"Fermat's Last Theorem" Disproved?
in [Math]
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From: cjcountess on 8 Aug 2010 15:02 Wow Virgil, where's your common sense? I realy don't have to add anything to this conversation, just shorten it and pick out 2 important points. You really can learn something about your own subject by studying other subject, simply because everything is interelated and do share a common ground state reality. As such something should have told you before I did that numbers have dimensions. Take a lesson from reading descriptions. If something exist it has to have dimensions. To describe something is to outline its dimensions even if not explicitly stated and the more rigprous the description, the more outlined its dimensions. Having said that, if you arrange these numbers of any triplet claimed to represent Fermat's Last Theore, no matter what their form, in groups of like numbers, for instence 3x3, 4x4, and 5x5, they form three squares that fit together to form a triangle. And just as geometry arrise from these numbers, they must arise from geometry. I am sorry Virgil, but you cannot win this one. I revolutionized physics and now without even trying I revolutionized math. You see that is why I don't get too mad at you guys when you act childish You must fell like a child in my midst Just joking fellas. I must do that sometimes to keep from getting mad at some of you. I cannot believe that it took me to come along and reveal that. If Fermat"s theorem depends on "dimensionless intergers", I accomplished what I originaly set out to do, see if I could find a disproff. But if it does'nt so be it . It was an interesting ride. Conrad J Countess
From: bert on 8 Aug 2010 15:20 On 8 Aug, 20:02, cjcountess <cjcount...(a)yahoo.com> wrote: > If Fermat"s theorem depends on "dimensionless intergers", I > accomplished what I originaly set out to do, see if I could find a > disproff. > > But if it does'nt so be it . It was an interesting ride. What to you seems a convincing proof - or disproof - is to everybody else a load of non-mathematical handwaving and wordplay. --
From: Virgil on 8 Aug 2010 15:29 In article <befc7208-a7f5-4e77-9af6-f5e49609f513(a)5g2000yqz.googlegroups.com>, cjcountess <cjcountess(a)yahoo.com> wrote: > Wow Virgil, where's your common sense? My common sense says that anyone who does not spell any better than you do is unlikely to accomplish much. > > If Fermat"s theorem depends on "dimensionless INTERGERS", I > accomplished what I originaly set out to do, see if I could find a > DISPROFF. EMPHASIS added!
From: cjcountess on 9 Aug 2010 10:38 On Aug 8, 3:29 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <befc7208-a7f5-4e77-9af6-f5e49609f...(a)5g2000yqz.googlegroups.com>, > > cjcountess<cjcount...(a)yahoo.com> wrote: > > Wow Virgil, where's your common sense? > > My common sense says that anyone who does not spell any better than you > do is unlikely to accomplish much. > > > > > If Fermat"s theorem depends on "dimensionless INTERGERS", I > > accomplished what I originaly set out to do, see if I could find a > > DISPROFF. > > EMPHASIS added! My inperfect spelling is not going to save you. Bad spelling or not, I am correct. There are no such things that exist, physicaly or mentaly, which do not have dimensions. Conrad J Countess
From: cjcountess on 9 Aug 2010 12:31
The idea that geometrically, a cube is the square of a square really fascinates me, even though it may be technically wrong. Somehow I know that I am correct in a certain sense. After all, if one squares a line by extending it into the 90 degree angular direction, the same measure, if one lays that same square on its side and repeats the exact same process of extending the entire square into the 90 degree angular direction the same measure, one obtains a cube. In so called dimensionless integers, 1 x 1 equals 1, but in geometry, a line in the horizontal direction of 1 unit x a line of equal measure in the 90 degree angular direction, equals a square unit. So what is the square of that square? The square of the 1x1=1 square would still be 1, but the square of the geometrical square must be more. If it is 2 squares because of the sqrt2 being the measure of the diagonal than it is less in volume and area than if it were placed on its side and squared in the 90 degree angular direction to create a cube as I suggested. The cube as geometrically being, the square of a square, on the very fundamental level of 1 square x 1 square fascinates me. As a matter of fact, when one considers the natural unite "c" or the speed of light, c^2 produce a 3D geometrical spherical particle, as far as the empirical, geometrical, and mathematical, evidence suggest. Also suggesting that (c^2 = c^3) on the quantum level. This would even mean that if (c^2 + c^2 = 2c^2) than (c^3 + c^3 = 2c^3), shattering "Fermat's Last Theorem", on quantum level. It would also mean that (E=mc^2) = (E=mc^circled) = (E=mcSphered) = (E=mc^3) = (E=mc^triangled as conforming to Pythagorean Theorem yet shattering The Fermat's Last Theorem), on the quantum level also That to me is also fascinating Conrad J Countess |