p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p
-representation
q
Rational base numeration system for p-adic
numbers
Karel Klouda1 ,2
[email protected]
Joint work with C. Frougny
1 FNSPE,
2 LIAFA,
CTU in Prague
Université Paris 7
Digital expansions, dynamics and tilings
April 4-11, 2010, Aussois
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
Outline
n,∞
X
ai p i
,
q q
k =−`
ai ∈ Ap = {0, 1, . . . , p − 1},
p > q ≥ 1 co-prime
1. Brief introduction to p-adic numbers
2. Representation of positive integers [Akiyama, Frougny,
Sakarovitch, 2008]
3. Representation of negative integers
4. Representation of rational numbers
5. Finite representations
6. Representations of p-adic numbers
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p-adic absolute value
Definition
Let p be a prime number.
The p-adic valuation vp : Z \ {0} → R is given by
n = pvp (n) n0
with
p - n0 .
The extension to the set of rational numbers, for x =
vp (x) = vp (a) − vp (b).
a
b
∈Q
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p-adic absolute value
Definition
Let p be a prime number.
The p-adic valuation vp : Z \ {0} → R is given by
n = pvp (n) n0
with
p - n0 .
The extension to the set of rational numbers, for x =
vp (x) = vp (a) − vp (b).
The p-adic absolute value on Q:
(
0
|x|p =
p−vp (x)
if x = 0,
otherwise.
a
b
∈Q
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p-adic absolute value – examples
Example
v3 (6) = 1
⇒
|6|3 = 3−1 , |1/6|3 = 31
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p-adic absolute value – examples
Example
v3 (6) = 1
vp (pn ) = n
⇒
⇒
|6|3 = 3−1 , |1/6|3 = 31
|pn |p = p−n → 0 as n → ∞
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p-adic absolute value – examples
Example
v3 (6) = 1
vp (pn ) = n
⇒
⇒
|6|3 = 3−1 , |1/6|3 = 31
|pn |p = p−n → 0 as n → ∞
p = r1j1 r2j2 · · · rkjk , q co-prime to p ⇒
(
n −ji n
p if r = ri , i ∈ {1, 2, . . . , k },
= ri
q ≥ 1 otherwise.
r
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p
-representation
q
Qp – the set of p-adic numbers I
The set of p-adic numbers Qp is defined as a completion of Q with
respect to | |p .
Lemma
The p-adic absolute value | |p is non-Archimedean, i.e., for all
x, y ∈ Qr the following holds:
|x + y |p ≤ max{|x|p , |y |p }.
Theorem (Ostrowski)
Every non-trivial absolute value on Q is equivalent to the classical
absolute value | | or to one of the absolute values | |p , where p is
prime.
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
Qp – the set of p-adic numbers II
Archimedean
Q2
||
| |2
Q
Q3
| |3
| |5
Q5
| |p
Qp
non-Archimedean
R
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p
-representation
q
Standard representation of p-adic numbers I
Theorem
Every x ∈ Qp can be uniquely written as
x
= b−k0 p−k0 + · · · + a0 + a1 p + a2 p2 + · · · + ak pk + · · ·
X
=
ak p k
k ≥−k0
with ak ∈ Ap = {0, 1, . . . , p − 1} and k0 = −vp (x).
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p
-representation
q
Standard representation of p-adic numbers I
Theorem
Every x ∈ Qp can be uniquely written as
x
= b−k0 p−k0 + · · · + a0 + a1 p + a2 p2 + · · · + ak pk + · · ·
X
=
ak p k
k ≥−k0
with ak ∈ Ap = {0, 1, . . . , p − 1} and k0 = −vp (x).
Algorithm
Let s ∈ Z, s 6= 0. Put s0 = s and for all i ∈ N define si+1 and
ai ∈ Ap by
si = psi+1 + ai .
Return a = · · · a2 a1 a0 .
