Re: how

On 4/24/2024 5:40 PM, Moebius wrote:

Am 25.04.2024 um 02:17 schrieb Moebius:

Am 24.04.2024 um 22:36 schrieb Chris M. Thomasson:
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I can say that, well let create a symbol... (XYZ) hold all of the natural numbers. Therefore (XYZ) + 1 is already in (XYZ), fair enough?

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I have to admit that I don't know what you are talking about.
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We have the symbol "IN" to denote the set of (all) natural numbers.
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Now: For all n e IN: n + 1 e IN.
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Is that what you mean? :-)
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How about:
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"I can say that, well let introduce a symbol... [that denotes the set of all natural number, say "IN".] [The set] IN hold all of the natural numbers. Therefore n + 1 (with n in IN) is already in IN, fair enough?"
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?
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Yeah, fair enough. :-)
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Though it's not about the symbol (say "IN"), but about the set (it denotes).

 On second thought... We might DEFINE
      IN + k := {n + k : n e IN}  (k e IN).
 In this case we might indeed state:
      IN + 1 c IN
 (since it's true).
 So we might state: "I can say that, well let introduce a symbol that denotes the set of all natural number, say "IN". Then the set IN contains all natural numbers and the set IN + 1 (sse definition above) is a subset of IN, fair enough?"
 Yeah.
 

1, 1.25, 1.5, 1.75, 2
I this finite section, 1 and 2 are natural, and the other three are not. However, they are within the set of natural numbers, but three of them cannot be directly represented.
Although, they can be indexed using natural numbers, zero aside for a moment:
[0] = 1
[1] = 1.25
[2] = 1.5
[3] = 1.75
[4] = 2
There is an infinity between 1 and 2... They can be indexed using naturals?
;^) Just having some fun with numbers here. Try not to flame me off into a dry lake bed. :^)