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On 06.09.2024 18:41, Jim Burns wrote:Which are yet to be defined.On 9/6/2024 8:12 AM, WM wrote:Between 0 and your defined x or epsilon, not between 0 and everyOn 05.09.2024 21:57, Jim Burns wrote:On 9/5/2024 9:59 AM, WM wrote:Between 0 and x there are more than 0 unit fractionsNo.NUF(x) increases from 0 to more....at 0.
NUF(x) counts the unit fractions between 0 and x.
Between 0 and 0 there is no unit fraction. NUF(0) = 0.
possible x.
Enough for me.Between 0 and x there are more.than.any.k<ℵ₀ unit fractions.Betwee 0 and every epsilon you can define.
First, not last.ε = 1. But that is not related to the topic.NUF(0) = 0.∀ᴿx ∈ (-1,0]: ⌈x⌉ = 0 ∀ᴿx ∈ (0,1]: ⌈x⌉ = 1
There is a first increase in linear order.
Is there a first ε at which 5⋅⌈ε⌉ = 5 ?
So what?But here we have always 2^ℵo points where the function is constant0 is smaller than all that.Increases and decreases involve nearby points from which increases and
Therefore there is no increase at 0.
decreases increase and decrease.
before it increases by 1.
Can you answer whether f(x) increases or decreases at 0?It is constant 0.
What about the negative of Dirac’s delta?No, you can't answer without information about nearby points.I have information about nearby points on the negative axis. Only nearby
points less than x are relevant for judging about an increase in x.
Nearby points larger than x are irrelevant.
You in particular should know the reals are dense. There is noBut there is an x immediately after 0. What else should be there?Therefore your x is not the least one posiible.Yes.
x is NOT the first point > 0 after more.than.any.k<ℵ₀ unit.fractions.
Generalize.We do not know anything. But we know that all unit fractions differ from
What do we know about x ?
x > 0 What else?
each other. That is sufficient to know that NUF(x) increases by 1 only
at any unit fraction.
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