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p
-representation
q
Standard representation of p-adic numbers I
Theorem
Let x ∈ Qp . Then the standard representation of x is
1. uniquely given,
2. finite if, and only if, x ∈ N,
3. eventually periodic if, and only if, x ∈ Q.
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
MD algorithm
Algorithm (MD algorithm)
Let s be a positive integer. Put s0 = s and for all i ∈ N:
qsi = psi+1 + ai .
Return qp -expansion of s: < s > p = · · · a2 a1 a0 .
q
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p
-representation
q
MD algorithm
Algorithm (MD algorithm)
Let s be a positive integer. Put s0 = s and for all i ∈ N:
qsi = psi+1 + ai .
Return qp -expansion of s: < s > p = · · · a2 a1 a0 .
q
2
p a0
p
a1 p a0
s = s0 = s1 +
= s2
+
+
= ···
q
q
q
q q
q
k kX
−1
n
X
p
ai p i
ai p i
· · · = sk
+
= ··· =
q
q q
q q
i=1
i=1
p-adic numbers
Positive integers
Negative integers
p
q -expansion
Rationals
Finite case
– properties
• for q = 1 we get classical representation in base p,
p
-representation
q
p-adic numbers
Positive integers
Negative integers
p
q -expansion
Rationals
Finite case
– properties
• for q = 1 we get classical representation in base p,
L p = {w ∈ A∗p | w is
q
p
-expansion of some s ∈ N}
q
p
-representation
q
p-adic numbers
Positive integers
Negative integers
p
q -expansion
Rationals
Finite case
– properties
• for q = 1 we get classical representation in base p,
L p = {w ∈ A∗p | w is
q
p
-expansion of some s ∈ N}
q
• L p is prefix-closed,
q
• any u ∈ A+
p is a suffix of some w ∈ L p ,
q
• L is not context-free (if q 6= 1),
p
q
• π : A+
p 7→ Q the evaluation map. If v , w ∈ L p , then
q
v w
⇔
π(v ) ≤ π(w).
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
T qp – tree of nonnegative integers
1
2
2
0
2
1
G0
1
2
G1
0
4
3
G2
0
1
7
6
5
G3
2
0
1
11
10
9
8
G4
Children of the vertex n are given by q1 (pn + a) ∈ N, a ∈ Ap .
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
T qp – tree of nonnegative integers
1
2
2
0
2
1
G0
1
2
G1
0
4
3
0
1
G2
7
6
5
G3
2
0
1
11
10
9
8
G4
Children of the vertex n are given by q1 (pn + a) ∈ N, a ∈ Ap .
G0 = 1,
Gi+1 =
lp
q
Gi
m
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
MD algorithm – the negative case
Let s be a negative integer. The qp -expansion of s is
< s > p = · · · a2 a1 a0 from the MD algorithm:
q
s0 = s,
qsi = psi+1 + ai .
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
MD algorithm – the negative case
Let s be a negative integer. The qp -expansion of s is
< s > p = · · · a2 a1 a0 from the MD algorithm:
q
s0 = s,
qsi = psi+1 + ai .
Properties of (si )i≥0 :
(i) (si )i≥1 is negative,
p−1
(ii) if si < − p−q
, then si < si+1 ,
p−1
p−1
(iii) if − p−q
≤ si < 0, then − p−q
≤ si+1 < 0.
− p−1
p−q
0
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p
-representation
q
In which fields does it work?
k kX
−1
ai p i
p
+
s = sk
q
q q
i=1
We want:
kX
−1
p k
ai p i s −
= |sk |r →0
q q q i=1
r
r
as k → ∞
Hence, if p = r1j1 r2j2 · · · rkjk , the qp -expansion of s “works” only in
Qri , i = 1, . . . , k . The speed of convergence is then ≈ r −ji k .
