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| Guest |
 Posted: Mon Feb 21, 2005 2:09 pm |
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In sci.math stephen@nomail.com wrote:
: 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.
I was also confusing myself about what I meant
the codomain to be. I wrote RxR, but I was thinking
of subsets of R with size 2. Needless to say
this resulted in what I actually said being largely
nonsensical.
Stephen |
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| Arturo Magidin |
| Posted: Mon Feb 21, 2005 2:13 pm |
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Guest
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In article <1109030715.288735.287950@f14g2000cwb.googlegroups.com>,
<matt271829-news@yahoo.co.uk> wrote:
[.snip.]
[quote:1eed826b12]http://mathworld.wolfram.com/MultivaluedFunction.html seems quite happy
with multivalued inverse functions in cases such as this. I checked
this site before I posted as I usually find it pretty good...
[/quote:1eed826b12]
Mathworld is using a different set of conventions. Those are fine, but if you
allow for multivalued functions, then the basic equations
f(f^{-1}(x)) = x for all x in domain(f^{-1})
f^{-1}(f(x)) = x for all x in domain(f)
are no longer valid. That is, they are not part of the conventions in
which you allow multivalued functions and multivalued inverses. In
this instance, the error is the mixing of conventions.
Multivalued functions are fine, but you have to realize that once you
allow for the possibility, you have to go back and "clean up" a lot of
the stuff that had been done under the assumption that functions have
single values. The cancellation of inverses is one such thing.
--
======================================================================
"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 |
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| Guest |
| Posted: Mon Feb 21, 2005 6:30 pm |
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Arturo Magidin wrote:
[quote:19218aeff4]In article <1108861793.081073.222950@o13g2000cwo.googlegroups.com>,
matt271829-news@yahoo.co.uk> wrote:
[.snip.]
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.
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.
[/quote:19218aeff4]
Yeah, that looks fine too... but if I DID want to do it the way I
originally tried, what would be the correct way?
In other words, I start with
f(x) = 1
then I want something like
f^-1[f(x)] = f^-1(1)
leading to something not unadjacent to
x = f^-1(1)
In the case that f(x) = x^2, what is the correct way to lay this out so
as to arrive at the answer of x = +/-1? |
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| Guest |
| Posted: Mon Feb 21, 2005 8:17 pm |
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Arturo Magidin wrote:
[quote:1066893c05]In article <1109030715.288735.287950@f14g2000cwb.googlegroups.com>,
matt271829-news@yahoo.co.uk> wrote:
[.snip.]
http://mathworld.wolfram.com/MultivaluedFunction.html seems quite
happy
with multivalued inverse functions in cases such as this. I checked
this site before I posted as I usually find it pretty good...
Mathworld is using a different set of conventions. Those are fine,
but if you
allow for multivalued functions, then the basic equations
f(f^{-1}(x)) = x for all x in domain(f^{-1})
f^{-1}(f(x)) = x for all x in domain(f)
are no longer valid. That is, they are not part of the conventions
in
which you allow multivalued functions and multivalued inverses. In
this instance, the error is the mixing of conventions.
Multivalued functions are fine, but you have to realize that once you
allow for the possibility, you have to go back and "clean up" a lot
of
the stuff that had been done under the assumption that functions have
single values. The cancellation of inverses is one such thing.
[/quote:1066893c05]
Yeah, you're right (as usual). Let me try again.
f(x) = 1
then, allowing multivalued inverses as necessary, just go straight to
x = f^-1(1)
where "=" means that x can be any value that the rhs can take.
In this specific case the multivalued inverse is +/-sqrt(x), so
x^2 = 1
straight to
x = +/-sqrt(1)
Any better? |
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| Matt Gutting |
| Posted: Tue Feb 22, 2005 1:04 pm |
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Guest
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Luc The Perverse wrote:
[quote:cc9638e5e4]"Jim Spriggs" <jim.sprigs@ANTISPAMbtinternet.com.invalid> wrote in message
news:421905A7.8C113142@ANTISPAMbtinternet.com.invalid...
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.
"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.
[/quote:cc9638e5e4]
1) If "The state of being positive or being negative can only be intuitively
[whatever that means] applied to numbers in the real domain", and i is not in
the real domain, then how can you claim that "i is positive"?
