Item 8. Constructing the Square Area of Two Proving No Irrationals. 6 Total Pages

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1 Item 8 Constructing the Square Area of Two Proving No Irrationals 6 Total Pages 1

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3 We want to start with Pi. How Geometry Proves No Irrations They call Pi the ratio of the circumference of a circle to its diameter, even though they have no ratio. Someone had proved that there could not be one. Is that really true? Our claim is that the correct Pi has a ratio 201/64, as a decimal Geometry gives a way to prove that this is true. We are going to use one as our diameter, so that one times Pi becomes our circumference length, so that the number for Pi and our circumference length are one and the same. If our diameter is one, our radius is 1/2, that can construct a hexagon within our circumference putting 6 points on our circumference giving us six equal lengths. This would mean that the number for Pi that is the same as our circumference length would have to be dividable by 1, 2, 3, 4, 5, and 6 and it is. This is true for , yet impossible for the present day calculation for Pi. You can t even list it, yet alone divide it into 6 equal divisions. You can also use two as our diameter length so that our circumference length is here again our radius of one can construct the six lengths of our straight line hexagon, where again, we put 6 points on our circumference line, so that is also dividable by 1,2 3, 4, 5, and 6, and it is. Thanks to geometry we have a correct ratio for Pi. Applying Quantum Math First we apply quantum math of reducing our ratio to a single integer by addition to show how it is related to our geometry so that 201/64 = 3/10 = 3/1 = 3 our hexagon length. Then, for our decimal part = 21 = 3, again our hexagon, thanks to quantum math. If we use our listing of countable numbers starting with one as our diameter lengths, our ever increasing multiplication by Pi where we apply quantum math of reducing to a single integer by addition we will get this infinite repeating set 3, 6, 9, 3, 6, 9... infinite. 3

4 It appears that 3, 6, and 9 represent infinity. If you look at item 15 we show our three columns of integers for our twin primes, the only 3 integers is our central column between our twin primes is 3, 6, and 9, that total 18. For our first twin prime we also had only 3 integers 2, 5, and 8 that totals 15, column one. Then for our second prime column, column three, our three integers were 1, 4, and 7, that totaled 12. These 3 columns account for all of our single quantum integers one through nine, that total 45 our 34 curved lines plus our 11 time lines of our Poincare one geometry. When we form our quantum fraction from our dual number 45 we have 4/5 the positive parts of 31/32 that reduces to 4/5 from our probability model (A) M-Theory that produced our probability models (B), (C), (D), (E), and (F). thanks to the relativity of numbers. Later we will show how our 3 infinite sets of integers 2, 5, and 8, then 3, 6, and 9 plus 1, 4, and 7 proves that our Poincare Geometry are true. In these 3 sets our primes total 17, the 17 curved lines for each half of our Poincare one geometry. Then our single time integers 1, 2, 4, 5, 7, and 8 total 27 that represents the lowest state for bosons that produced the mass structure of our universe, protons, and neutrons. Whole Number Squares with Square Roots While we can form whole number and fraction squares that have square roots, we want to just stay with our whole number applications, as it was here that false applications began. By using our countable numbers we can establish what squares we can establish that have whole number square roots. We can do so by squaring our whole numbers so that one squared equals one that has a square root of one. Next we square two so that two squared equals 4, so that the square root of 4 = 2. We can square 3 so that 3 squared = 9 as the square root of 9 is 3. Then we have 4 squared = 16 with its square root of 4. For 5 we have 5 squared equals 25 with its square root of 5. 4

5 Then, we have 6 squared and so on. We can then make a list of squares that have square roots that can exist 1, 2, 4, 9, 16, 25, 36, 49, 64, 81, and so on where all of the dual numbers have application of our two geometries and our probability models. We will show, there exist no other squares between these squares that have a square root. What happened with mathematics when they thought they proved that the square root of two was irrational. They began to say that the square root of three was irrational, also the square root of 5, 6, 7, and so on, were irrational when the truth of this matter was that these squares did not exist. Their lived a man who once said every house is constructed by someone, however, he who constructed all things is God. How Geometry Proves No Irrational Numbers Look at our first page, where we show how to construct the square area of two, by two square root triangles of one half base times height. For illustration two we show our square area of two within our two by two square. Here it is easy to see that the square area of two is exactly ½ of our two by two square, as the parts that lie outside of our area of two square can represent our vacuum space (equal representation) of our Poincare one geometry another part of P = NP. If you cut out our square area of two and separate our two square root triangle by cutting along our base line we have two separated square root triangles. Now, cut our height line of each that will give 4 total parts where you can form two squares of area one, with two sets of square roots where the square root of one is one. Next, we want to show importance of listing the squares that exist. You can look up the golden ratio that they feel is associated with the Fibonacci Sequence. They start with a line segment and label its two end points (A) and (C). then, they added a point (B) between (A) and (C). as a result they claimed a ratio equation of AB to BC. 5

6 I think that they forgot that they constructed (A) to (C) by the motion of forming that line of (A) to (C). we can also use motion to form point (B) by going backwards from (C) then forward again to our end point (C). What does our motion (no dought related to time) produce, one line length from (A) to (B) then 3 lengths of lines from (B) to (C) showing that they can t have the ratio that they claim. Next, they claim that they have an exact value for their claim of the golden ratio. They start with one (our additional one of Fibonacci Sequence) so that we have one plus the square root of five over two. Here is the important part. There exist no square of area five, deleting the square root of five. The only thing that we have left is our additional one from our Fibonacci Sequence that when divided by two goes on forever without ever reaching one, as we have ½, ¼, 1/8, 1/16, and so on so that the golden ratio does not exist, fictitious. Thanks, Richard Eicholtz 6