Archivo para abril 2015

On the demonstration and refutation of Fermat’s last theorem and the Pythagorean’s one   Leave a comment

I consider Fermat’s last theorem is true to the same extent that the Pythagoras’s theorem is false. But it could be said too they both are wrong, or even that Fermat’s Last theorem is at the same time right and wrong depending on the perspective of the observer.

When we create a square area we make it taking as spatial reference the coordinates X and Y. With them, we create a referential length, the side of our square. If we trace a diagonal from those XY coordinates of our created square, that is preserving our first referential length of measure and spatial coordinates, we are creating a new referential length with new referential coordinates. The Z coordinate related to X and Y does not have the same referential length than them and it is of a different nature. To create a new referential Z coordinate is similar to create a new prime number each time we find a disproportion between numbers. Here, the disproportion between coordinates is created by displacing the XY coordinates toward the right or left side, forming an angle smaller than 90 degrees. It is not possible, if we want to preserve the rationality of our mathematical creations, to compare thighs of different nature, related to different referential lengths without taking consideration of their differences and making them proportional in some extent.

The fact that the area got raising the hypotenuse to the square has the same space than the addition of the areas got raising the legs of the right triangle to the square, is not enough to say that they are equal. They can be coincident spaces but never equal. It is logically impossible that they are equal because the addition of (leg) a + (leg) b is not equal to (hypotenuse) h. Our coincident areas are created from different referential lengths and spatial coordinates, and are of a different nature.

When we trace a diagonal from X or Y we are moving, displacing spatially those coordinates. If we preserve that XY coordinates as static references and introduce a displacement represented by Z, Y becomes Z, it supposes to transform our referential space introducing a new plane on a same space shared by XY and Z.


The square area of the hypotenuse is a hybrid area, a multi-planar area created with different referential lengths.

In this sense the Pythagorean theorem is not right.

And it is confirmed by the fact that the Fermat’s last theorem is true (i will clarify this affirmation later). I think it is not necessary to know a lot of mathematics to verify (or to refute) the Fermat’s last theorem. It is only necessary to use the common sense, and to think in a sensible way rejecting any irrational solution. Most of the current mathematics have been developed upon the base on irrationality, upon the impossibility to find out a rational and simple solution to the discovering of the disproportion created when comparing the diameter and the perimeter on the circumference, the hypotenuse and the legs on the right triangle. This elemental mathematical mistake still sustains all the scientific development of our western civilization.

So we think that a + b is not = c but a^2 + b^2 = c^2 but a^3 + b^3 is not = c^3. There is obviously an incongruence (or various ones).

When we create a cubic area (n = 3, n different than 2 in the Fermat theorem), we introduce new spatial displacements. We preserve the referential and plane coordinates XY and the created Z, but we displace those coordinates toward the new vertical plane too. That variation does not imply the transformation of referential lengths when it consists on 90 degrees. But creating the cubic figure we displace 45 degrees the Y coordinate which becomes X’, becoming coincident with the Z coordinate. We are combining planes. And if we relate in a same figure the existent Z with the created X’ or the existent Y with the created Z’, we will combine different referential planes of different dimensions (the square dimension and the cubic one) in a derivative way.


In the picture above we can look at the larger cube of the hypotenuse from the perspective of the XY coordinates (which rule the square area of the legs) or from the perspective of the created X’Y’ coordinates (which rule the area of the hypotenuse).

In both cases the perspective of the cube is different but the cube does not vary at all. The cubic volume is the same independently the angle we are observing it from.

But the problem is the perspective we had when we started to building our larger cube. Because this initial perspective will determine the way which we will build our cube in.

When we traced the hypotenuse of the first right triangle, we did not do any transformation. But we make a transformation when we create the side L2 (which provides the perspective) on the smaller cubes of the legs. That side is exactly the same thing that our first (and untransformed) hypotenuse, it is a diagonal traced from the square coordinates and placed on the Z coordinate. But here we need to transform it for making it proportional to our referential coordinates XY. So I compared the length of the side of the square area with the length of its hypotenuse (we are still in the smaller squares) and projected its intersection point on the referential X or Y coordinates. And I use the result of that projection – which I called “d” for calculating the side in perspective L2. L2 is formed by L1 – d.

Now look at the larger cube. We are looking at it from the perspective of the XYZ coordinates but because its referential side is displaced toward the right we see it in a different way. For examining the cube on its own and natural perspective – in a similar way than the cubes of the legs – we need to displace the coordinates toward right. If you move the picture in that way you will see this different perspective:


The side on perspective L2′ (related to the coordinates XYZ) becomes L1′ on the coordinates X’Y’). The hypotenuse on Z becomes on X’. We have now totally new coordinates X’Y’, equal to XY but in a different plane, a displaced one.

