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Topic:
Fermat's Last Thm. A short proof
Replies:
1
Last Post:
Feb 6, 2012 1:24 PM




Re: Fermat's Last Thm. A short proof
Posted:
Feb 6, 2012 1:24 PM


I don't see that XX, YY and ZZ are necessarily integers, nor these factors m, k, j ... . Rational numbers can be as small as you please. An integer has infinitely many rational "factors." So where is the contradiction?
On Mon, Feb 6, 2012 at 11:38 AM, Michael C Weir <discussions@mathforum.org> wrote: > Fermat's Last Theorem ? simple proof. > > Let's assume X^n + Y^n = Z^n for X,Y,Z being rational numbers, with n > 2 and n= whole number. > > For ease of demonstrating the argument, let n = 3. It will be easy to translate the argument for any n later. > > The argument goes like this: > > Because we assume X^n + Y^n = Z^n, we can make certain statements that are true about X, X^2, Y, Y^2, Z, Z^2. With these statements, we can solve Z in terms of a mix of numbers, including Y, and also X. > > Most of the numbers are rational, due to assumptions or calculations. There is one number that it is impossible to determine if it is rational or irrational, specifically (m^3 + 1)^1/3 for the case n =3. Examples can be found where this number is either rational or irrational. > > If it is irrational, then we are finished, as either Y or Z or both are irrational (m^3+1)^1/3. This is a contradiction with original assumption. > > If it is rational, then we can rewrite the original equation X^3 + Y^3 = Z^3 as > (XX)^3 + (YY)^3 = (ZZ)^3, where XX, YY, and ZZ are all smaller than their counterparts X, Y, and Z by a factor of (m^3 + 1)^1/3, which we have chosen to be rational and which is larger than 1. > > If we go through the same reasoning with this new equation, we again come to a new conclusion that (ZZ)/(YY) = (k^3 + 1)^1/3 , which while still larger than 1, the k in the equation is smaller than m. > > If (k^3 + 1)^1/3 is irrational, then we are finished, as either ZZ or YY, or both are irrational. > > But if (k^3 + 1) is rational, then this too can be factored out of the equation, and a new solution could be found, say j. (j^3 + 1)^1/3 is now rational or else there is a contradiction. Each time the factor is getting smaller, ie m> k > j. > > It is not even necessary for there ever to be a stop to this process. There would be an infinite number of factors of Z, and you could never determine if there was even a repeating set of numbers, as the factors are continually getting smaller. But isn't this just a variation of the definition of an irrational number? If the factors of a number are infinite and different from each other, can there ever be a repeating set of numbers? There's always another factor waiting in the wings to disprove any collection of digits. > > That is the argument. Here is the math. > > If X^3 + Y^3 = Z^3, then X + Y = A*Z where 1<A<2^1/2 and X^2 +Y^2 = B*Z^2 where > 1<B<A<2^1/2 > > > > > reason ; X is chosen to be smaller of X,Y. Maximum value of X is less than that of Y, addition of both is less than 2 Y. If Z is greater than 2^1/12 Y, then Z^2 is greater than 2Y^2, and for every n>2 thereafter. > > So, we have 3 equations now > > 1.X + Y = A*Z > 2.X^2 + Y^2 = B*Z^2 > 3.X^3 + Y^3 = Z^3 > > Because X is smaller than Y (because we chose X to be smaller), let X = mY where 0<m<1. > > Then our 3 equations become > > 1 m*Y + Y = A*Z > 2 m^2*Y^2 = B*Z^2 > 3 m^3*Y^3 + Y^3= Z^3 > > 3 becomes (m^3 + 1) Y^3 = Z^3 > > or further Z^3 / Y^3 = m^3 + 1 > > or Z/Y =(m^3 + 1)^1/3 > > If (m^3 + 1)^1/3 is irrational, then either Y, Z or both are irrational. So assume it is rational, and the argument above applies.



