Mathematics of Cryptography
... divide both sides by 7 to find the equation 3x + 2y = 5. Using the extended Euclidean algorithm, we find s and t such as 3s + 2t = 1. We have s = 1 and t = −1. The solutions are Particular: x0 = 5 × 1 = 5 and y0 = 5 × (−1) = −5 General: x = 5 + k × 2 and y = −5 − k × 3 ...
... divide both sides by 7 to find the equation 3x + 2y = 5. Using the extended Euclidean algorithm, we find s and t such as 3s + 2t = 1. We have s = 1 and t = −1. The solutions are Particular: x0 = 5 × 1 = 5 and y0 = 5 × (−1) = −5 General: x = 5 + k × 2 and y = −5 − k × 3 ...
Lie Groups and Lie Algebras
... This definition is more general than what we will use in the course, where we will restrict ourselves to so-called matrix Lie groups. The manifold will then always be realised as a subset of some Rd . For example the manifold S 3 , the three-dimensional sphere, can be realised as a subset of R4 by t ...
... This definition is more general than what we will use in the course, where we will restrict ourselves to so-called matrix Lie groups. The manifold will then always be realised as a subset of some Rd . For example the manifold S 3 , the three-dimensional sphere, can be realised as a subset of R4 by t ...
