F. Poyo

Apr. 29, 1996

In the following text, we try to intuitively understand p-adic algebraic field (simular concepts to base-p digit system) without decimal point or minus symbol.

Here we define the distance between a rational number v=ax(p^n)/b (a,b,p are relatively prime) and the origin "0" by

|v|p=1/(p^n).

Thus, the definition means that a number v is the more divided by p, the nearer from "0". For example,

We call the distance from "0" "the vicinity of 0".p=3, v=5 |v|p=|5|3=1/(3^0) since v=5x(3^0)/1 p=3, v=1/2 |v|p=|1/2|3=1/(3^0) since v=1x(3^0)/2 p=3, v=9 |v|p=|9|3=1/(3^2) since v=1x(3^2)/1 p=3, v=5/18 |v|p=|5/18|3=1/(3^(-2)) since v=5x(3^(-2))/2

For example, 123.45 in base 10 is written byinfty i v= Sum v p 0 < v < p-1 k is an integer i=k i = i =

-2 -1 0 1 2 3 123.45 = 5x10 + 4x10 + 3x10 + 2x10 + 1x10 + 0x10 + ...

where k=-2, all the coefficients v (i>2) are 0. i

We can not represent a negative number by the above definition, but in the case that p is a prime number, we can do. For example,

Suppose we represent 0.5 in base p.

0.5 = a + bx7 + cx(7^2) + dx(7^3) + ...

Since we know 2 is represented by

it follows that2 = 2 + 0x7 + 0x(7^2) + ...,

2x0.5 = 1 = 2xa + 2xbx7 + ...

By using the knowledge that 2xy type integer is one of

2x1=2, 2x2=4, 2x3=6, 2x4=8=1+7, 2x5=10=3+7, 2x6=12=5+7,

we obtain

a=4. (0)

We can also write the relationship with congruence form by

1 = 2xa mod 7 (1)

Formally we rewrite (1) by

1/2 = a mod 7 (1')

Thus, (0) is a solution of (1'). Note the left hand side of (1') is 1/2=0.5.

The carry to the position of 10 is given by

(2xa - 1)/7 = 1 (surplus is trimmed). (2)

To obtain b, taking the carry 1 from a into account, we write

0 = 2xb+1 mod 7 (3)

Formally we rewrite (3) by

-1/2 = b mod 7, (3')

where

(1/2-a)/7=(1/2-4)/7=-1/2 (2')

b is given by

b=3

as a solution of (3'). We can compute the carry to the position of 100 by

(2xb + 1 - 0)/7=1 (4)

as well. To obtain c with the carry 1, we would get the solution of

0 = 2xc + 1 mod 7. (5)

However, this is the same equation as (3). Therefore,

c=3

we can obtain the digits of the other positions repeatedly by

d=3, e=3, f=3, ...

Now we have the representation of 0.5 in base 7 under our definition by

0.5 = 4 + 3x7 + 3x(7^2) + ... + 3x(7^n) + ...

The series looks divergent, but since we have "the vicinity of 0",

lim 3x(7^n) = 0, n->infty

it is not divergent. Therefore we finally obtain the representation of 0.5 in base 7 (7-adic number) by

0.5= ....333334

Generally we can obtain a p-adic representation of v by the formalization of the above (1'), (2'), (3'), thus

- Get a by v = a mod p
- Get B by B = (v-a)/p
- Get b by B = b mod p
- Get C by C = (B-b)/p
`...`

**v=-1 (p=7):**-
- a=6 by -1 = a mod 7
- B = (-1-6)/7 = -1
- b=6 by -1 = b mod 7
`(Since This is the same equation as that in (A), we can obtain c=6, d=6, e=6, ... as well.)`

`-1 = ....66666`

Recall 0.5=...33334 by 7-adic representation, let us do the multiplication 2X0.5 on paper by the representation.

