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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:17ff291e39]Let
x = -sqrt(y)
and now consider y=1.
Then a solution is x=-1, but x=1 is also a solution.
Right?
[/quote:17ff291e39] |
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| Henry |
| Posted: Sat Feb 19, 2005 11:11 am |
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Guest
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On Sat, 19 Feb 2005 18:56:39 -0000, "Mike Terry"
<news.dead.person.stones@darjeeling.plus.com> wrote:
[quote:1542c94f96]If P ===> Q, this does not mean that Q ===> P.
[/quote:1542c94f96]
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:70c7ebf4f4]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.[/quote:70c7ebf4f4]
Regards,
Rick |
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| Guest |
| Posted: Sat Feb 19, 2005 1:34 pm |
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Rick Decker wrote:
[quote:0e6eb4d2ff]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.
[/quote:0e6eb4d2ff]
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:faa90f6cca]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?
[/quote:faa90f6cca]
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:e1656cadc9]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?
[/quote:e1656cadc9]
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:77b680b05b]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?
[/quote:77b680b05b]
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:3b45c85ed7]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?
[/quote:3b45c85ed7]
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:5d84cbdb8b]in article <1108838089.897054.324070@g14g2000cwa.googlegroups.com>,
[/quote:5d84cbdb8b]
<deleted>
[quote:5d84cbdb8b]
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".
[/quote:5d84cbdb8b]
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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Henry <se16@btinternet.com> writes:
[quote:db70a71484]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.
[/quote:db70a71484]
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:82b0cdedc6]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?
[/quote:82b0cdedc6]
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:128ea6a495]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:128ea6a495]
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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| Guest |
| Posted: Sat Feb 19, 2005 6:43 pm |
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David C. Ullrich wrote:
[quote:ac5dec16eb]On 19 Feb 2005 10:34:49 -0800, 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.
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.
[/quote:ac5dec16eb]
And ultimately the problem is that the inverse function of f(x) = x^2
is not single-valued (hence the temptation to assume that sqrt(x) is
not single-valued). As another example,
x = 0
therefore
sin(x) = sin(0)
therefore x = pi is a solution (wrong because the inverse sine function
is not single-valued, except by "convention"). |
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