THE CONGRUENT NUMBER PROBLEM 1. Introduction
... Example 3.4. Since Fermat showed 1 and 2 are not congruent numbers, there is no arithmetic progression of 3 rational squares with common difference 1 or 2 (or, more generally, common difference a nonzero square or twice a nonzero square). We now can explain the origin of the peculiar name “congruent ...
... Example 3.4. Since Fermat showed 1 and 2 are not congruent numbers, there is no arithmetic progression of 3 rational squares with common difference 1 or 2 (or, more generally, common difference a nonzero square or twice a nonzero square). We now can explain the origin of the peculiar name “congruent ...
Non-Homogenizable Classes of Finite Structures
... A relational structure with a countable domain is called homogeneous if it is highly symmetric in the precise technical sense that any isomorphism between any two of its finite induced substructures extends to an automorphism of the whole structure. In many areas of combinatorics, logic, discrete ge ...
... A relational structure with a countable domain is called homogeneous if it is highly symmetric in the precise technical sense that any isomorphism between any two of its finite induced substructures extends to an automorphism of the whole structure. In many areas of combinatorics, logic, discrete ge ...
34(2)
... by Daniel C, Fielder and Paul S, Brackman Members^ The Fibonacci Association In 1965, Brother Alfred Brousseau, under the auspices of The Fibonacci Association, compiled a twovolume set of Fibonacci entry points and related data for the primes 2 through 99,907. This set is currendy available from Th ...
... by Daniel C, Fielder and Paul S, Brackman Members^ The Fibonacci Association In 1965, Brother Alfred Brousseau, under the auspices of The Fibonacci Association, compiled a twovolume set of Fibonacci entry points and related data for the primes 2 through 99,907. This set is currendy available from Th ...
Introduction to topological vector spaces
... is convex, balanced, and linearly open, the set {c ∈ R | cv ∈ C} is an open interval around 0 in R. Hence if v lies in C , there exists (1 + ε)v ∈ C also, and therefore kvkC ≤ 1/(1 + ε) < 1. Conversely, suppose kvkC = r < 1. If r = 0, then cv lies in C for all c in R. Otherwise, v/r is on the bounda ...
... is convex, balanced, and linearly open, the set {c ∈ R | cv ∈ C} is an open interval around 0 in R. Hence if v lies in C , there exists (1 + ε)v ∈ C also, and therefore kvkC ≤ 1/(1 + ε) < 1. Conversely, suppose kvkC = r < 1. If r = 0, then cv lies in C for all c in R. Otherwise, v/r is on the bounda ...
