Show that number is integer 5-adic, find 5 positions...

[evinda]

Hello! (Wave)

I want to conclude that the number $\frac{1}{2}$ is an integer $5-$ adic and to calculate the first five positions of its powerseries.

In order to conclude that $\frac{1}{2}$ is an integer $5-$ adic, do I have to use this definition?

Let $p \in \mathbb{P}$.
The set of the integer $p$-adic numbers is defined like that:

$$\mathbb{Z}_p= \{ (\overline{x_n})_{n \in \mathbb{N}_0} \in \Pi_{n=0}^{\infty} \frac{\mathbb{Z}}{p^{n+1} \mathbb{Z}} | x_{n+1} \equiv x_n \pmod {p^{n+1}} \}$$

or is there also an other way to do this? (Thinking)

 

[RLBrown]

1/2 is in lowest terms and
5 does not divide 2.
What does that tell you about the ord of |1/2|₅ ?1/2 is in fact a 5-adic integer which expands as
View attachment 3394

 

[mathbalarka]

What's $1/2$ modulo $5$? What's $1/2$ modulo $5^2$? What's $1/2$ modulo $5^3$? Can you compute $1/2$ modulo $5^n$ in general? What's the corresponding $5$-adic representation is supposed to be then?

 

[evinda]

What's $1/2$ modulo $5$? What's $1/2$ modulo $5^2$? What's $1/2$ modulo $5^3$? Can you compute $1/2$ modulo $5^n$ in general? What's the corresponding $5$-adic representation is supposed to be then?

Could you explain me how we can calculate $\frac{1}{2} \pmod 5, \dots , \frac{1}{2} \pmod {5^n}$ ? (Worried) (Sweating)

 

[mathbalarka]

I am sure you are familiar with modular arithmetic in $(\Bbb Z_n)^\times$. By $1/2$ modulo $n$ we mean the inversion of $2$ modulo $n$.

(Of course this is possible if and only if $(2, n) = 1$, but $2$ and $5$ are coprime so this is always possible)

 

[evinda]

I am sure you are familiar with modular arithmetic in $(\Bbb Z_n)^\times$. By $1/2$ modulo $n$ we mean the inversion of $2$ modulo $n$.

(Of course this is possible if and only if $(2, n) = 1$, but $2$ and $5$ are coprime so this is always possible)

So are we looking for a $x \in \{ 1, \dots, 4 \}$ such that $x \cdot 2 \equiv 1 \pmod 5$ ?

So, is it $\frac{1}{2} \pmod 5 \equiv 3 \pmod 5$ ? (Thinking)

 

[mathbalarka]

Right.

 

[evinda]

Right.

Nice! And how can I calculate now the first five positions of its powerseries? (Thinking)

Also, could it happen that a rational number isn't an integer 5-adic number? If so, in which case? (Worried)

 

[RLBrown]

Nice! And how can I calculate now the first five positions of its powerseries? (Thinking)

Also, could it happen that a rational number isn't an integer 5-adic number? If so, in which case? (Worried)

1) Hensel Lifting
2) No

PS: you should get...
View attachment 3403

 

[evinda]

But, what would we do if we had a rational like $\frac{3}{8}$ ? Would we have to find the inverse of $8$ modulo $5$ and multiply it by $3$ ? (Worried)

 

[RLBrown]

But, what would we do if we had a rational like $\frac{3}{8}$ ? Would we have to find the inverse of $8$ modulo $5$ and multiply it by $3$ ? (Worried)

You are solving a first degree polynomial congruence.

0 $\equiv$ 8x-3 mod $5^n$ as n $\rightarrow \infty$

 

[evinda]

You are solving a first degree polynomial congruence.

3 $\equiv$ 8x mod 5

Could you explain me why we solve this congruence, in order to show that $\frac{3}{8}$ is 5-adic? (Worried)

 

[RLBrown]

Could you explain me why we solve this congruence, in order to show that $\frac{3}{8}$ is 5-adic? (Worried)

Sorry, forgot general case, that was first digit only.
I edited it:(

 

[evinda]

Sorry, forgot general case, that was first digit only.
I edited it:(

Could you explain me why we solve this congruence, to find the solutions of $\frac{3}{8}$ $\pmod {5^n}$ ? (Worried)

 

[RLBrown]

Also, could it happen that a rational number isn't an integer 5-adic number? If so, in which case? (Worried)

I miss read this part above.
A rational number will always be a 5-adic number.

However you are correct
A rational number will NOT always be an INTEGER 5-adic number!

I convert the number to an integer (multiply by 5^m) before p-adic expansion using lifting. Then shift the decimal on the result to the left m places.

 

[mathbalarka]

Also, could it happen that a rational number isn't an integer 5-adic number? If so, in which case?

Yes. Take $\frac15$. On the other hand $\Bbb Q$ is rather included in the fractional completion of $\mathbf{Z}_p$, namely, $\mathbf{Q}_p$.

Nice! And how can I calculate now the first five positions of its powerseries?

Why don't you just forget about the power series for a second? It makes no sense to produce a power series in the general archimedean sense, so I'd rather prefer you to look at the algebraic definition. Calculate the corresponding tuple in $\prod \Bbb Z/p^i \Bbb Z$.

- - - Updated - - -

Could you explain me why we solve this congruence, in order to show that $\frac{3}{8}$ is 5-adic? (Worried)

Recall the algebraic inverse limit definition of $p$-adics.

