TWO VERY SPECIAL PYTHAGOREAN TRIANGLES
... Pythagorean theory of numbers, Triangular numbers and Pentagonal numbers played very important role. Rana and Darbari [1] obtained special Pythagorean Triangles, with their legs to be consecutive, in terms of Triangular Numbers and Darbari [2] found special Pythagorean Triangles, with their legs to ...
... Pythagorean theory of numbers, Triangular numbers and Pentagonal numbers played very important role. Rana and Darbari [1] obtained special Pythagorean Triangles, with their legs to be consecutive, in terms of Triangular Numbers and Darbari [2] found special Pythagorean Triangles, with their legs to ...
Proofs
... Theorem: There exists a bijection from A= [0,1] to B= [0, 2]. Proof: We build two injections and conclude there must be a bijection without ever exhibiting the bijection. Let f be the identity map from A to B. Then f is an injection (and we conclude that | A | | B | ). Define the function g from ...
... Theorem: There exists a bijection from A= [0,1] to B= [0, 2]. Proof: We build two injections and conclude there must be a bijection without ever exhibiting the bijection. Let f be the identity map from A to B. Then f is an injection (and we conclude that | A | | B | ). Define the function g from ...
13(3)
... Mathematics Department University of Santa Clara, Santa Clara, California 95053* All checks ($12.00 per year) should be made out to the Fibonacci Association or the Fibonacci Quarterly. Two copies of manuscripts intended for publication in the Quarterly should be sent to Verner Ee Hoggatt, Jr«, Math ...
... Mathematics Department University of Santa Clara, Santa Clara, California 95053* All checks ($12.00 per year) should be made out to the Fibonacci Association or the Fibonacci Quarterly. Two copies of manuscripts intended for publication in the Quarterly should be sent to Verner Ee Hoggatt, Jr«, Math ...
37(2)
... Also, it is clear that G is bijective and x < G(x) for all x > 0. Thus, G~l exists and is also strictly increasing with G~l(x) < x. Let u = G_1(x). Then G(u) = x and u = 3x + v8x 2 +1. Since u < x, we have u = 3x - V8x2 4-1. Also, since 8(G_1(x))2 +1 = (8x - 3V8x2 +1) 2 is a perfect square, it follo ...
... Also, it is clear that G is bijective and x < G(x) for all x > 0. Thus, G~l exists and is also strictly increasing with G~l(x) < x. Let u = G_1(x). Then G(u) = x and u = 3x + v8x 2 +1. Since u < x, we have u = 3x - V8x2 4-1. Also, since 8(G_1(x))2 +1 = (8x - 3V8x2 +1) 2 is a perfect square, it follo ...