2) i is not defined as "the solution which returns the positive square root of
negative one" - such a definition would be absurd in light of your comments; see
(1). i could be defined as "one solution, arbitrarily chosen, to the solution
x^2 = -1". Better would be to define the complex numbers as ordered pairs of
reals (a,b) with (a,b) + (c,d) = (a+c,b+d) and (a,b) * (c,d) = (ac - bd,ad + bc)
and then define i = (1,0).
Matt |
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| Matt Gutting |
| Posted: Tue Feb 22, 2005 1:12 pm |
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jstevh@msn.com wrote:
[major snip]
[quote:2806831528]
Your pet beliefs when shown to be wrong, you keep.
Your pet "proofs" when shown to be false, you keep claiming they are
true.
[/quote:2806831528]
Hmm ... sounds familiar.
Matt
[quote:2806831528]
James Harris
[/quote:2806831528] |
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| Arturo Magidin |
| Posted: Thu Mar 03, 2005 6:58 am |
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In article <20050303012643.023b621f.kevin@hotmail.com>,
Kevin <kevin@hotmail.com> wrote:
[quote:4f308f7579]I learned that the complex numbers aren't ordered (or well
ordered, I don't know the difference actually), but anyway, I
thought I learned that you can't order complex numbers, so i > 0
is nonsense.
[/quote:4f308f7579]
What you learned, presumably, is that it is impossible to define an
ordering on the complex numbers ->that is compatible with the
operations<-.
What this means is that it is impossible to define an order relation >
on the complex numbers such that:
(i) For each complex number a, one and only one of a<0, a=0, and a>0
holds.
(ii) a<0 if and only if (-1)*a > 0
(iii) If a > 0 and b > 0 then a+b >0 and ab>0.
To see why, you first note that -1<0; for if (-1) > 0, then by (ii)
you woul dhave -(-1) = 1 < 0, but by (iii) you would also have
(-1)*(-1) >0. So -1 < 0.
Now take i; if i>0, then i*i = -1 >0, but that is impossible. If i<0,
then -i>0, and again we have that (-i)*(-i)>0, which contradicts that
-1<0.
HOWEVER, it is quite possible to define an ordering on the complex
numbers. For example, a very easy and natural ordering on the complex
numbers is to define
(a + bi) < (c + di) if and only if a<c or (a=c and b<d).
This is the "lexicographic order", and on the complex plane it is the
same as saying that being to the left of, or on the same vertical line
and below, is the same as being "smaller than". In this definition, i
is indeed bigger than 0.
This is a perfectly fine ordering that has a lot of good uses. It is
not compatible with the operations, however, so it also fails to be
any good for a number of (algebraic) applications. That does not mean
you cannot "define an ordering on the complex numbers."
(Oh, and if you accept the Axiom of Choice, then of course the complex
numbers ->can<- be well ordered, though we cannot write down a rule
for this ordering).
--
======================================================================
"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 |
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| Bill Dubuque |
| Posted: Fri Mar 04, 2005 4:24 pm |
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Arturo Magidin <magidin@math.berkeley.edu> wrote:
[quote:4ee2ce0025]Kevin <kevin@hotmail.com> wrote:
I learned that the complex numbers aren't ordered (or well
ordered, I don't know the difference actually), but anyway,
I thought I learned that you can't order complex numbers,
so i > 0 is nonsense.
What you learned, presumably, is that it is impossible to define
an ordering on the complex numbers ->that is compatible with the
operations<-.
What this means is that it is impossible to define an order relation
on the complex numbers such that:
(i) For each complex number a, one and only one of a<0, a=0, and a>0
holds.
(ii) a<0 if and only if (-1)*a > 0
(iii) If a > 0 and b > 0 then a+b >0 and ab>0.
To see why, you first note that -1<0; for if (-1) > 0, then by (ii)
you woul dhave -(-1) = 1 < 0, but by (iii) you would also have
(-1)*(-1) >0. So -1 < 0.
Now take i; if i>0, then i*i = -1 >0, but that is impossible. If i<0,
then -i>0, and again we have that (-i)*(-i)>0, which contradicts that
-1<0.
[/quote:4ee2ce0025]
Your axioms for an ordered field F are a little bit redundant.
It requires only a subset P such that: P+P,P*P<P, P\/-P = F\0
which induces the order: x>y <=> x-y in P (here P = positives).
Furthermore it is easy to show that a field may be ordered iff
-1 is not a sum of squares in F (see Artin-Schreier theory).
--Bill Dubuque |
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