MHB P-adic Expansion: Find $\frac{1}{2}$ in $\mathbb{Q}_3$

[evinda]

Hi! (Wave)

I want to write the $p-$ adic expansion of the number $\frac{1}{2}$ in the field $\mathbb{Q}_3$.

How can I do this? (Thinking)

 

[mathbalarka]

We've already gone through similar problems before, so I won't give out any hints. Recall the algebraic definition of $3$-adics.

 

[evinda]

We've already gone through similar problems before, so I won't give out any hints. Recall the algebraic definition of $3$-adics.

That's what I have tried:

$ x=\frac{1}{2} \pmod 3 \Rightarrow 2x \equiv 1 \pmod 3 \Rightarrow x\equiv 2\pmod 3$
$ 2x \equiv 1 \pmod{3^2} \Rightarrow x \equiv 5 \pmod 9$
$2x \equiv 1 \pmod {3^3 } \Rightarrow x\equiv 14 \pmod{27}$
$2x \equiv 1 \pmod {3^4} \Rightarrow x \equiv 41 \pmod{3^4}$
$2x \equiv 1\pmod{3^5} \Rightarrow x \equiv 122 \pmod{3^5}$Using the formula $x_n=\sum_{i=0}^{\infty} a_i 3^i $, I found that $a_0=2, a_1=1,a_2=1, a_3=1$.

Is there a formula, that we could use, in order to find the coefficients of the infinite series?
(Thinking)

 

[mathbalarka]

Yes, there is a very generic solution to $2x = 1 \pmod{3^n}$. It's fundamental number theory stuff.

Again, I won't give out any hint. Try to solve it by yourself.

 

[evinda]

Yes, there is a very generic solution to $2x = 1 \pmod{3^n}$. It's fundamental number theory stuff.

Again, I won't give out any hint. Try to solve it by yourself.

You mean that I have to use the formula $x=\frac{3^n+1}{2}$ ?

But.. isn't there also a theorem, that says the solution we find are periodic? Or am I wrong? (Thinking)

 

[mathbalarka]

You mean that I have to use the formula ... ?

Yes.

But.. isn't there also a theorem, that says the solution we find are periodic? Or am I wrong?

Yes, prove it. You can always have theorems of your own, you know ;)

 

[evinda]

Yes, prove it. You can always have theorems of your own, you know ;)

How could I do this? (Worried)

 

[mathbalarka]

Well, you need to find coefficients $a_i$ in

$$\frac{1 + 3^n}{2} = \sum_{i = 0}^n a_i 3^i$$

How would you do that? Just a suggestion : I'd have used induction to prove that $a_0 = 2$ and $a_k = 1$ for all $k \geqslant 1$.

 

[evinda]

Well, you need to find coefficients $a_i$ in

$$\frac{1 + 3^n}{2} = \sum_{i = 0}^n a_i 3^i$$

How would you do that? Just a suggestion : I'd have used induction to prove that $a_0 = 2$ and $a_k = 1$ for all $k \geqslant 1$.

So, we have seen that a solution of the congruence $2x_n \equiv 1 \pmod {3^n}$ is this: $x_n=\frac{1+3^n}{2}$.

Then, we conclude that: $\frac{1+3^n}{2}=2+\sum_{i=1}^{n-1} 3^i$.

But.. could you explain me why this: $2+\sum_{i=1}^{n-1} 3^i$ is then the power series of $\frac{1}{2}$ ? I am confused now... (Worried)

 

[RLBrown]

Hi! (Wave)

I want to write the $p-$ adic expansion of the number $\frac{1}{2}$ in the field $\mathbb{Q}_3$.

How can I do this? (Thinking)

Here is a different look at this problem. This should not be considered rigorous, but it provided me with some insights. As an engineer myself, I like an algebraic approach using digits. I will not use congruence, Hensel's Lema, nor any heavy lifting.

An french engineer, Simon Stevin of Bruges (1548 - 1620), introduced the repeating decimal numbers as popular numerals to represent the rational numbers in applied math.

Let's use base 3 Stevin style numerals to represent rationals, with a twist. A bracket is used to indicate an infinite number of digits repeating that bracketed value. In this discussion we will require a bracket on BOTH ends of the numeral. Our goal is zero within the bracket on the RHS of the number.
We will be using long arithmetic in base 3. Natural numbers without a sub-script are base 10.

STEP 1:
First, in STEP 1, let's begin with an observation that goes back to Euler (1707 – 1783) -- long before Hensel (1861 – 1941)
0 = [2].[2]₃
because,
Multiplication by 3 only moves the point (.) one place to the left, leaving this number unchanged.
3 times [2].[2]₃ = [2].[2]₃
and if 3x = x, there are two solutions, x = infinite is one,
but it amused Euler to consider x=0, as will we.

STEP 2:
Notice that
[2].[2]₃ = [2].[0]₃ + [0].[2]₃
and
[0].[2]₃ = 1
therefore
[2].[0]₃ = -1

STEP 3:
Dividing by 2
[1].[0]₃ = -1/2

STEP 4:
Add 1
1 + [1].[0]₃ = +1/2
[1]2.[0]₃ = +1/2 done

It is fun to think how close Euler came to defining the rational p-adic form and considering the Cauchy Limit for p-adic digital numbers that are not truncated. If Newton's (1642 – 1727) work was known to Euler, he might have noticed that the Newton rational iterates for a polynomial zero (like x^2-2) will form a unique Cauchy sequence when you begin with a correct guess (the first digit) and then write the rational iterates in a p-adic form. Convergence of digits from right to left would have delighted him.

 

[RLBrown]

That's what I have tried:

Using the formula $x_n=\sum_{i=0}^{\infty} a_i 3^i $, I found that $a_0=2, a_1=1,a_2=1, a_3=1$.

Is there a formula, that we could use, in order to find the coefficients of the infinite series?
(Thinking)

Once you believe that:
x = ...1111111112₃
You can use the same technique as for repeating decimals, but of course, in base 3
3x = ...11111111120₃
Subtracting:
3x-x = 1₃
2x =1
x = 1/2 VERIFIED