Infinity Numbers (Expanded P-Adic Numbers)

[Lokalgott]

Hello,

I'm new here and I'm looking to talk about p-adic numbers.
It's not a specific school problem .
I'm trying to expand p-adic numbers into something that I call "Infinity Numbers".

Is it possible here to discuss topics like that?

Kind regards

 

[phinds]

Hello,

I'm new here and I'm looking to talk about p-adic numbers.
It's not a specific school problem .
I'm trying to expand p-adic numbers into something that I call "Infinity Numbers".

Is it possible here to discuss topics like that?

Kind regards

Only if it is a part of an existing canon of math, with published articles on the subject (published in reputable journals). If it is a personal theory, then no. And by the way, your introduction post is supposed to be in the introduction section. Good idea to read the rules.

 

[martinbn]

Hello,

I'm new here and I'm looking to talk about p-adic numbers.
It's not a specific school problem .
I'm trying to expand p-adic numbers into something that I call "Infinity Numbers".

Why don't you call it what everyone else calls it?

Is it possible here to discuss topics like that?

Kind regards

If you give enough details. Right now it is completely unclear what you want to discuss.

 

[Lokalgott]

I want to talk about the arithemtics of p-adic numbers:

Check this video - introduces the stuff I want to talk about:

P-Adic Numbers - Interactions

Kind regards

 

[martinbn]

I want to talk about the arithemtics of p-adic numbers:

You can discuss p-adic numbers.

Check this video - introduces the stuff I want to talk about:

P-Adic Numbers - Interactions

Kind regards

I haven't watched it, i'd say just talk about what you want to talk about.

 

[Lokalgott]

Well if you watch it then you know what I'm talking about so it might be more difficult for you to understand then

I'd like to have people who dealt with the arithmetics of p-adic numbers because then it's more easy for me and for them more easier to follow my thoughts.

But I want to talk about these arithmetics from p-adic numbers.

Is there a formula editor here so that I can show more easy what I mean?

First off I want to talk about the transformation of the geometric progression

f(x)= 1+x^1+x^2+x^3+x^inf = 1/(1-x)

First off:
I want to talk about that numbers x>1 do make sense

Like if you insert x=10

I say that

f(10)=11111..111...=-1/9 makes sense

And second off:

I want to talk about that numbers like

f(10)=111111..111...

and

9*f(10)=99999...999...

are not just the same infinity instead they are different numbers
and actually are equivalents to these numbers that you get when inserting them into

1/(1-x)

so for

f(10)=1/(1-10) =-1/9

and for

9*f(10)=9* 1/(1-10)=-1

So these infinite numbes can turn out to be some form of negative numbers?

What do you and others think of that?

Best regards

 

[berkeman]

Thread closed temporarily for Moderation...

 

[berkeman]

After a Mentor discussion, the thread is reopened provisionally.

Is there a formula editor here so that I can show more easy what I mean?

I just now sent you a Private Message (PM) with tips on using LaTeX to post equations.

 

[fresh_42]

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

In general, the series that represents a p-adic number is not convergent in the usual sense, but it is convergent for the p-adic absolute value.

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.

 

[Lokalgott]

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

 

[fresh_42]

You have some serious problems here!
1) $p$-adic numbers cannot have powers of $p$ as coefficients.
2) $p$ is prime, so $p=10$ is not allowed.
$$
\dfrac{1}{1-x}=\sum_{n=0}^\infty x^n
$$
is the series expansion around zero. Hence $x=10$ doesn't make sense.

The $p$-adic numbers are difficult. One has to distinguish numbers from formal expressions, formal expressions from functions, functions from the topological spaces they are defined on, and the topological spaces, in particular $\mathbb{Q},\mathbb{R},\mathbb{Q}_p$ must properly be distinguished.

Read the Wikipedia article, and the English one is in this case far better than e.g. the German one. This is not always the case, but here it is. Another reasonable, because a more professional source would be O'Meara,
Introduction to Quadratic Forms.

It is important that you learn to distinguish between all these concepts. You cannot write
$$
\dfrac{1}{1-x}=\sum_{n=0}^\infty x^n
$$
replace $x$ by any number you want, and call this a p-adic number. That's not how it works. I have tried to summarize the concept in
https://www.physicsforums.com/insig...l-number-systems-that-we-have/#p-adic-Numbers
but this cannot replace a good book. And, to be honest, neither can Wikipedia or Veritasium. Especially the latter sells the imagination of understanding, not truly understanding. The English Wikipedia entry on $p$-adic numbers is not bad. There is also an entry on $p$-adic analysis, but I haven't checked that one. It is a rather strange branch of mathematics and I didn't want to complicate things further.

If you have problems understanding $p$-adic numbers, then open a new thread with your questions, but not without providing an answer to the question of what exactly you did not understand. We can talk about $\mathbb{Q}\subseteq \mathbb{R}$, $\mathbb{Q}\subseteq \mathbb{Q}_p$, the p-adic evaluation, but not mixing all these.

This thread is closed now.