Re: Does the number of nines increase?

On 6/27/2024 5:09 AM, WM wrote:

Le 26/06/2024 à 20:46, "Chris M. Thomasson" a écrit :

On 6/26/2024 12:15 AM, WM wrote:

 

9.999... has one 9 less after the decimal point than 0.999... .

>
NO! Think of: .(9) * 10 = 9.(9) = 10

 Wrong.
 10*0.999...999 = 9.999...990

Huh? Where is that last zero coming from? Your very strange darkness? Anyway:
Are you suggesting a prefix and a postfix, a header and a footer setup?
[prolog][infinity][epilog]
Fwiw, this can be modeled in a difference between two n-ary points.
take two 4d points:
p0 = (1, 2, 3, 4)
p1 = (-1, -.42, 2, .001)
dif = p1 - p0
mid = p0 + dif / 2
Anyway, back to a way base 10 can represent infinitely repeating decimals:
0.999... = 1
0.(9) = 1
9.(9) = 10
10 * (.999...) = 10 * 1 = 10
9.(9) * .(9)   = 10 * 1 = 10
Take .(1122) where 101/900 is good enough for that precision, 4 digits. Now take:
.(4269123) = 474347/1111111
You must be familiar with:
355/113

 ==> 10*0.999...999 < 9.999...999
 ==> 9*0.999...999 < 9
 ==> 0.999... < 1
 

There are still infinite nines... :^)

 If the set ℕ is actually complete, then the set ℕ_0 has one element more.

You cannot add another natural number to N because its already there. N is all of them. So, N + 1 = N