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Aiya-Oba
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 Posted: Tue Jul 22, 2008 6:43 pm Post subject: Rational Hypotenuse Axiom.By Aiya-Oba |
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Other than the hypotenuse, no ratio of the sum of the squares of all
the sides of non isosceles right triangle, to the square of one side
is rational.-Aiya-Oba
Thus, a^2 + b^2 + c^2/c^2 = 2
Such that,
3^2 + 4^2 + 5^2/3^2 = 5.555555556
3^2 + 4^2 + 5^2/4^2 = 3.125
and,
3^2 + 4^2 + 5^2/5^2 = 2.
Q E D |
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Jesse F. Hughes
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| Posted: Wed Jul 23, 2008 5:11 pm Post subject: Re: Rational Hypotenuse Axiom.By Aiya-Oba |
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Aiya-Oba <aaiyaoba@rcc.mass.edu> writes:
[quote]Other than the hypotenuse, no ratio of the sum of the squares of all
the sides of non isosceles right triangle, to the square of one side
is rational.-Aiya-Oba
Thus, a^2 + b^2 + c^2/c^2 = 2
Such that,
^^^^^^^^^ ???
3^2 + 4^2 + 5^2/3^2 = 5.555555556
[/quote]
50 / 9 is rational.
[quote]
3^2 + 4^2 + 5^2/4^2 = 3.125
[/quote]
50 / 16 is rational.
[quote]
and,
3^2 + 4^2 + 5^2/5^2 = 2.
Q E D
^^^^^[/quote]
Those letters don>t mean "End of post", you know. They>re supposed to
follow a proof, not an example (and certainly not a counterexample).
I suspect you meant that in every case, the ratios
(a^2 + b^2 + c^2)/a^2 and (a^2 + b^2 + c^2)/b^2
are non-integers when c^2 = a^2 + b^2 and a != b, i.e., that the
ratios
2(a^2+b^2)/a^2 and 2(a^2 + b^2)/b^2
are non-integers when a != b. Of course, this is also false. If the
length a is any multiple of the length b (say, b = 1 and a = 2), then
the second ratio is an integer. Let a = kb (where k is an integer)
and we have
2 ( k^2 b^2 + b^2 ) / b^2 = 2 ( k^2 + 1 ),
which is obviously a counterexample.
So even a "fixed" statement of your "axiom" (another mis-used term) is
false.
--
"I>d step through arguments in such detail that it was like I was
teaching basic arithmetic and some poster would come back and act like
I hadn>t said anything that made sense. For a while I almost started
to doubt myself." -- James S. Harris, so close and yet.... |
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Aiya-Oba
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| Posted: Wed Jul 23, 2008 7:32 pm Post subject: Re: Rational Hypotenuse Axiom.By Aiya-Oba |
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On Jul 23, 5:11 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
[quote]Aiya-Oba <aaiya...@rcc.mass.edu> writes:
Other than the hypotenuse, no ratio of the sum of the squares of all
the sides of non isosceles right triangle, to the square of one side
is rational.-Aiya-Oba
Thus, a^2 + b^2 + c^2/c^2 = 2
Such that,
^^^^^^^^^ ???
3^2 + 4^2 + 5^2/3^2 = 5.555555556
50 / 9 is rational.
3^2 + 4^2 + 5^2/4^2 = 3.125
50 / 16 is rational.
and,
3^2 + 4^2 + 5^2/5^2 = 2.
Q E D
^^^^^
Those letters don>t mean "End of post", you know. They>re supposed to
follow a proof, not an example (and certainly not a counterexample).
I suspect you meant that in every case, the ratios
(a^2 + b^2 + c^2)/a^2 and (a^2 + b^2 + c^2)/b^2
are non-integers when c^2 = a^2 + b^2 and a != b, i.e., that the
ratios
2(a^2+b^2)/a^2 and 2(a^2 + b^2)/b^2
are non-integers when a != b. Of course, this is also false. If the
length a is any multiple of the length b (say, b = 1 and a = 2), then
the second ratio is an integer. Let a = kb (where k is an integer)
and we have
2 ( k^2 b^2 + b^2 ) / b^2 = 2 ( k^2 + 1 ),
which is obviously a counterexample.
So even a "fixed" statement of your "axiom" (another mis-used term) is
false.
--
"I>d step through arguments in such detail that it was like I was
teaching basic arithmetic and some poster would come back and act like
I hadn>t said anything that made sense. For a while I almost started
to doubt myself." -- James S. Harris, so close and yet....
[/quote]
Hi Jesse F. Hughes:
Thanks for your interest in Rational Hypotenuse Axiom.
However, you did not observe that I specifically made sure that the
right triangle at issue here is the non-isosceles right triangle,
which mean Rational Hypotenuse Axiom, is all about the famous
Pythagorean Triplet, hence the word axiom, to indicate the already
well-known and accepted relation ( a^2 + b^2 = c^2 ) of the
Pythagorean Triple. In other words, wherever the Pythagorean Triplet
holds, Rational Hypotenuse Axiom is true. -Aiya-Oba (Poet/
Philosopher) . |
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Jesse F. Hughes
Guest
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| Posted: Thu Jul 24, 2008 1:14 am Post subject: Re: Rational Hypotenuse Axiom.By Aiya-Oba |
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[quote]Hi Jesse F. Hughes:
Thanks for your interest in Rational Hypotenuse Axiom.
However, you did not observe that I specifically made sure that the
right triangle at issue here is the non-isosceles right triangle,
which mean Rational Hypotenuse Axiom, is all about the famous
Pythagorean Triplet, hence the word axiom, to indicate the already
well-known and accepted relation ( a^2 + b^2 = c^2 ) of the
Pythagorean Triple.
[/quote]
It>s amazing how many of those words above don>t seem to mean what you
think they mean.
[quote]In other words, wherever the Pythagorean Triplet
holds, Rational Hypotenuse Axiom is true. -Aiya-Oba (Poet/
Philosopher) .
[/quote]
Great. Another self-described philosopher.
If anyone asks, I>m a short order cook.
--
Jesse F. Hughes
"It>s easy folks. Just talk about my approach to your favorite
mathematician. If they can>t be interested in it, they>ve
demonstrated a lack of mathematical skill." -- James Harris |
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Aiya-Oba
Guest
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| Posted: Sat Jul 26, 2008 6:11 pm Post subject: Re: Rational Hypotenuse Axiom.By Aiya-Oba |
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On Jul 23, 4:14�pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
[quote]Hi Jesse F. Hughes:
Thanks for your interest in Rational Hypotenuse Axiom.
However, you did not observe that I specifically made sure that the
right triangle at issue here is the non-isosceles right triangle,
which mean Rational Hypotenuse Axiom, is all about the famous
Pythagorean Triplet, hence the word axiom, to indicate the already
well-known and accepted relation ( a^2 + �b^2 � = � c^2 ) �of �the
Pythagorean Triple.
It>s amazing how many of those words above don>t seem to mean what you
think they mean.
In other words, wherever the Pythagorean Triplet
holds, Rational Hypotenuse Axiom is true. -Aiya-Oba (Poet/
Philosopher) .
Great. �Another self-described philosopher.
If anyone asks, I>m a short order cook.
--
Jesse F. Hughes
"It>s easy folks. �Just talk about my approach to your favorite
mathematician. �If they can>t be interested in it, they>ve
demonstrated a lack of mathematical skill." -- James Harris
[/quote]
Thanks,
Jesse F.Hughes: The glory of every medium is in
its message.
A philosopher that does not seem a fool before the Wise, is otherwise.-
Aiya-Oba. |
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