fresh_42 said:
I think you should read
https://en.wikipedia.org/wiki/P-adic_number.
The problem with p-adic numbers as series ##\sum_{i=k}a_ip^i## with ##k\in \mathbb{Z}## is the convergence. If we only write p-adic numbers as such sums, then it is a purely formal construction. Also a formal construction is ##\dfrac{1}{1-x}= \sum_{i=0}^\infty x^i .## Filling in a number for the variable ##x## and allowing coefficients ##0\leq a_i < p## yields equations like
$$
\ldots + 3\cdot 5^3+3\cdot 5^1+3\cdot 5^{-1}= \ldots 3030_5,3=\overline{30}_5,3=\overline{0_5,3}=-\dfrac{1}{40}_{dec}
$$
in a ##5##-adic notation. The crucial point is
The ##5##-adic number ##\overline{0_5,3}=\sum_{i=-1}^\infty a_i 5^i ## with ##a_{2n-1}=3 ## and ##a_{2n}=0## for all ##n\in \{0,1,2,3,\ldots\}## is not convergent in the usual, Archimedean sense. It is a formal construction. In order to converge, we need the ##5##-adic evaluation, the p-adic absolute value. p-adic numbers are the completion of the rational numbers according to the p-adic absolute value. The rational numbers are just specific p-adic numbers. A general p-adic number is difficult (for me) to imagine. But I understand the topological concept of a completion: add all limits of converging series, and consider convergence based on the p-adic absolute value (in order to define what "getting closer and closer, converging" means).
This means: Infinity is not a self-explaining term anymore. You must say what exactly you mean by it.
Yes I agree. Infinity is not a self-explaining term anymore!
I watched a video and was thinking about that in the following geometric progression the p-adic number ##p## itself
$$ p^0+p^1+p^2+p^\infty$$
always stands for ##-1##
because if you add ##1## to that number all ciphers clap over to ##0##
example
2-adic:
$$ 2^0+2^1+2^2+2^\infty =222...222...$$
$$222...222...+1=1000...000...$$
zeroes everywhere with a ##1## that will never constructed
and since this is zero the infinite p-adic number, here ##222...222...## must be ##-1##
3-adic
$$ 3^0+3^1+3^2+3^\infty =333...333...$$
$$333...333...+1=1000...000...$$
5-adic:
$$ 5^0+5^1+5^2+5^\infty =555...555...$$
$$555...555...+1=1000...000...$$
etc.
Now what about 10-adic?
If I insert ##10## into the geometric progression:
10-adic:
$$ 10^0+10^1+10^2+10^\infty =111...111...$$
$$111...111...+1=111...111...112$$
which this time isn't ##-1##
So since we know that in 10 base system:
$$999...999...=m |*10$$
$$999...999...990=10m$$
substract both equations
$$999...999...999=m
-999...999...990=10m$$
$$9=-9m$$
$$-1=m$$
so
$$m=999...999...=-1$$
We know that:
$$999...999.../9=-1/9$$
$$111...111=-1/9$$
So in base 10:
The geometric progression with p=10 turns out to be ##-1/9## instead of ##-1##
Now I'm asking does that with prime numbers only work to be ##-1## for the geometric progression when you insert ##p## itself?
And does the transformation really gives us also the real value then?
Because if we insert ##10## into
$$ \sum_{n=0}^\infty x^n=1/(1-x)$$
$$1/(1-10)=-1/9$$
We also get that same value we calculated in a different way.
So it seems to be right and maybe true
And for that that in base 10:
$$111...111...+1=111...111...112$$
Must be
$$-1/9+1=10/9$$
So the other interesting stuff is that on one boundary an infinite number is positive (over ##111...111..##) and on the other side it becomes negative (under ##111...111...112##)
So infinite Numbers include positive and negative numbers and also rationals.
Now you also could express numbers like irrational numbers like ##\pi## inform of infinite numbes . Really funny
