote:
[quote:080c8cc92f]TROLL
[/quote:080c8cc92f]
There's no need to keep saying that. We've all figured it out long ago.
--
--Tim Smith |
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| guenther.vonKnakspott@gmx |
| Posted: Sun Feb 20, 2005 2:28 am |
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jstevh@msn.com wrote:
[quote:71a013251a]Nevertheless, accepted or not, it's wrong.
Here's another example:
z = sqrt(x) - sqrt(y)
and if x=1, and y=1, z = 2 IS in fact, a solution.
Or do any of you disagree?
How many possible solutions are there for z, given x=1, and y=1?
And what are their values?
Just curious.
James Harris
[/quote:71a013251a]
For Christ's sake Harris, before you start ranting about core problems
in mathematics. The DEFINITION of sqrt(x) is the positive real root of
x when x is a positive real. sqrt(x) is a function not an equation.
Don't get stupid about this one, perhaps you should tell us what part
of DEFINITION is confusing you. By the way Harris:
Nihil ex nihilo.
Wake up Harris you have got nothing. |
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| Jesse F. Hughes |
| Posted: Sun Feb 20, 2005 4:32 am |
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jstevh@msn.com writes:
[quote:65292c5e65]Nevertheless, accepted or not, it's wrong.
Here's another example:
z = sqrt(x) - sqrt(y)
and if x=1, and y=1, z = 2 IS in fact, a solution.
Or do any of you disagree?
How many possible solutions are there for z, given x=1, and y=1?
And what are their values?
[/quote:65292c5e65]
There is one solution and it is z = 0.
By definition, sqrt(x) is that *non-negative* real number w such that
w^2 = x. That's what the sqrt function means. It is utterly
unambiguous.
--
Jesse F. Hughes
"My experience indicates that the people who post on this newsgroup
are about at the level of a 10 year old in the year 2060."
-- More wisdom from James Harris, time traveler |
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| Al Bundy |
| Posted: Sun Feb 20, 2005 11:23 am |
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Ah...that pesky sign thing. You, of course refer to "the law of the
signs", which is a two element cyclic group with the plus sign as the
identity element. This should clue you in on two things immediately...
1. The plus sign and the negative sign are not equivalent, hence the
plus axis and negative axis are not equivalent. and..
2. This sign convention is really only correctly applied to linear
problems (even then it works badly...see 1.)....ie its a one
dimensional cyclic group.
On 19 Feb 2005 13:28:06 -0800, jstevh@msn.com wrote:
[quote:033e16250a]What happens with 1 and -1 though is that people *decide* that they are
basically the same number except for this pesky minor sign thing, and
believe you can remove the ambiguity by fiat.[/quote:033e16250a] |
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| Guest |
| Posted: Sun Feb 20, 2005 11:43 am |
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Jesse F. Hughes wrote:
[quote:7af300ec11]jstevh@msn.com writes:
Nevertheless, accepted or not, it's wrong.
Here's another example:
z = sqrt(x) - sqrt(y)
and if x=1, and y=1, z = 2 IS in fact, a solution.
Or do any of you disagree?
How many possible solutions are there for z, given x=1, and y=1?
And what are their values?
There is one solution and it is z = 0.
By definition, sqrt(x) is that *non-negative* real number w such that
w^2 = x. That's what the sqrt function means. It is utterly
unambiguous.
[/quote:7af300ec11]
Well, let's check that answer.
z = sqrt(x) - sqrt(y), so
z + sqrt(y) = sqrt(x), and squaring gives
z^2 + 2z sqrt(y) + y = x
and grouping gives
(z^2 + y - z) = -2z sqrt(y)
and squaring (keeping things simple for now) gives
(z^2 + y - x)^2 = 4z^2 y
so I have a complicated expression there, but luckily I'm just checking
x = y = 1, which gives
z^4 = 4z^2, which is
z^2(z^2 - 4) = 0
so you have *three* solutions, which are z = 0, z=2, and z=-2.