If we created our larger cubic volume of the hypotenuse from its own X’Y’ coordinates, we will create it in the same way than we created the smaller cubes of the legs. So we will make proportional the diagonal to the square coordinates X’y’ for calculating the side of Z’ on perspective. Here the operation will be L2′(X’Y’) = L1(X’Y’) – d’.

In this case I calculated d’ displacing the hypotenuse – which has the role of referential length – toward the new larger hypotenuse which is coincident with the X coordinate. I am not considering the red dots on the cube built from this perspective.

If we turn the figure toward the left again we will see the cube with the other previous perspective again.

But if we build the cube from this other perspective, we need to make the transformation of the hypotenuse in a different way. Because here our referential length is the hypotenuse itself in its diagonal role related to XY, that is to say as Z coordinate. So here we cannot transform the hypotenuse in the way I did on the smaller cubes (or in the larger cube from its own X’Y’ coordinates); in those cases I calculated the side on perspective L2 or L2′, making the hypotenuse proportional to the square coordinates by subtracting d from the referential XY (or X’Y’) length. Now, from this other perspective, it is necessary the opposite transformation,to make the square XY lengths proportional to the referential diagonal (the hypotenuse) length. So it is necessary to add d to the length of the sides placed on the Y coordinate.

The cube built in this way is larger than the previous one and its volume includes the red dots. It is already not a simple question of perspectives. I have changed the actual volume of the cube depending on the perspective I had when I built it.

So, what perspective are we building the cubes of the hypotenuse? The cubes on the Fermat’s last theorem are cubes built from its own and natural referential coordinates X’Y’? in that case, its volume will be different than the addition of the two other cubes because all those cubes were built on the base of a proportional referential length, XY, X’Y’ which is inconsistence with the Pythagorean Theorem in which the square areas are built on the base of non proportionated referential lengths and coordinates.

But if we build the larger cube adding d to L1′ in the explained way, adding the red dots to the volume, we will have preserved the coincident disproportion existent between the square areas of the hypotenuse and the legs in the Pythagorean theorem (I made that statement without actually measure those volumes but I think it should be right because the larger cube is now consistent with the original disproportion existent in the Pythagorean theorem between the legs and the hypotenuse, between the referential lengths of XY and Z).

So we can refute or demonstrate in both cases the Fermat theorem depending on our referential perspective when building the cubes.

The Pythagorean theorem appears incongruent in any case, although its resultant square areas can be used knowing they are only coincident, not equal, and why: the square area of the hypotenuse is a multiplanar, multidimensional, space.

A right triangle is not a square divided by 2, it’s a square which has become a triangle because of the displacement of the Y coordinate. If we want to use it in terms of square we need to take account that displacement.

The Pythagorean’s theorem does not compare square areas, although they result coincident. It compares triangular areas, which is different because the referential length is neither the length 1 nor the referential square 1. The reference used on the Pythagorean comparison is a new prime length formed by the mentioned displacement of the coordinate Y, the reference is not the side of the square (1 in its prime representation) but the hypotenuse.


The square of 1×1 sides is the unique prime area that we have. It is equivalent to the primary number 1. When it comes to numbers, each time we find a disproportion we invent a new prime number which is proportionated with the number 1. So, number 5, for example, is neither proportionated with 2 nor 3, but it’s proportionated with 1; and we can reestablish the lost proportion fixing the disproportion emerged, by creating a new prime number.

But in the case of the areas there is the problem that the disproportion appears inside the primary area of 1 itself when we tracing the diagonal inside; When it comes to areas we lose our prime reference for reestablishing the disproportion.

So, We can invent a new prime square area related to the hypotenuse instead of the side, which is equivalent to invent a new number 1 different than the existent one. But we need to be willing to accept there are infinite different number ones and be able to related them (with all their derivative numbers, lengths and areas).

Or we can try to make that irrational reference created by the hypotenuse – irrational as disproportionated with our prime reference 1 – by making it proportionated with our referential square taking consideration how that disproportion was created.

Looking at how the disproportion created by the hypotenuse is reestablished in the Pythagorean theorem, we could deduce easily how reestablish the disproportion created in prime square area of sides 1 when tracing the hypotenuse inside of the square.

The square of length 1 placed on the coordinate Y save the disproportion acting as the rational side of a prime rational square of sides 1, and as the rational hypotenuse of an irrational square with irrational sides less than 1. For creating that prime irrational square with a rational hypotenuse is necessary to displace the square coordinates used as reference for building the rational square.

So we could reestablish the irrational disproportion created by the hypotenuse relating tits square area with another square area with irrational sides but a rational hypotenuse inside.


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