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PROOF OF 1:3 PUBLISHED TODAY
Posted:
Sep 21, 2013 12:26 PM


International Journal of Applied Mathematical Research, 2 (4) (2013) 452454 ©Science Publishing Corporation www.sciencepubco.com/index.php/IJAMR Pythagoras 1:3, an expression of the finite universe of mathematics Vinoo Cameron Hope research, Athens, Wisconsin, USA *Corresponding author Email: frontier.com Abstract The author has published several papers which are may be hard to understand and stand as a testament to the discovery of 1:3 mathematics which is an absolute accommodation of numbers, prime numbers placement , and precise angles. The author has published a continuous Prime sieve at the divergence of (5/6, 1/6) and (1/3 and 2/3), a final attempt is made here to communicate the simple mathematics of the finite universe of mathematics to the dead current mathematics, that in the authors opinion is dead to the absolute reality of mathematics, as a form of approximate mathematics and has offered this manuscript for review at any mathematical venue. This discussion is about 1:3 Pythagoras (?1+?9=?10 hypotenuse) and points to the clear mathematical relationship between 19 and 3^2. The span of the divergence at 1:3 is clearly2+ 1/ (3^2) from the base of 2. Current mathematics is currently operating in the approximation of their myriad current theory, and they seem content. Keywords: Pythagorean 1:3, configuration of 1(5/6, 1/6:1/3, 2/3), Arabic numerical (1^23^2). 1 Introduction This is expressed by the ten published papers referenced below, the divergence of this angle is in the configuration of a cone in two planes, one of these planes is represented by the 1/3 + 2/3=1, and the other plane is represented by 1/6 +5/6.=1 As shown previously the 1:3 divergences accommodates the numbers naturally, and the equalization of numbers is at 4:12(1:3) as shown in the last paper on end calculus. 5+11+17=9+11+13 (175=12; 139=4) 2 Mathematics The Arabic numerical define numerals from 1^23^2(19). Pythagoras at 1:3 represents the squared expression of the Arabic numerals (1^2+3^2) =? (10). There is not much explaining to be done here here as the author does not feel obliged to explain the simplicity of his work any further for a mathematics that excels at approximations (whole numbers are absolute). 3^2=9, a square with sides of 3 each Reciprocal Value = 1/ (3^2) =0.111111111111. Now in applying the Pythagoras theorem to the square, a linear advancement of 3 is in one plane, followed by ascension of 1 in the perpendicular plane, which yields a diagonal hypotenuse of ? (10). 1 19 2. 2. 1 : 3 2 3 1 9 n n ? ? ? ? ? ? ? ?? ?? [? (10)]^2[? (9)] ^2=1 10^29^2=19 19/19=1 (



