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Sign conventions

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Guest

Posted: Sun Feb 20, 2005 5:53 pm

step...@nomail.com wrote:
[quote:dc08d6b3d9]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".[/quote:dc08d6b3d9]

I refuse to deny the existence of multivalued functions, and my
statement is perfectly clear. See e.g.
http://mathworld.wolfram.com/MultivaluedFunction.html.

Luc The Perverse

Posted: Sun Feb 20, 2005 8:10 pm

Guest

"Jim Spriggs" <jim.sprigs@ANTISPAMbtinternet.com.invalid> wrote in message
news:421905A7.8C113142@ANTISPAMbtinternet.com.invalid...
[quote:57b5f53fa0]Luc The Perverse wrote:

i is the positive square root of -1 by definition. Problem solved.

i cannot be said to be positive or negative.
[/quote:57b5f53fa0]
"i" does not exist! The state of being positive or being negative can only
be intuitively applied to numbers in the real domain.

"i" is by definition the solution to the function which returns the positive
square root of negative one. Therefore i is positive. -i is negative, and
both, as would be assumed, when squared equal -1.

--
LTP

When the llama speaks you listen. Unfortunately the llama hasn't spoken
yet.

Luc The Perverse

Posted: Sun Feb 20, 2005 8:16 pm

Guest

"Jim Spriggs" <jim.sprigs@ANTISPAMbtinternet.com.invalid> wrote in message
news:42193151.FBA85E9D@ANTISPAMbtinternet.com.invalid...
[quote:7fd94dfcb0]i is the positive square root of -1 by definition. Problem solved.

Perhaps he meant that i*i = (-i)*(-i) = -1

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:7fd94dfcb0]
Well my universe has been shaken

--
LTP

When the llama speaks you listen. Unfortunately the llama hasn't spoken
yet.

Guest

Posted: Sun Feb 20, 2005 10:42 pm

jstevh@msn.com wrote:
[quote:740d38bace]Let

x = -sqrt(y)

and now consider y=1.

Then a solution is x=-1, but x=1 is also a solution.

Right?
[/quote:740d38bace]
What is your definition of "sqrt"?

Guest

Posted: Sun Feb 20, 2005 10:42 pm

jstevh@msn.com wrote:
[quote:226b315301]Let

x = -sqrt(y)

and now consider y=1.

Then a solution is x=-1, but x=1 is also a solution.

Right?
[/quote:226b315301]
What is your definition of "sqrt"?

Guest

Posted: Sun Feb 20, 2005 10:42 pm

jstevh@msn.com wrote:
[quote:8a806e5e32]Let

x = -sqrt(y)

and now consider y=1.

Then a solution is x=-1, but x=1 is also a solution.

Right?
[/quote:8a806e5e32]
What is your definition of "sqrt"?

Gib Bogle

Posted: Mon Feb 21, 2005 1:21 am

Guest

David Kastrup wrote:

[quote:effe64c242]It is a veritable thicket of bad signs that is looming around James,
and he really needs to cut the branches.

[/quote:effe64c242]
Maybe he was ... Born Under a Bad Sign ... (and just about to lose his
mind.)

Gib Bogle

Posted: Mon Feb 21, 2005 1:24 am

Guest

David Kastrup wrote:

[quote:dca3f96fee]It is a veritable thicket of bad signs that is looming around James,
and he really needs to cut the branches.
[/quote:dca3f96fee]
Maybe he was Born Under a Bad Sign.

GOMEZ ROJAS DAVID MAXIMIL

Posted: Mon Feb 21, 2005 2:38 am

Guest

jstevh@msn.com wrote:
[quote:349b9afce7]x = -sqrt(y)
and now consider y=1.
Then a solution is x=-1, but x=1 is also a solution.
Right?
No. By definition, given non-negative y, the sqrt function
returns
the non-negative real x such that x^2 = 1.
x=1 is a solution to (-x)^2 = 1, but it is not a solution to
x = -sqrt(y).
[/quote:349b9afce7]
Don't try to think about sqrt() as the function that solves equations,
'cause, as it's said up there, sqrt() is a real-valued function, i.e.
it takes ONE number and returns ONE number. For a given number x, what
number is sqrt(x)? The definition is: the _positive_ real number w
such that w^2 = x. It can't be wrong, it can't be correct: it's just
a definition.
Now, let's go to the context of quadratic equations: when you have,
let's say, to solve x^2 = 4, you can get ONE of the 2 answers by saying
x = sqrt(4) = 2. If you want to do 'the complete work', you must remember
that it's not true for all x that sqrt(x^2) = x. Following the definition
of sqrt(), actually we have that sqrt(x^2) = abs(x), i.e.

sqrt(3^2) = 3 and sqrt((-3)^2) = 3, too.

Let's say it this way: when you have x^2 = 4, you have two solutions:
the positive real number sqrt(4), and the negative real number -sqrt(4),
i.e. x = 2 and x = -2. sqrt() is not multi-valued.

--
DvD Gomez
dgomezNOSPAM@ing.uchile.cl

Guest

Posted: Mon Feb 21, 2005 7:11 am

mhochster@gmail.com wrote:
[quote:53b6894cf4]jstevh@msn.com wrote:
Let

x = -sqrt(y)

and now consider y=1.

Then a solution is x=-1, but x=1 is also a solution.

Right?

What is your definition of "sqrt"?
[/quote:53b6894cf4]
The square root operator returns that number which when multipled times
itself gives the square root argument e.g. sqrt(4) returns 2 and -2, as
*both* multiply times themselves to give 4.

Mathematicians have naively and arbitrarily thought they could remove
the inherent ambiguity by fiat--a childish effort--and I can not only
point out in small ways why that's a huge mistake, but also in big
ways.

