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Assume that there is an x e IR such that NUF(x) = 1. Let x0 e IR such that NUF(x0) = 1. This means that there is exactly one unit fraction u such that u < x0. Let's call this unit fraction u0. Then (by definition) there is a (actually exactly one) natural number n such that u0 = 1/n. Let n0 e IN such that u0 = 1/n0. But then (again by definition) 1/(n0 + 1) is an unit fraction which is smaller than u0 and hence smaller than x0. Hence NUF(x0) > 1. Contradiction!We can reduce the interval (0, x) c [0, 1] such that x converges to 0.
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