 

[evinda]

Yes. Take $\frac15$. On the other hand $\Bbb Q$ is rather included in the fractional completion of $\mathbf{Z}_p$, namely, $\mathbf{Q}_p$.

So, in general, how can we conclude if a rational number is a p-adic integer number? (Thinking)

Recall the algebraic inverse limit definition of $p$-adics.

How can we use the algebraic inverse limit definition of $p$-adics? (Worried)

 

[RLBrown]

Could you explain me why we solve this congruence, to find the solutions of $\frac{3}{8}$ $\pmod {5^n}$ ? (Worried)

Let's use 3/8 as an example:
(LHS can be verified by pasting RHS into WolframAlpha)

First digit?
  2 = Mod[8 0-3,5]
  0 = [URL="http://www.wolframalpha.com/input/?i=Mod%5B8+1-3%2C5%5D"]Mod[8 1-3,5][/URL]
  3 = Mod[8 2-3,5]
  1 = Mod[8 3-3,5]
  4 = Mod[8 4-3,5]
     Answer: First digit = 1

Second digit?
  5 = Mod[8(1+0 5^1)-3,25]
 20 = Mod[8(1+1 5^1)-3,25]
 10 = Mod[8(1+2 5^1)-3,25]
  0 = [URL="http://www.wolframalpha.com/input/?i=Mod%5B8%281%2B3+5%5E1%29-3%2C25%5D"]Mod[8(1+3 5^1)-3,25][/URL]
 15 = Mod[8(1+4 5^1)-3,25]
     Answer: Second digit = 3

Third digit?
  0 = [URL="http://www.wolframalpha.com/input/?i=Mod%5B8+%281+%2B+3+5%5E1+%2B+0+5%5E2%29+-+3%2C+5%5E3%5D"]Mod[8 (1 + 3 5^1 + 0 5^2) - 3, 5^3][/URL]
 75 = Mod[8 (1 + 3 5^1 + 1 5^2) - 3, 5^3]
 25 = Mod[8 (1 + 3 5^1 + 2 5^2) - 3, 5^3]
100 = Mod[8 (1 + 3 5^1 + 3 5^2) - 3, 5^3]
 50 = Mod[8 (1 + 3 5^1 + 4 5^2) - 3, 5^3]
     Answer: Third digit = 0

If continued the digits repeat as
View attachment 3405

 

[evinda]

1) Hensel Lifting

View attachment 3403

How can we apply the Hensel Lifting? (Thinking)

 

[RLBrown]

How can we apply the Hensel Lifting? (Thinking)

In mathematics, Hensel's lemma, also known as Hensel's lifting lemma, named after Kurt Hensel, is a result in modular arithmetic, stating that if a polynomial equation has a simple root modulo a prime number p, then this root corresponds to a unique root of the same equation modulo any higher power of p, which can be found by iteratively "lifting" the solution modulo successive powers of p.

Depending upon the derivative of the polynomial being solved, a number of "next digits" can be found in one lift application.

OR more simply, you can lift by guessing the next digit (by trying each candidate in the congruence relationship) as I did in the above "3/8 as a 5-adic" example. Simply pick the digit 0,1,2,3 or 4 that makes the LHS=0.

 

[mathbalarka]

How can we use the algebraic inverse limit definition of $p$-adics? (Worried)

State the inverse limit definition first before using it. What are $p$-adic integers?

 

[evinda]

State the inverse limit definition first before using it. What are $p$-adic integers?

Let $p \in \mathbb{P}$.

The set of the integer p-adic numbers is defined as:

$$\mathbb{Z}_p:= \{( \overline{x_n})_{n \in \mathbb{N}_0} \in \Pi _{n=0}^{\infty} \frac{\mathbb{Z}}{p^{n+1} \mathbb{Z}} | x_{n+1} \equiv x_n \pmod {p^{n+1}}\}$$

The inverse limit is:

$$\mathbb{Z}_p=\lim_{\overleftarrow{n}} \frac{\mathbb{Z}}{p^n \mathbb{Z}}$$

How can I use the above? (Sweating)

 

[mathbalarka]

What's the natural embedding of $1/2$ in $\Bbb Q$ onto $\Bbb Z/5Z$? Now lift it up to $\Bbb Z/5^k \Bbb Z$ for $k > 1$ using Hensel and take the inverse limit of the corresponding sequence to embed it in $\mathbf{ Z}_5$. Does that make sense?

 

[evinda]

What's the natural embedding of $1/2$ in $\Bbb Q$ onto $\Bbb Z/5Z$? Now lift it up to $\Bbb Z/5^k \Bbb Z$ for $k > 1$ using Hensel and take the inverse limit of the corresponding sequence to embed it in $\mathbf{ Z}_5$. Does that make sense?

Could you explain me how we could apply the Hensel Lemma in this case? I haven't uderstood it... (Sweating)

 

[mathbalarka]

I've already given an answer in here and evinda figured out how to Hensel this out in the comment below.

 

[evinda]

I've already given an answer in here and evinda figured out how to Hensel this out in the comment below.

Thanks a lot! (Clapping)

 

[evinda]

What's the natural embedding of $1/2$ in $\Bbb Q$ onto $\Bbb Z/5Z$? Now lift it up to $\Bbb Z/5^k \Bbb Z$ for $k > 1$ using Hensel and take the inverse limit of the corresponding sequence to embed it in $\mathbf{ Z}_5$. Does that make sense?

Which is the easiest way to find, for example, the $x$, such that $2x \equiv 1 \pmod {5^3}$ ? (Worried)