If you were real mathematicians then that's all that would be
necessary, but many of you seem to believe that mathematics is just
some little toy you build or break at will.

Emotion RULES math society.

Now I can objectively step through mathematical arguments, but if you
people can't accept math and logic, then what good does it do?

Notice that you are math students and professors, people who post on
math newsgroups, but many of you abhor mathematics itself, clearly
having pet beliefs.

Your pet beliefs when shown to be wrong, you keep.

Your pet "proofs" when shown to be false, you keep claiming they are
true.

And it seems to me you do that out of a contempt for mathematics
itself, having taken on the title of mathematicians without any respect
for logic or mathematical rules.

You are a style over substance society--a fashion show--made up of
people who try to play a role for your own benefit in terms of titles
and prestige.

Someone like me is most foreign to you as you can't understand how I
can continue going under the barrage of social humiliation you try to
toss on me.

That proves my point as much as anything else as post after post after
post, you people cite social issues, as if a real mathematician would
pause to consider whether people would like them anymore or not, before
they put forward a mathematical argument.

So some of you track my posts obsessively replying year after year
after year because you are NOT real mathematicians and you have no
concept of how a real mathematician is motivated, and you can't
understand that ridicule is just some oddity to an actual
mathematician, while it's everything to people playing at being one.

James Harris

Justin

Posted: Mon Feb 21, 2005 11:45 am

Guest

jstevh@msn.com wrote:

: The square root operator returns that number which when multipled times
: itself gives the square root argument e.g. sqrt(4) returns 2 and -2, as
: *both* multiply times themselves to give 4.

No. The square root *function* returns the nonnegative real number whose
square is the argument.

Idiotic rambling snipped.

Justin

Shmuel (Seymour J.) Metz

Posted: Mon Feb 21, 2005 11:57 am

Guest

In <cvb152$20jb$2@msunews.cl.msu.edu>, on 02/20/2005
at 09:57 PM, stephen@nomail.com said:

[quote:844f1f8265]If you want multi-valued functions you have to consider functions
whose domains are sets of reals.
[/quote:844f1f8265]
No. You're confusing domains with ranges.

[quote:844f1f8265]For example you could define
a function f: R -> R x R where f(x)={sqrt(x), -sqrt(x)}
[/quote:844f1f8265]
The domain is R, not 2^R or a subset.

[quote:844f1f8265](we would have to modify this a little to handle 0).
[/quote:844f1f8265]
No. {0,0} = {0}.

[quote:844f1f8265]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,
[/quote:844f1f8265]
How are, e.g., {0}, {2,-2} not sets?

[quote:844f1f8265]and it could contain three values.
[/quote:844f1f8265]
FSVO three equal to two or one.

--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>

Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply to
domain Patriot dot net user shmuel+news to contact me. Do not
reply to spamtrap@library.lspace.org

Jim Spriggs

Posted: Mon Feb 21, 2005 12:07 pm

Guest

jstevh@msn.com wrote:
[quote:12787a6380]
The square root operator returns that number which when multipled times
itself gives the square root argument e.g. sqrt(4) returns 2 and -2, as
*both* multiply times themselves to give 4.
[/quote:12787a6380]
You contradict yourself. You write "that number" which implies that
there is only one of them but then, in your example, you give two such
numbers.

Arturo Magidin

Posted: Mon Feb 21, 2005 12:12 pm

Guest

In article <1108861793.081073.222950@o13g2000cwo.googlegroups.com>,
<matt271829-news@yahoo.co.uk> wrote:

[.snip.]

[quote:97341732bb]By convention, sqrt(x) means the positive root (presumably because
people want expressions involving sqrt(x) to be unambiguous and
single-valued). I personally find this convention counterintuitive and
often unhelpful, but there you are, you have to live with it. It's the
accepted definition.

Unfortunately - and this is possibly the root (ho-ho) of your problem -
this definition means that sqrt(x) is not the (complete) inverse
function of x^2. When you solve something like

x^2 = 1

you want to "undo" the x^2 by applying the inverse function, which one
naturally assumes to be sqrt. This gives

sqrt(x^2) = sqrt(1)

or, given the conventional definition, apparently

x = 1

thereby losing the solution x = -1. In fact, the required inverse
function is +/-sqrt, which gives

+/-sqrt(x^2) = +/-sqrt(1)

or

x = +/-1

which is correct.
[/quote:97341732bb]
Actually, the correct way to do this is to remember that sqrt(x^2)
(for real number x) is not equal to x, but to the absolute value of x;
that is,

For every real number x, sqrt(x^2) = |x|.

If you do that, then from

x^2 = 1

taking square roots you have

sqrt(x^2) = sqrt(1)

so

|x| = 1,

which gives the two valid solutions to the original equation.

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

Arturo Magidin
magidin@math.berkeley.edu

Guest

Posted: Mon Feb 21, 2005 1:42 pm

In sci.math "Shmuel (Seymour J.) Metz" <spamtrap@library.lspace.org.invalid> wrote:
: In <cvb152$20jb$2@msunews.cl.msu.edu>, on 02/20/2005
: at 09:57 PM, stephen@nomail.com said:

:>If you want multi-valued functions you have to consider functions
:>whose domains are sets of reals.

: No. You're confusing domains with ranges.

Yes. That was a typo. I meant codomain.

:>For example you could define
:>a function f: R -> R x R where f(x)={sqrt(x), -sqrt(x)}

: The domain is R, not 2^R or a subset.

The codomain is R x R.

:> (we would have to modify this a little to handle 0).

: No. {0,0} = {0}.

Which is why I said 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,

: How are, e.g., {0}, {2,-2} not sets?

That should have been "is now a set". Another unfortunate typo.

Stephen

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