The math is easy, human ignorance is not.
People will argue, argue, argue this point as if by fiat you can
destroy a mathematical absolute, but if you do the math, the real
answer pops up anyway, and the people be damned.
That's what makes math great.
Your opinion doesn't matter.
James Harris |
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| Jim Spriggs |
| Posted: Sun Feb 20, 2005 11:48 am |
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Luc The Perverse wrote:
[quote:56b06adbe4]
i is the positive square root of -1 by definition. Problem solved.
[/quote:56b06adbe4]
i cannot be said to be positive or negative. |
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| Henry |
| Posted: Sun Feb 20, 2005 12:36 pm |
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On Sat, 19 Feb 2005 23:30:55 +0100, David Kastrup <dak@gnu.org> wrote:
[quote:afc4f8ec67]Henry <se16@btinternet.com> writes:
On Sat, 19 Feb 2005 18:56:39 -0000, "Mike Terry"
news.dead.person.stones@darjeeling.plus.com> wrote:
If P ===> Q, this does not mean that Q ===> P.
Indeed not, but what is true is either P ===> Q or Q ===> P.
So which of the two following is true: x odd ===> x>0,
or x>0 ===> x odd ?
[/quote:afc4f8ec67]
For x odd and positive both are true;
for x even and positive the first is true;
for x odd and non-positive the second is true;
for x even and non-positive both are true.
So in every case either the first or the second is true, as I said
originially.  If you don't like that then
JOIN THE CAMPAIGN FOR REPEAL OF THE LAW OF EXCLUDED MIDDLE! |
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| David Kastrup |
| Posted: Sun Feb 20, 2005 12:39 pm |
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jstevh@msn.com writes:
[quote:a92cfb7f03]Jesse F. Hughes wrote:
jstevh@msn.com writes:
Nevertheless, accepted or not, it's wrong.
Here's another example:
z = sqrt(x) - sqrt(y)
and if x=1, and y=1, z = 2 IS in fact, a solution.
Or do any of you disagree?
How many possible solutions are there for z, given x=1, and y=1?
And what are their values?
There is one solution and it is z = 0.
By definition, sqrt(x) is that *non-negative* real number w such that
w^2 = x. That's what the sqrt function means. It is utterly
unambiguous.
Well, let's check that answer.
z = sqrt(x) - sqrt(y), so
z + sqrt(y) = sqrt(x), and squaring gives
z^2 + 2z sqrt(y) + y = x
[/quote:a92cfb7f03]
So what? Squaring turns -1 = 1 into 1 = 1. Does that prove that
-1 = 1?
[quote:a92cfb7f03]The math is easy, human ignorance is not.
[/quote:a92cfb7f03]
You bet. Whoever told you that you could do stuff like squaring an
equation without any effect on the set of possible solutions.
[quote:a92cfb7f03]People will argue, argue, argue this point as if by fiat you can
destroy a mathematical absolute, but if you do the math, the real
answer pops up anyway, and the people be damned.
That's what makes math great.
Your opinion doesn't matter.
[/quote:a92cfb7f03]
So you think -1 = 1?
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum |
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| Jesse F. Hughes |
| Posted: Sun Feb 20, 2005 1:51 pm |
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jstevh@msn.com writes:
[quote:7a4584b48b]Jesse F. Hughes wrote:
jstevh@msn.com writes:
Nevertheless, accepted or not, it's wrong.
Here's another example:
z = sqrt(x) - sqrt(y)
and if x=1, and y=1, z = 2 IS in fact, a solution.
Or do any of you disagree?
How many possible solutions are there for z, given x=1, and y=1?
And what are their values?
There is one solution and it is z = 0.
By definition, sqrt(x) is that *non-negative* real number w such that
w^2 = x. That's what the sqrt function means. It is utterly
unambiguous.
