Fermat's Last Theorem simple proof impossible?
in [Math]
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From: James Waldby on 24 May 2010 17:08 On Mon, 24 May 2010 15:47:50 -0400, DRMARJOHN wrote: [ > Gerry Myerson wrote ] <-- snipped attribution unsnipped >> On May 22, 11:01 pm, DRMARJOHN ... wrote: >> > The question of FLT is posed for n=3 and more. >> >> Yes - and that means that any proposed solution must refer, >> at some point, to the assumption that n is at least 3. >> Nothing in your exposition referred at all to the value of n=3 > > I always refer3 to FLT. Does not that mean that N is 3 or more? What do you mean by "I always refer3 to FLT" ? Please proofread, or have a friend do so, before you post. Also, please don't snip the attribution lines. Perhaps you mean that when you are writing about Fermat's Last Theorem you assume n > 2 in the a^n + b^n = c^n equation. If you make this assumption in an attempted proof, you should state the assumption, and show that it matters in the proof. Here is a restatement of what GM told you: 1. What you wrote made no distinction between n > 2 and other n. 2. Thus, if what you wrote were correct, the result would apply to all n. 3. But the result does not apply when n=2. [Example: 3^2 + 4^2 = 5^2.] 4. Therefore, what you wrote does not apply to all n. 5. Because (4) contradicts (2), what you wrote is incorrect. [GM wrote:] >> and as a consequence, if it were valid, it would apply as well >> to n = 2. Thus, your exposition is invalid. -- jiw
From: J. Clarke on 24 May 2010 17:52 On 5/24/2010 3:47 PM, DRMARJOHN wrote: >> On May 22, 11:01 pm, DRMARJOHN<MJOHN...(a)AOL.COM> >> wrote: >> >>> The question of FLT is posed for n=3 and more. >> >> Yes - and that means that any proposed solution must >> refer, >> at some point, to the assumption that n is at least >> 3. >> Nothing in your exposition referred at all to the >> value of n=3 > > I always refer3 to FLT. Does not that mean that N is 3 or more? No, it doesn't. You have to show that your argument does _not_ apply for N=2. You can't just claim it, you have to _show_ it. >> and as a consequence, if it were valid, it would >> apply as well >> to n = 2. Thus, your exposition is invalid. >> >>> N of 3 is exponential. X^2 is straight line. >> >> Hold it right there. What in the name of Gauss is >> that supposed >> to mean? What does X^2 mean to you, what does >> "straight line" >> mean to you, and in what sense is X^2 a straight >> line? >> >> Humpty Dumpty said, "When I use a word, it means what >> I want it >> to mean. It's just a question of who's to be master." >> I think >> we have wandered through the looking glass. >> -- >> GM
From: DRMARJOHN on 26 May 2010 05:15 > On 5/24/2010 3:47 PM, DRMARJOHN wrote: > >> On May 22, 11:01 pm, DRMARJOHN<MJOHN...(a)AOL.COM> > >> wrote: > >> > >>> The question of FLT is posed for n=3 and more. > >> > >> Yes - and that means that any proposed solution > must > >> refer, > >> at some point, to the assumption that n is at > least > >> 3. > >> Nothing in your exposition referred at all to the > >> value of n=3 > > > > I always refer3 to FLT. Does not that mean that N > is 3 or more? > > No, it doesn't. You have to show that your argument > does _not_ apply > for N=2. You can't just claim it, you have to _show_ > it. > > >> and as a consequence, if it were valid, it would > >> apply as well > >> to n = 2. Thus, your exposition is invalid. > >> > >>> N of 3 is exponential. X^2 is straight line. > >> > >> Hold it right there. What in the name of Gauss is > >> that supposed > >> to mean? What does X^2 mean to you, what does > >> "straight line" > >> mean to you, and in what sense is X^2 a straight > >> line? > >> > For those who are teachers: if a student asks you why does not the equation for solving x^2 + y^2 = z^2 apply to x^3 + y^3 = z^3 ? How do you explain that? Some questions seem to arise because how I have constructed the graph is unclear. The y-axis goes from 0 to 1. Placed on this axis is (0< R <1), with high points: .1,.2,.3...9. For (A^n)+(B^n)=1, A^n is [.5 to 1), B^n is (0 to .5], thus there is .5+.5=1, .4+.6=1, .3+.7=1, etc. On the y-axis are the exponents from 3: 3,4,5...approaching infinity. The ordinate for A of A^n at .5 is not placed at .5. For the exponent 3, the cube root of .5 is .79370..., the ordinate above 3 is placed at .79370..., this is the base, the A of A^n for exponent 3 for .5. Then, for exponent 4, the root for .5 is .840896415...Thus the ordinate for exponent 4 is placed at .8408....The graphing for A^n is for the succession of exponents. This then compares the differences of A as the exponents increase. A different graph then represents , for example, the expansion of A and B, for instance, for A^6 & B^6 for A = .891 and B = .890, the A and B just different than .890898718...the irrational base of .5. An illustration of the approach for A^n + B^n = C ^n, with C^n = 1 A^6 + B^6 = 1 A = .891 B = .890 A^2 = .793881 B^2 = .792100 A^3 = .707347971 B^3 = .704969000 A^4 = .630247042161 B^4 = .62742241000 A^5 = .561550114565451 B^5 = .55840594490000 A^6 = .500341152097816841 B^6 = .496981290961000000 A^6 + B^6 = .997322442958816841 Between A & B, between .891 and .890, is an irrational number, .890898718... This irrational number is the 6th root of .5000. Tis illustrates the statement that a pair of rational numbers, the A & B, are above and below an irrational root. The .891 is > .89089..., the .890 is< .89089... . The A ends in 1 at the third decimal, the B ends in a 0 at the 3rd decimal; they are one digit apart. The rational roots are minutely different than the irrational root. A ends in an odd number, therefore A^6 is also an odd number. B is an even number, therefore B^6 is an even number. If .8909 and .8908 were used, the 9 at the fourth decimal of A would be 1 at A^6. For B, the 8 (of the fourth decimal place), would be 4 at B^6. Again the odd 1 of A^ 6 and the even 4 at B^6 add to form an odd number, which cannot be the last digit of 1.000.
From: spudnik on 26 May 2010 15:39 so, did you show that your "results" did not occur for n=2? it is hard to read what you mean, but it might just be your English. so, just try to read Shakespeare, "or else." the very first element of "proof" is from Liebniz, to show "necessity & sufficiency," although showing just one of them is also good. so, you just have to set up a wordproblemma, using those two words. or else, put it into Sanskrit, firstly, if you are classically literate in that language. > Some questions seem to arise because how I have constructed the graph is unclear. The y-axis goes from 0 to 1. Placed on this axis is (0< R <1), with high points: .1,.2,.3...9. For (A^n)+(B^n)=1, A^n is [.5 to 1), B^n is (0 to .5], thus there is .5+.5=1, .4+.6=1, .3+.7=1, etc. On the y-axis are the exponents from 3: 3,4,5...approaching infinity. The ordinate for A of A^n at .5 is not placed at .5. For the exponent 3, the cube root of .5 is .79370..., the ordinate above 3 is placed at .79370..., this is the base, the A of A^n for exponent 3 for .5. Then, for exponent 4, the root for .5 is .840896415...Thus the ordinate for exponent 4 is placed at .8408....The graphing for A^n is for the succession of exponents. This then compares the differences of A as the exponents increase. > > A different graph then represents , for example, the expansion of A and B, for instance, for A^6 & B^6 for A = .891 and B = .890, the A and B just different than .890898718...the irrational base of .5. > > An illustration of the approach for A^n + B^n = C ^n, with C^n = 1 > > A^6 + B^6 = 1 > > A = .891 B = .890 > A^2 = .793881 B^2 = .792100 > A^3 = .707347971 B^3 = .704969000 > A^4 = .630247042161 B^4 = .62742241000 > A^5 = .561550114565451 B^5 = .55840594490000 > A^6 = .500341152097816841 B^6 = .496981290961000000 > > A^6 + B^6 = .997322442958816841 > > Between A & B, between .891 and .890, is an irrational number, .890898718.... > This irrational number is the 6th root of .5000. Tis illustrates the statement that a pair of rational numbers, the A & B, are above and below an irrational root. The .891 is > .89089..., the .890 is< .89089... . The A ends in 1 at the third decimal, the B ends in a 0 at the 3rd decimal; they are one digit apart. The rational roots are minutely different than the irrational root. A ends in an odd number, therefore A^6 is also an odd number. B is an even number, therefore B^6 is an even number. If .8909 and .8908 were used, the 9 at the fourth decimal of A would be 1 at A^6. For B, the 8 (of the fourth decimal place), would be 4 at B^6. Again the odd 1 of A^ 6 and the even 4 at B^6 add to form an odd number, which cannot be the last digit of 