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Am Fri, 10 Jan 2025 22:38:51 +0100 schrieb WM:The natural numbers are an infinite set. For all of them it is true,On 10.01.2025 21:06, joes wrote:Good. It is not true for the infinite sets.Am Fri, 10 Jan 2025 20:17:37 +0100 schrieb WM:Of course. Otherwise you would have to find a counterexample.On 10.01.2025 19:28, joes wrote:But it is true for every naturalAm Fri, 10 Jan 2025 18:21:43 +0100 schrieb WM:>No, I expect it is true for all natural numbers, none of which isI have no expectations about cardinality. I know that for everyYou wrongly expect this to hold in the infinite.
finite initial segment the even numbers are about half of the
natural numbers.
This does not change anywhere. It is true up to every natural
number.
infinite.
No, that is only a habit of the last century.Mathematics is all about formalising.(if you formalise it correctly)!Irrelevant.
Find an element of N or E that is not covered by the equation.∀n ∈ ℕ: |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| = 2.Those are not N and E.
Every n is finite.No. For n->oo,That doesn't make it true for N and G.I am not interested in these letters but only in all natural numbers.
All natural numbers are twice as many as all even natural numbers. If
your N and G denote all natural numbers and all even numbers, then 2 is
true also for them.
G is both the set {2, 4, ..., 2n} and {2, 4, ..., 42n};And all of them can be denoted by n. If these sets are not fixed, then there is no bijection possible.
indeed, {2, 4, ..., 2kn} for every k e N.
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