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Message |
| Guest |
 Posted: Sat Feb 19, 2005 11:11 am |
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Let
x = -sqrt(y)
and now consider y=1.
Then a solution is x=-1, but x=1 is also a solution.
Right? |
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| Joseph Fagan |
| Posted: Sat Feb 19, 2005 11:11 am |
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Guest
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sqrt() means the positive square root.
In your example, x=-1;
x=1 is not a solution.
<jstevh@msn.com> wrote in message
news:1108829501.431405.246240@f14g2000cwb.googlegroups.com...
Quote: Let
x = -sqrt(y)
and now consider y=1.
Then a solution is x=-1, but x=1 is also a solution.
Right?
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| James Dolan |
| Posted: Sat Feb 19, 2005 11:11 am |
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Guest
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in article <1108838089.897054.324070@g14g2000cwa.googlegroups.com>,
<jstevh@msn.com> wrote:
|Rick Decker wrote:
|> 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?
|> >
|> No, not under the usual interpretation of the sqrt function.
|>
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|Well,
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|x = -sqrt(y), and squaring both sides
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|x^2 = y, so y = x^2, and x = 1 IS a solution.
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|What gives?
there's an old trick question that goes:
"who's buried in grant's tomb?"
the answer is supposed to be "grant, of course", and you're supposed
to feel like an idiot the first time that you hear the question,
because you actually have to stop and think about it for a second or
two before realizing how obvious it is. "grant's tomb" means,
roughly, "the place where grant is buried" and so the question is
asking "who's buried in the place where grant is buried?" or "whose
tomb is grant's tomb?".
the parallel math question is:
"what's the square root of x^2?"
and again you might feel a bit stupid if you have to stop and think
for a second or two before realizing that the question is asking
"whose square is x's square?" and so of course the answer is x.
except that x is not really _the_ answer. after people got really
sick of the trick question "who's buried in grant's tomb?", someone
eventually realized that there's an extra trick to it, because grant
isn't the only person in the tomb; mrs grant is there too. and of
course it turns out that x isn't the only one in x's tomb, either; -x
is there too.
so what's the lesson of all this? i hope that some of it is obvious.
you have to be appropriately careful about some things in order to
avoid falling for some of the oldest tricks in the book, like:
"grant is the occupant of grant's tomb. also mrs grant is the
occupant of grant's tomb. therefore grant = mrs grant. but this
contradicts the known fact that grant's beard was much thicker than
mrs grant's. therefore the original assumption that x^n+y^n=z^n
must have been false, or maybe i've discovered a fundamental rip in
the fabric of the universe- not quite sure which yet."
so how careful do you have to be? i guess careful enough to avoid
falling for these sorts of tricks. sometimes you should be
hyper-careful about the distinction between "the occupant of grant's
tomb" and "an occupant of grant's tomb" or "the first occupant of
grant's tomb" or whatever, or between "the square root of 17" and "a
square root of 17" or "the positive square root of 17", and so forth.
if you don't know how to be appropriately careful about these sorts of
things, then it just shows how empty the words are when you go around
saying things like "it's just simple algebra, people! you're telling
me that you don't believe in algebra!".
i think that this also shows what a bad idea it is to fall for the
stupidity that being a mathematician is like being an inhuman freak
who just mechanically cranks out formal proofs and accepts their
conclusions without pausing to see whether they violate common sense;
the judicious and fully human application of common sense is an
essential (if unfortunately uncommon) part of being a (human)
mathematician.
by the way, the ambiguity in phrases like "the square root of x" is in
a way at the heart of galois theory. for example the "unsolvability
by radicals of the general quintic polynomial" is established by
showing that no formula using only radicals plus the standard rational
operations +,-,*,/ can have "the right kind" of ambiguity to match the
ambiguity inherent in the phrase "the root of the general quintic
polynomial".
--
[e-mail address jdolan@math.ucr.edu] |
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| Henry |
| Posted: Sat Feb 19, 2005 11:11 am |
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On Sat, 19 Feb 2005 18:56:39 -0000, "Mike Terry"
<news.dead.person.stones@darjeeling.plus.com> wrote:
Quote: If P ===> Q, this does not mean that Q ===> P.
Indeed not, but what is true is either P ===> Q or Q ===> P. |
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| Rick Decker |
| Posted: Sat Feb 19, 2005 12:32 pm |
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Guest
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jstevh@msn.com wrote:
Quote: Let
x = -sqrt(y)
and now consider y=1.
Then a solution is x=-1, but x=1 is also a solution.
Right?
No, not under the usual interpretation of the sqrt function.
Regards,
Rick |
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| Guest |
| Posted: Sat Feb 19, 2005 1:34 pm |
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Rick Decker wrote:
Quote: 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?
No, not under the usual interpretation of the sqrt function.
Well,
x = -sqrt(y), and squaring both sides
x^2 = y, so y = x^2, and x = 1 IS a solution.
What gives? |
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| David Kastrup |
| Posted: Sat Feb 19, 2005 1:39 pm |
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Guest
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jstevh@msn.com writes:
Quote: Rick Decker wrote:
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?
No, not under the usual interpretation of the sqrt function.
Well,
x = -sqrt(y), and squaring both sides
x^2 = y, so y = x^2, and x = 1 IS a solution.
What gives?