1.000. thusNso: Death to the lightconeheads; long-live Minkowsksi! thusNso: "real-valued time" is why, we have quaternions; it's the "scalar" in Hamilton's lingo of vectors. now, you mentioned tensors, and that is apropos, because it is used for stress & strain, which are clearly irrereversible; perhaps, that is one of the first math-physics examples of it. thusNso: I haven't read _Disquisitiones_ in Latin, either, but there are good translations & it is highly recommended by the LaRouchies ... they should put it on their website, like they have *Les OEuvres du Fermatttt*, but you can look at some cool tutorials, in the meantime, at wlym.com. thusNso: I never read a word about Palin's hubbie's Seccesh "movement" in the Liberal Media (Owned by consWervatives) and that is sort-of the issue in AZ. I'm all for kids whose parents managed to sneak across the border & give birth, but I was taken aback by the "sense of entitlement" that the older kids have, about college (the DREAM Act; I stated to a group of them, that crossing the border is essentially a Mexican "rite of passage," and it is certainly not very dangerous as a proper hike, if you check the FAQs and maps & so forth from the Mexican goment (and those advocacy/ haven groups in the USA). well, it's either that or college *in* Mexico, or you'll probably be made to join a gang. La Raza d'Atzlan are openly racist, not just by their title; at least, that's the impression that I got, attending one of their meetings at UCLA, two or three years ago -- it's in their God-am constitution. of course, teh real problem is "free trade," and this is already here to roost; the little spill in the Gulf is being used by British Petroleum -- which is also the #1 driller in the Alaska North Slope, that Ted Palin works for -- to creata an "outsourcing" mandate to solve the problem, because we can't do it with our post-industrial cargo cult. well, screw it; read LaRouche, if you want to know the history with Lincoln and his "Spot Resolutions;" Cinco de Mayo should be a pan-american holiday! thusNso: Dear AG candidate Kelly; no change from Jerry Brown's '69 "platform," eh? it is intolerably stupid, insofar as we do need "fossilized fuels TM (sik)," to not get our share from our own "reserves." really, though, it is merely biomass, and the techniques have progressed since '69. Dubya's bro's ban offshore of Florida (and Louisiana) seemed like a tactical maneuver to support the oilcos' scarcity programme in our state. (why O why O why do folks believe, that the oilcos did not support the Kyoto Protoccol, which was just another cap'n'trade "free trade" nostrum, that Dubya'd have undoubtdely signed, if he had been told?) British Petroleum, the balls-out advocate of cap'n'trade, "Beyond Petroleum," is also the biggest company in the Alaska North Slope -- doesn't any body wonder, why no-one asked Palin about her BP-employed hubbie, and his Seccesionist ideals? one must take into consideration, with all of the hype about it, that oil comes out of the ground underwater in "seeps," under pressure. so, how much would come out, if BP et al ad vomitorium were not pumping like crazy? Waxman's current cap'n'trade bill just mandatorizes the huge, voluntary cap'n'trade since 2003 -- tens of billions in hedging per annum. what the Liberal Media (Ownwd by consWervative) don't talk about, is that he brought the first cap'n'trade bill in '91, under HW (who worked with Gore on the Kyoto cap'n'trade). what it amounts to, as Waxman basically admitted to, when he was at UCLA, is "let the arbitrageurs raise the price of energy, as much as they can in the 'free market' -- free beer, freedom!" a small, adjustable carbon tax would achieve the same ends -- as I even read "in passing" in a guest editorial in the WSUrinal, as well as from an "expert" in a UCLA seminar, but who said that it was (some how) "politically impossible" -- without being the Last Bailout of Wall Street (and the City of London). --mister Kelly, please, take me off of your list, Brian H. --Light: A History! http://wlym.com
From: hagman on 26 May 2010 17:17