Well, let's check that answer.
z = sqrt(x) - sqrt(y), so
z + sqrt(y) = sqrt(x), and squaring gives
z^2 + 2z sqrt(y) + y = x
and grouping gives
(z^2 + y - z) = -2z sqrt(y)
and squaring (keeping things simple for now) gives
(z^2 + y - x)^2 = 4z^2 y
so I have a complicated expression there, but luckily I'm just checking
x = y = 1, which gives
z^4 = 4z^2, which is
z^2(z^2 - 4) = 0
so you have *three* solutions, which are z = 0, z=2, and z=-2.
[/quote:7a4584b48b]
I want to try to follow your reasoning here. You start with an
equation, square both sides and solve. And then you conclude that the
solutions for the squared equation are also solutions for the original
equation, right?
Let x + 4 = 1. Squaring both sides gives x^2 + 8x + 16 = 1. Thus,
(x + 5)(x + 3) = 0 and hence x = -5 or x = -3.
We may therefore conclude that x = -5 is a solution to x + 4 = 1.
Fascinating reasoning there.
Who told you that the solutions to an equation (f(x))^2 = a^2 are
also solutions to f(x) = a? It's not always true.
[quote:7a4584b48b]The math is easy, human ignorance is not.
[/quote:7a4584b48b]
Well, ignorance comes easier for some than others, I guess.
--
Jesse F. Hughes
"How come there's still apes running around loose and there are
humans? Why did some of them decide to evolve and some did not? Did
they choose to stay as a monkey or what?" -Kans. Board of Ed member |
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| Gib Bogle |
| Posted: Sun Feb 20, 2005 2:16 pm |
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jstevh@msn.com wrote:
[quote:86ae26ace1]Well, let's check that answer.
z = sqrt(x) - sqrt(y), so
z + sqrt(y) = sqrt(x), and squaring gives
z^2 + 2z sqrt(y) + y = x
and grouping gives
(z^2 + y - z) = -2z sqrt(y)
and squaring (keeping things simple for now) gives
(z^2 + y - x)^2 = 4z^2 y
so I have a complicated expression there, but luckily I'm just checking
x = y = 1, which gives
z^4 = 4z^2, which is
z^2(z^2 - 4) = 0
so you have *three* solutions, which are z = 0, z=2, and z=-2.
The math is easy, human ignorance is not.
People will argue, argue, argue this point as if by fiat you can
destroy a mathematical absolute, but if you do the math, the real
answer pops up anyway, and the people be damned.
That's what makes math great.
Your opinion doesn't matter.
[/quote:86ae26ace1]
Here is a clue, James. "A implies B" does not imply that "B implies A".
"a = b" ==> "a^2 = b^2", but "a^2 = b^2" /=> "a = b".
Note that "a = b" and "a = -b" both lead to "a^2 = b^2". Whichever
equality you start from is the one you must return to when you undo the
operation of squaring both sides. (This has echoes of about Form 4,
i.e. age 14, mathematics).
This is how we are able to avoid The End of Mathematics As We Know It
which would result from 1 = -1.
Gib |
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| David Kastrup |
| Posted: Sun Feb 20, 2005 2:27 pm |
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Gib Bogle <bogle@ihug.too.much.spam.co.nz> writes:
[quote:7161c774df]jstevh@msn.com wrote:
x = y = 1, which gives
z^4 = 4z^2, which is
z^2(z^2 - 4) = 0
so you have *three* solutions, which are z = 0, z=2, and z=-2.
The math is easy, human ignorance is not.
People will argue, argue, argue this point as if by fiat you can
destroy a mathematical absolute, but if you do the math, the real
answer pops up anyway, and the people be damned.
That's what makes math great.
Your opinion doesn't matter.
Here is a clue, James. "A implies B" does not imply that "B implies A".
"a = b" ==> "a^2 = b^2", but "a^2 = b^2" /=> "a = b".
