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<2^1/2 and X^2 +Y^2 = B*Z^2 where
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
X + Y = A*Z
X^2 + Y^2 = B*Z^2
X^3 + Y^3 = Z^3
Because X is smaller than Y (because we chose X to be smaller), let X = mY where 0
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.
"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."
ReplyDeleteIt is such vague statements that are the reason that no mathematician is taking your attempts seriously. It is easily possible for there to be an infinitude of different rational numbers to have rational product. Eg:
a_n:=(n^2+2n)/(n^2+2n+1) is a sequence of rational numbers and the product of a_n from n=1 to infinity is 1/2.
The whole point of a rigorous proof is that every logical step must be beyond doubt, not just something that seems plausible. Being self-critical is one of the most important skills a mathematician can have.
Yes, you are right about a number made up of rationals will approach a limit that is rational number. But that is not being discussed. We are talking about 3 rational numbers raised to the same rational power and that we test the first 3 numbers in that equality to see if all 3 numbers can be factored . If the 3 numbers can be factored, then divide the 3 numbers by the common factor, and start the algorithm from the beginning again. Each time you do the test for a rational number
DeleteIf the test proves that there is a rational number that is a common factor, then you would factor out that rational number from the equation. Each time you would do this, the factor would get smaller, but so would the resulting 3 numbers in the equation. It is kind of like Russian nesting dolls.
Let us now look at a statistical approach to this whole system. Since we do not know the outcome of the test, let us assign the probability of the first factor being a rational number as ½ (there are many more irrational numbers than rational, but there are at lease ½ that are irrational) . And for the second set of numbers, that probability is also ½. So the probability of the second set of numbers and the first set of numbers having a rational factor is 1/2^2 or 1/4.
Every time you get a rational factor for the equation, you have test the next equation to see if it is also rational. So, really we are looking for the solution of 1/2^n as n goes to infinity. The limit is 0. So there is zero probability of there being an infinite string of rational numbers that will satisfy the equation when is 3.
And you can generate this same argument for n >3.
Firstly the probability of a real number being rational is zero is zero with respect to any probability measure on R as the rationals are countable.
DeleteSecondly the probability of an equation being satisfied being zero does not imply that it is never satisfied. The probability of a pair of real numbers satisfying x=y is 0. The probability of an arbitrary integer being prime is 0. If your logic was valid we could deduce that primes don't exist.
Even if your "algorithm" works and you get a sequence of decreasing rational factors you can "factor out" (which is trivially true for ALL homogenous equations), I don't know what you expect to achieve.
It is easy to also construct a sequence of decreasing rational solutions to the equation x^2+y^2=z^2 from any given rational solution, and this process does not preclude the possibility of such an initial solution existing.
First
DeleteThe probability of a real number being rational is not zero, as there exist rational numbers. In any case, it helps the argument that you will fewer tests to reach the conclusion that you have an irrational factor.
Second
You are comparing apples with oranges and grapefruits.
The probability of any number being prime is near zero, not zero. You do a procedure for a number n, dividing it by2,3, ...n-1. If none of the sequence divides the number, it is prime. If the probability was zero, you would never get any primes, which you obviously do get a large number of them..
If you do a procedure to test if any x = y, where x and y are different numbers, of course the answer is always there is no match. In that case, the probability is zero that there will ever be a match.
If you do a procedure to test whether the resulting pair of numbers is rational or not. If the answer is yes and you have a rational number, then you can trivially divide the homogeneous equation by this rational number. The resulting equation can then be run through the simple math and again produce another number pair, and same test can then be preformed.
Because you can perform this same test over and over again (if you continually get a rational number) you will have to multiply the individual probabilities together. Of course once you get an irrational result, all the number pairs that preceded this result has one number is an irrational number, a contradiction with our original assumption that all numbers are rational.
In order to say that there exists rational numbers in the whole chain of reasoning means that you have to multiply the individual probabilities an infinite number of times.
So, even if the probability is 1/2 ( which you so correctly pointed out, is too large), even then the limit is 0.
Finally, you can construct as many sequences as you want. all these sequences will be missing the key that makes this whole argument work.
That missing ingredient is the 3rd. statement that comes from the equation x^n + y^= z^n. If you only have n = 2, there is no 3rd statement.
There is a significant difference between a set being empty and of measure zero.
DeleteNothing you say here comes close to a coherent mathematical argument. If you believe otherwise, go ahead and get yourself published, famous and rich!
I am done here.