"Fermat's Last Theorem" Disproved?
in [Math]
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From: cjcountess on 28 Jul 2010 19:36 Kermit, I see your point as well as Fredericks, but looking at this geometrically, I cannot help but to notice that the "a b and c" that are to be squared, refer to geometrical unite lengths, that are not just dimensionless integers. This seems to me to make a difference. Even if, as one poster said, the theorem does not apply to triangle with diagonal of square as hypotenuse, because "sqrt2", is not non zero positive integer, this tell me that if theory applies to all but exception to the rule, than the rule is not universal. And if its roots are in dimensional geometry as opposed to dimensionless non zero positive integers, as I suspect, than I must still question the theorem. Conrad J Countess
From: Virgil on 28 Jul 2010 21:19 In article <41c9de7e-adc7-4e06-805a-53aa993bbe60(a)t2g2000yqe.googlegroups.com>, cjcountess <cjcountess(a)yahoo.com> wrote: > Kermit, I see your point as well as Fredericks, > > but looking at this geometrically, I cannot help but to notice that > the "a b and c" that are to be squared, refer to geometrical unite > lengths, that are not just dimensionless integers. This seems to me to > make a difference. > > Even if, as one poster said, the theorem does not apply to triangle > with diagonal of square as hypotenuse, because "sqrt2", is not non > zero positive integer, this tell me that if theory applies to all but > exception to the rule, than the rule is not universal. > > And if its roots are in dimensional geometry as opposed to > dimensionless non zero positive integers, as I suspect, than I must > still question the theorem. > > Conrad J Countess Since you still do not seem to understand the distinction between real numbers, which represent arbitrary lengths, and integers, which don't, and FLT is about integers, you are merely confused. To disprove FLT is way beyond your capacity.
From: Gerry Myerson on 29 Jul 2010 00:06 In article <Virgil-B659A4.19195628072010(a)bignews.usenetmonster.com>, Virgil <Virgil(a)home.esc> wrote: > In article > <41c9de7e-adc7-4e06-805a-53aa993bbe60(a)t2g2000yqe.googlegroups.com>, > cjcountess <cjcountess(a)yahoo.com> wrote: > > > Kermit, I see your point as well as Fredericks, > > > > but looking at this geometrically, I cannot help but to notice that > > the "a b and c" that are to be squared, refer to geometrical unite > > lengths, that are not just dimensionless integers. This seems to me to > > make a difference. > > > > Even if, as one poster said, the theorem does not apply to triangle > > with diagonal of square as hypotenuse, because "sqrt2", is not non > > zero positive integer, this tell me that if theory applies to all but > > exception to the rule, than the rule is not universal. > > > > And if its roots are in dimensional geometry as opposed to > > dimensionless non zero positive integers, as I suspect, than I must > > still question the theorem. > > > > Conrad J Countess > > Since you still do not seem to understand the distinction between real > numbers, which represent arbitrary lengths, and integers, which don't, > and FLT is about integers, you are merely confused. > > To disprove FLT is way beyond your capacity. To disprove FLT is beyond anyone's capacity (given certain plausible consistency assumptions). -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: Virgil on 29 Jul 2010 00:19 In article <gerry-370F8C.14061829072010(a)mx01.eternal-september.org>, Gerry Myerson <gerry(a)maths.mq.edi.ai.i2u4email> wrote: > In article <Virgil-B659A4.19195628072010(a)bignews.usenetmonster.com>, > Virgil <Virgil(a)home.esc> wrote: > > > In article > > <41c9de7e-adc7-4e06-805a-53aa993bbe60(a)t2g2000yqe.googlegroups.com>, > > cjcountess <cjcountess(a)yahoo.com> wrote: > > > > > Kermit, I see your point as well as Fredericks, > > > > > > but looking at this geometrically, I cannot help but to notice that > > > the "a b and c" that are to be squared, refer to geometrical unite > > > lengths, that are not just dimensionless integers. This seems to me to > > > make a difference. > > > > > > Even if, as one poster said, the theorem does not apply to triangle > > > with diagonal of square as hypotenuse, because "sqrt2", is not non > > > zero positive integer, this tell me that if theory applies to all but > > > exception to the rule, than the rule is not universal. > > > > > > And if its roots are in dimensional geometry as opposed to > > > dimensionless non zero positive integers, as I suspect, than I must > > > still question the theorem. > > > > > > Conrad J Countess > > > > Since you still do not seem to understand the distinction between real > > numbers, which represent arbitrary lengths, and integers, which don't, > > and FLT is about integers, you are merely confused. > > > > To disprove FLT is way beyond your capacity. > > To disprove FLT is beyond anyone's capacity (given certain plausible > consistency assumptions). Unless Wiles proof proves flawed it certainly is beyond anyone's capacity, but even if Wiles proof proves to be flawed (which I very much doubt), cjcountess would clearly be in in way over his/her/its head, at least without several years of hard study, in attempting a disproof.
From: bert on 29 Jul 2010 05:21
On 29 July, 00:36, cjcountess <cjcount...(a)yahoo.com> wrote: > And if its roots are in dimensional geometry as opposed to > dimensionless non zero positive integers, as I suspect, then > I must still question the theorem. Well, if they were, you well might; but they aren't. The theorem about a^n, b^n and c^n is in dimensionless numbers, whose squares, cubes, fourth powers and so on are also dimensionless numbers. Your geometrical view of it is simply misleading you. -- |