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On 05.09.2024 21:57, Jim Burns wrote:On 9/5/2024 9:59 AM, WM wrote:No. NUF(x) counts the unit fractions between 0 and x. Between 0 and 0NUF(x) increases from 0 to more....at 0.
there is no unit fraction. NUF(0) = 0.
Impossible. NUF(0) = 0. There is a first increase in linear order.It cannot increase to 2 or more before having accepted 1.NUFᵈᵉᶠ(x) cannot increase to 2 without having already been ≥ 2
Therefore no x can be the least unit fraction.0 is smaller than all that. Therefore there is no increase at 0.It cannot increase to ℵo without having accepted 1, 2, 3, ...x > 0 ⇒
0 < ... < ⅟⌊4+⅟x⌋ < ⅟⌊3+⅟x⌋ < ⅟⌊2+⅟x⌋ < ⅟⌊1+⅟x⌋ < x
x is larger than all that. Therefore your x is not the least one
posiible.
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