Note that "a = b" and "a = -b" both lead to "a^2 = b^2". Whichever
equality you start from is the one you must return to when you undo
the operation of squaring both sides. (This has echoes of about Form
4, i.e. age 14, mathematics).
This is how we are able to avoid The End of Mathematics As We Know
It which would result from 1 = -1.
[/quote:7161c774df]
Of course it would. This is a fundamental deficiency with signs that
has escaped mathematicians over all the centuries since negative
numbers have been invented.
I mean, let's not stop at the fundamental deficiency of algebraic
integers which suffer from the fatal shortcoming that they are only a
subset of the algebraic numbers. Let's also show that the positive
numbers have the fatal deficiency that they are equal to the negative
ones after squaring.
It takes a revolutionary like James to figure out this sort of thing
and correct math books all over the world. Nobody has yet understood
properly that -1 = +1 since (-1)^2 = (+1)^2.
It is a veritable thicket of bad signs that is looming around James,
and he really needs to cut the branches.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum |
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| Jim Spriggs |
| Posted: Sun Feb 20, 2005 3:25 pm |
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[quote:880a2a4ef8]"Jim Spriggs" <jim.sprigs@ANTISPAMbtinternet.com.invalid> wrote in message
news:42193151.FBA85E9D@ANTISPAMbtinternet.com.invalid...
Which demonstrates the problem: the product of two positive numbers
should be positive. So "positive" and "negative" don't apply to complex
numbers (with non-zero real part).
[/quote:880a2a4ef8]
Sorry, I meant "with non-zero imaginary part." |
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| Guest |
| Posted: Sun Feb 20, 2005 4:57 pm |
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In sci.math matt271829-news@yahoo.co.uk wrote:
: And ultimately the problem is that the inverse function of f(x) = x^2
: is not single-valued
This is a contradiciton. By definition a function is "single-valued".
The function f(x)=x^2 does not have an inverse unless x is restricted
to the positive reals.
If you want multi-valued functions you have to consider functions
whose domains are sets of reals. For example you could define
a function f: R -> R x R where f(x)={sqrt(x), -sqrt(x)} (we
would have to modify this a little to handle 0). Of course if
you do that you have to do some work defining what f(x)+f(x)
is, because f(x) is not a set, and it could contain three values.
And what is f(x)*f(x)? That is actually a tricky question.
Stephen |
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| David Kastrup |
| Posted: Sun Feb 20, 2005 5:45 pm |
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stephen@nomail.com writes:
[quote:6f1ed7c9a2]In sci.math matt271829-news@yahoo.co.uk wrote:
: And ultimately the problem is that the inverse function of f(x) = x^2
: is not single-valued
This is a contradiciton. By definition a function is "single-valued".
The function f(x)=x^2 does not have an inverse unless x is restricted
to the positive reals.
[/quote:6f1ed7c9a2]
Or to the negative reals. Or to union of the non-negative rationals
and the negative irrationals.
And a number of other possibilities...
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum |
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| David Kastrup |
| Posted: Sun Feb 20, 2005 5:49 pm |
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Henry <se16@btinternet.com> writes:
[quote:4928467a78]On Sat, 19 Feb 2005 23:30:55 +0100, David Kastrup <dak@gnu.org> wrote:
Henry <se16@btinternet.com> writes:
On Sat, 19 Feb 2005 18:56:39 -0000, "Mike Terry"
news.dead.person.stones@darjeeling.plus.com> wrote:
If P ===> Q, this does not mean that Q ===> P.
Indeed not, but what is true is either P ===> Q or Q ===> P.
So which of the two following is true: x odd ===> x>0,
or x>0 ===> x odd ?
For x odd and positive both are true;
for x even and positive the first is true;
for x odd and non-positive the second is true;
for x even and non-positive both are true.
[/quote:4928467a78]
If? Protector of this damned strumpet, talk'st thou to me of "if"s?
Thou art a traitor, off with his head.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum |
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