1 = -1, and squaring both sides
1 = 1, so 1 = 1, and 1 = -1 IS a solution. According to your
appalling logic.
In other words: squaring is not an equivalence operation.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum |
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| Mike Terry |
| Posted: Sat Feb 19, 2005 1:56 pm |
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Guest
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<jstevh@msn.com> wrote in message
news:1108838089.897054.324070@g14g2000cwa.googlegroups.com...
Quote: Rick Decker wrote:
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?
No, not under the usual interpretation of the sqrt function.
Well,
x = -sqrt(y), and squaring both sides
x^2 = y, so y = x^2, and x = 1 IS a solution.
What gives?
If P ===> Q, this does not mean that Q ===> P.
Regards,
Mike. |
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| fishfry |
| Posted: Sat Feb 19, 2005 2:37 pm |
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Guest
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In article <1108838089.897054.324070@g14g2000cwa.googlegroups.com>,
jstevh@msn.com wrote:
Quote: Rick Decker wrote:
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?
No, not under the usual interpretation of the sqrt function.
Well,
x = -sqrt(y), and squaring both sides
x^2 = y, so y = x^2, and x = 1 IS a solution.
What gives?
If you square both sides of -2 = 2 you get 4 = 4, what does that have to
do with anything?
By convention, sqrt(x) for real x denotes the positive square root. |
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| Jesse F. Hughes |
| Posted: Sat Feb 19, 2005 2:59 pm |
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Guest
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jstevh@msn.com writes:
Quote: Rick Decker wrote:
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?
No, not under the usual interpretation of the sqrt function.
Well,
x = -sqrt(y), and squaring both sides
x^2 = y, so y = x^2, and x = 1 IS a solution.
What gives?
Neat trick. Let me try it. Let x = 2. Squaring both sides, we get
x^2 = 4 and (-2)^2 = 4, so x = -2 is a solution to the equation
x = 2. Hence -2 = 2.
I don't see anything wrong with your reasoning.
--
Jesse F. Hughes
"Well, you know as soon as you have a new number I will be happy to
add it to the list. Don't try those childish tit-for-tat games with
me." -- Ross Finlayson on Cantor's theorem. |
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| Guest |
| Posted: Sat Feb 19, 2005 4:28 pm |
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James Dolan wrote:
Quote: in article <1108838089.897054.324070@g14g2000cwa.googlegroups.com>,
<deleted>
Quote:
by the way, the ambiguity in phrases like "the square root of x" is
in
a way at the heart of galois theory. for example the "unsolvability
by radicals of the general quintic polynomial" is established by
showing that no formula using only radicals plus the standard
rational
operations +,-,*,/ can have "the right kind" of ambiguity to match
the
ambiguity inherent in the phrase "the root of the general quintic
polynomial".
And I proved that it's not a mathematical issue.
That is, Galois Theory is about form rather than substance.
The ambiguity in square roots and radicals and solutions to polynomials
of degree greater than 1, in general, cannot be removed.
A simple way to understand it is to consider
x^2 + 3x + 2 = 0
as there is no convention to force x to have a single value.
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.
You cannot.
Attempts at removing it, fail.
It's actually easy to show as I did with x = -sqrt(y).
But denial of the truth, is harder to handle.
___JSH |
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| Schweinkolben |
| Posted: Sat Feb 19, 2005 4:45 pm |
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| David Kastrup |
| Posted: Sat Feb 19, 2005 5:30 pm |
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Guest
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Henry <se16@btinternet.com> writes:
So which of the two following is true: x odd ===> x>0,
or x>0 ===> x odd ?
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum |
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| David C. Ullrich |
| Posted: Sat Feb 19, 2005 6:31 pm |
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Guest
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On 19 Feb 2005 10:34:49 -0800, jstevh@msn.com wrote:
Quote: Rick Decker wrote:
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?
No, not under the usual interpretation of the sqrt function.
Well,
x = -sqrt(y), and squaring both sides
x^2 = y, so y = x^2, and x = 1 IS a solution.
What gives?
Here's a simpler example of what gives, so you may be able
to see the problem yourself:
Let's consider solutions to x = 1. The question is whether
x = -1 is a solution to x = 1. Squaring both sides gives
x^2 = 1, so x = -1 _is_ a solution.
************************
David C. Ullrich |
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| David C. Ullrich |
| Posted: Sat Feb 19, 2005 6:31 pm |
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Guest
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On Sat, 19 Feb 2005 23:30:55 +0100, David Kastrup <dak@gnu.org> wrote:
Quote: 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 ?
Seems like neither of those is true - noting that
"P ===> Q or Q ===> P" _is_ a tautology we have a
little conundrum.
The answer is that we're using two different notions
of ===>. In fact when someone says A -> B what is often
meant is (x)(A(x) -> B(x)), where "(x)" is "for all x".
So. _If_ we mean -> strictly as in propositional logic
then the answer to which of x odd ===> x>0,
or x>0 ===> x odd is true is "can't tell, depends
on the value of x".
Otoh if we're taking A -> B to mean an implicit
universal quantification then the answer is of
course "neither is true". The fact that
((A -> B) or (B -> A)) is a tautology makes
(x)((A -> B) or (B -> A)) a theorem, but it
doesn't make ((x)(A->B) or (x)(B->A)) a theorem.
************************
David C. Ullrich |
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