...3334 = 0.5 x)_______2 = 2 11 6 6 ... ---------- ...00001 = 1

Recall -1=...6666 in 7-adic representation. The distance between -1 and 0 should be

0 - (-1) = 1

is it correct in our definition, too? Let us see the confirmation on paper by adding (-1) and 1 as follows. Since

surely,...6666 =-1 +)_______1 = 1 ...0000 = 0,

-1 + 1 = 0.

***** An example of 9-adic number system *****

Suppose p=9=3x3, then the 9-adic representation of v=1/3 is calculated by

- (A)
- In order to solve 1/3 = a mod 9, we write

However, since`1 = 3 x a mod 9. (#)``3x1=3, 3x2=6, 3x3=0+9, 3x4=12=3+9, 3x5=15=6+9, 3x6=18=0+2x9, 3x7=21=3+2x9, 3x8=24=6+2x9,`any a which satisfy (#) does not exist.

***** An example of 6-adic number system *****

Suppose p=6=2x3, then the 6-adic representation of v=0.5 is calculated by

- (A)
- In order to solve 0.5 = a mod 6

However, since`1 = 2 x a mod 6 (##)``2x1=2, 2x2=4, 2x3=0, 2x4=2+6, 2x5=4+6`any a which satisfy (##) does not exist.

In the above examples, non-prime number p does not construct satisfiable decimal system, why? This is because when we calculate the multiplication of u in the both side of v=y mod z, thus

u x v = u x (y mod z),

we can do this only when z and u are relatively prime each other. Assuming z and u have a common divisor a, thus

z = a x m u = a x n,

since yu x y=a x n x yIn the both cases y=m x j + k and y=k (k=1,2,...m-1),

u x y=a x n x (m x j + k)=z x n x j + a x n x k u x y=a x n x k = a x n x kwe obtain

u x v = u x y mod z = a x n x k (in y=m x j + k, k).Specially, in the case of non-integer v=1/u (u, integer and u != 1),

u x v = 1 = u x y mod z = a x n x k = u x k (k=1,2,...,m)is not satisfied by any k. Therefore z and arbitrary u must be relatively prime. This means that z must be prime number.

p-adic number set in which the number of elements is finite.Here we define the vicinity of 0 by the different manner from the above sections. The definition is

|p^N|=0We do not take the numbers larger than p^N into account. (All the term above p^N is zero.)

Under this definition of the vicinity of 0, suppose p=2, then for example v=1/3 is calculated by

- 1/3 = a mod 2
1 = 3xa mod 2 a=1 by 3x1=1+2- B = (1/3-1)/2=-1/3
- -1/3 = b mod 2
0 = 3xb+1 mod 2 b=1- C = (-1/3 - 1)/2 = -2/3
-2/3 = c mod 2 0 = 3xc+2 mod 2 c=0- D = (-2/3 - 0)/2 = -1/3
d=1 This is the same as the result of (B).Therefore, we obtain

1/3 = ...10101011 (n digits).In order to confirm this, let n=6 (6bit), then it follows that

101011 = 1/3 x)____11 = 3 101011 101011 ----------- 10000001 ===> 000001v=-1 in 2-adic number system (binary) is calculated by

(A) Add 1 to both sides of -1 = a mod 2then0 = a+1 mod 2a=1As well as the above, each digit in each position is 1. Thus,

- B=(-1 - 1)/2=-1
- Think -1 = b mod 2 in the next, but this is the same equation as (A). Therefore, we obtain
b=1For the confirmation, let n=5, since-1 = 111111 (n digits).it follows that surely (-1)+1=0.11111 = -1 +) 1 = 1 ------------------- 100000 = 00000 (since 2^6=0),

ConclusionIn the case of infinite-number-of-digits system, p-adic number system is constructed by the vicinity of 0

where|v|p=1/(p^n)v=(a/b)(p^n)when and only when p is a prime number.

In the case of finite(n)-digits system, we can use the different definition in the vicinity of 0 as

If we can use "-" or decimal point ".", we can make p-adic system in arbitrary p (, which is not confirmed).|p^q|p=0=V(p^q) ( q > n-1 ).