Fermat's Last Theorem simple proof impossible?
in [Math]
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From: Ostap Bender on 26 May 2010 18:41 On May 26, 2:17 pm, hagman <goo...(a)von-eitzen.de> wrote: > On 26 Mai, 15:15, DRMARJOHN <MJOHN...(a)AOL.COM> wrote: > > > 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...... You forgot to substitute "2" for "n". Here is the fully substituted version: For FLT, when A^2 + B^2 = 1, (0<R<1) all real numbers between 0 and 1, there is an infinite # of solutions: they all have irrational A & B or A or B. Consider .5 + .5 = 1 as the first solution. When A^2 = .5, the root or base of .5 is A -- it is irrational. Both A & B are irrational. There are two sets of solutions. The first is with the bases A & B irrational. There is a first A^2 & B^2 above and below .5 that both have irrational bases (the roots), A & B. There is a second A^2 & B^2 above and below .5 that both have irrational bases, A & B. There is a third A^2 & B^2 above and below .5 that both have irrational bases, A & B. ....There is an Nth A^2 & B^2 above and below .5 that both have irrational bases, A & B. There are an infinite # of such pairs. The second repeats the first set, except only one of a pair A & B is irrational, the other is rational. The two sets of solutions comprise the complete set of solutions. Irrational roots have a terminus at an infinite point. This enables each pair of A^2 & B^2 to add to 1. (for the 1st set.) When a pair has a root that is truncated at any point before infinity, it loses its character that enables for success. Any such truncated root is a rational number. For the second set, when one rational base can pair with an irrational base, truncating that one irrational number, it loses its character that enables for success. A later illustration will clarify how the pair of irrational bases changing to one irrational base + one rational base is different than a construction of a rational + an irrational. For the beginning pair, the first solution of .5 + .5 = 1, both A & B have to be irrational bases. If the root of .5 is truncated at the 2 - 1000 place, it becomes a rational base. (note that all rational roots as they approach infinity, approach 1.) When the root of .5 is irrational, there is a solution. Any deviation from that becomes a rational base, and can not be a solution. The character of the irrational base enables success; the character of the truncated is non-success. Only irrational roots give solutions. All rational roots give non- solutions. Therefore x^2 + y^2 = z^2 has no solutions. ---------------------------
From: DRMARJOHN on 27 May 2010 15:30 > > 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 x-axis are the > exponents from 3: 3,4,5...approaching infinity. The > ordinate for A of A^n at .5 is placed at-- 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 > the 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. Martin Johnson
From: spudnik on 27 May 2010 22:00 you are relying on rationals, taht are decimals (in the base of ten, although some authorss will call rationals "decimals," in any integral base, which is one class of solutions (what ever it's called, in what ever we're talking about)). in any case, it is almost a standard, that one use the base that is associated with the prime exponent ... which is really the meaning of some of Fermat's theorems & challenges [*]. and that makes me very happy, then very sad ... because you're probably trying to find some guru/god/guy or some goddess, who already wrote this up in the hither & yon of Vedic psychorama ... which reminds me of A.C.Clarke and the Satellevator Daytrips, ba-doom. thusNso: Shell is about half British, but Netherlands is the big port o'call (also, the place to call when the windmill feathers, inapproprietly). why should I believe in your kind of free energy, and how could I measure it (sik) ??... maybe, it really is "free trade." > Then introduce free energy technology [ellipsis]. ------- * anyway, for those of you/us/them in need of "skills," I want to suggest Fermat's "reconstruction of Euclid's porisms;" they seem rather a r b i t r a r y , but that's just me, "you, idiota!" --Light: A History! http://wlym.com
From: DRMARJOHN on 5 Jun 2010 04:18 FLT 6-3-10 A^n + B^n = 1 This approach to structuring the problem of FLT is different than the traditional approach. First, instead of asking what A and B may yield a rational correct C, this approach begins with (A^n) and (B^n) as rational numbers derived from an unknown A and B, e.g., when (A^n) = .6 and (B^n) = .4, then .6+.4=1. It asks, what is the characteristic of the A & B that gives these rational answers. Second, this is not exploring the FLT for any certain exponent. Instead it begins with 5+.5=1 and illustrates the process for all exponents from 3 to infinity. Instead of starting with success as dependent on rational whole numbers, this begins with the statement that success of A^n and B^n comes from irrational numbers. This is a counter-intuitive approach, and thus, I would suggest, has been overlooked. 6-5-10 Martin J > you are relying on rationals, > taht are decimals (in the base of ten, although > some authorss will call rationals "decimals," in any > integral base, > which is one class of solutions (what ever it's > called, > in what ever we're talking about)). in any case, > it is almost a standard, that one use the base > that is associated with the prime exponent ... > which is really the meaning of some > of Fermat's theorems & challenges [*]. > > and that makes me very happy, then very sad ... > because > you're probably trying to find some guru/god/guy or > some goddess, > who already wrote this up in the hither & yon > of Vedic psychorama ... which reminds me of > A.C.Clarke and > the Satellevator Daytrips, ba-doom. > > thusNso: > Shell is about half British, but Netherlands is the > big port o'call > (also, the place to call when the windmill feathers, > inapproprietly). > > why should I believe in your kind of free energy, and > how could I measure it (sik) ??... maybe, > it really is "free trade." > > > Then introduce free energy technology [ellipsis]. > > ------- > * anyway, for those of you/us/them in need of > "skills," > I want to suggest Fermat's "reconstruction of > Euclid's porisms;" > they seem rather a r b i t r a r y , but that's > just me, > "you, idiota!" > > --Light: A History! > http://wlym.com
From: Gerry on 5 Jun 2010 08:58
On Jun 5, 10:18 pm, DRMARJOHN <MJOHN...(a)AOL.COM> wrote: > This approach to structuring the problem of FLT is different > than the traditional approach.... This is a counter-intuitive > approach, and thus, I would suggest, has been overlooked. I would suggest that posting this once is more than enough. What is the point in posting the exact same thing in three different places? -- GM |