On 26 Mai, 15:15, DRMARJOHN <MJOHN...(a)AOL.COM> wrote: > > On 5/24/2010 3:47 PM, DRMARJOHN wrote: > > >> On May 22, 11:01 pm, DRMARJOHN<MJOHN...(a)AOL.COM> > > >> wrote: > > > >>> The question of FLT is posed for n=3 and more. > > > >> Yes - and that means that any proposed solution > > must > > >> refer, > > >> at some point, to the assumption that n is at > > least > > >> 3. > > >> Nothing in your exposition referred at all to the > > >> value of n=3 > > > > I always refer3 to FLT. Does not that mean that N > > is 3 or more? > > > No, it doesn't. You have to show that your argument > > does _not_ apply > > for N=2. You can't just claim it, you have to _show_ > > it. > > > >> and as a consequence, if it were valid, it would > > >> apply as well > > >> to n = 2. Thus, your exposition is invalid. > > > >>> N of 3 is exponential. X^2 is straight line. > > > >> Hold it right there. What in the name of Gauss is > > >> that supposed > > >> to mean? What does X^2 mean to you, what does > > >> "straight line" > > >> mean to you, and in what sense is X^2 a straight > > >> line? > > For those who are teachers: if a student asks you why does not the equation for solving x^2 + y^2 = z^2 apply to x^3 + y^3 = z^3 ? How do you explain that? > > Some questions seem to arise because how I have constructed the graph is unclear. The y-axis goes from 0 to 1. Placed on this axis is (0< R <1), with high points: .1,.2,.3...9. For (A^n)+(B^n)=1, A^n is [.5 to 1), B^n is (0 to .5], thus there is .5+.5=1, .4+.6=1, .3+.7=1, etc. On the y-axis are the exponents from 3: 3,4,5...approaching infinity. The ordinate for A of A^n at .5 is not placed at .5. For the exponent 3, the cube root of .5 is .79370..., the ordinate above 3 is placed at .79370..., this is the base, the A of A^n for exponent 3 for .5. Then, for exponent 4, the root for .5 is .840896415...Thus the ordinate for exponent 4 is placed at .8408....The graphing for A^n is for the succession of exponents. This then compares the differences of A as the exponents increase. > > A different graph then represents , for example, the expansion of A and B, for instance, for A^6 & B^6 for A = .891 and B = .890, the A and B just different than .890898718...the irrational base of .5. > > An illustration of the approach for A^n + B^n = C ^n, with C^n = 1 > > A^6 + B^6 = 1 > > A = .891 B = .890 > A^2 = .793881 B^2 = .792100 > A^3 = .707347971 B^3 = .704969000 > A^4 = .630247042161 B^4 = .62742241000 > A^5 = .561550114565451 B^5 = .55840594490000 > A^6 = .500341152097816841 B^6 = .496981290961000000 > > A^6 + B^6 = .997322442958816841 > > Between A & B, between .891 and .890, is an irrational number, .890898718.... > This irrational number is the 6th root of .5000. Tis illustrates the statement that a pair of rational numbers, the A & B, are above and below an irrational root. The .891 is > .89089..., the .890 is< .89089... . The A ends in 1 at the third decimal, the B ends in a 0 at the 3rd decimal; they are one digit apart. The rational roots are minutely different than the irrational root. A ends in an odd number, therefore A^6 is also an odd number. B is an even number, therefore B^6 is an even number. If .8909 and .8908 were used, the 9 at the fourth decimal of A would be 1 at A^6. For B, the 8 (of the fourth decimal place), would be 4 at B^6. Again the odd 1 of A^ 6 and the even 4 at B^6 add to form an odd number, which cannot be the last digit of 1.000. What people have repeatedly tried to ask you is: What is wrong with the following argument? (It is a slightly modified quote of your argument and if this modified quote were correct, it would sho that x^2+y^2=z^2 has no positiv integer solutions) The y-axis goes from 0 to 1. Placed on this axis is (0< R <1), with high points: .1,.2,.3...9. For (A^n)+(B^n)=1, A^n is [.5 to 1), B^n is (0 to .5], thus there is .5+.5=1, .4+.6=1, .3+.7=1, etc. On the y- axis are the exponents from 2: 2,3,4,5...approaching infinity. The ordinate for A of A^n at .5 is not placed at .5. For the exponent 2, the square root of .5 is .70710678..., the ordinate above 3 is placed at .70710678..., this is the base, the A of A^n for exponent 2 for . 5. Then, for exponent 3, the root for .5 is .79370...Thus the ordinate for exponent 3 is placed at .79370... The graphing for A^n is for the succession of exponents. This then compares the differences of A as the exponents increase. A different graph then represents , for example, the expansion of A and B, for instance, for A^2 & B^2 for A = .707 and B = .708, the A and B just different than .70710678...the irrational base of .5. hagman |