p-adic numbers
Positive integers
Negative integers
p
q -expansions
Rationals
Finite case
p
-representation
q
of negative integers
Proposition
Let k be a positive integer, and denote B =
j
p−1
p−q
k
. Then:
(i) if k ≤ B, then < −k > 1 p = ω b with b = k (p − q),
q q
(ii) otherwise, < −k > 1 p = ω bw with w ∈ A+
p and b = B(p − q).
q q
p-adic numbers
Positive integers
Negative integers
p
q -expansions
Rationals
Finite case
p
-representation
q
of negative integers
Proposition
Let k be a positive integer, and denote B =
j
p−1
p−q
k
. Then:
(i) if k ≤ B, then < −k > 1 p = ω b with b = k (p − q),
q q
(ii) otherwise, < −k > 1 p = ω bw with w ∈ A+
p and b = B(p − q).
q q
2
−4
1
−3
2
0
0
2
1
2
−2
−1
0
1
0
2
3
4
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
T qp – tree of negative integers
1
T pq
2
2
ω
0
ω
1
ω
2
0
2
1
1
2
0
4
3
0
1
7
6
5
2
0
1
11
10
9
8
-1
-2
0
-3
1
-4
2
0
-5
-6
1
2
0
-7
-8
-9
T pq
1
2
0
1
-10
-11
-12
-13
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
T qp – tree of negative integers
2
ω
0
ω
1
ω
2
0
2
1
1
2
0
3
1
5
1
8
-1
-2
F0
0
-3
1
F1
-4
F2
2
0
-5
-6
F3
1
2
0
-7
-8
-9
F4
1
2
0
1
-10
-11
-12
-13
F5
F0 = −B,
Fi+1 =
lp
q
Fi
m
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
Trees T qp and T qp
11
7
2
p = 8, q = 5
6
7
ω
0
ω
3
ω
6
0
5
1
2
3
2
1
4
6
5
4
5
0
3
10
9
8
7
-1
-2
1
-3
4
-4
7
2
-5
-6
5
0
3
-7
-8
-9
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
Trees T qp and T qp
11
7
2
p = 8, q = 5
6
7
ω
0
ω
3
ω
6
0
5
1
2
3
2
1
4
6
5
4
5
0
3
10
9
8
7
-1
-2
1
-3
4
-4
7
2
-5
-6
5
0
3
-7
-8
-9
The trees T p and T p are isomorphic if and only if
q
q
p−1
∈ Z,
p−q
i.e., B(p − q) = p − 1
p
-representation
q
p-adic numbers
Positive integers
Negative integers
p
q -expansion
Rationals
Finite case
p
-representation
q
– properties
• for q = 1 and p prime we get the standard p-adic
representation,
L p = {w ∈ A∗p | ω bw is
q
p
-expansion of s ≤ −B, w0 6= b}
q
• L p is prefix-closed,
q
• any u ∈ A+
p is a suffix of some w ∈ L p ,
q
• L p is not context-free (if q 6= 1).
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
MD algorithm for rationals
Algorithm (MD algorithm)
Let x = st , s, t ∈ Z co-prime, s 6= 0, and t > 0 co-prime to p.
Put s0 = s and for all i ∈ N define si+1 and ai ∈ Ap by
q
si
si+1
=p
+ ai .
t
t
Return the qp -expansion of x: < x > p = · · · a2 a1 a0 .
q
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p
-representation
q
Examples
x
< x >1 p
q q
5
-5
11/4
11/8
11/5
5
-5
11/7
(si )i≥0 )
p = 3, q = 2
5, 3, 2, 1, 0, 0, . . .
-5,-3,-2,-2,-2, . . .
11,6,4,0,0, . . .
11,2,-4,-8,-8,-8, . . .
11,4,1,-1,-4,-6,-4,-6, . . .
p = 30, q = 11
11 25 5, 1, 0, 0, . . .
ω 19 8 5 -5,-2,-1,-1,. . .
ω (12 21 5) 23 13
11, 1, -5, -3, -6, -5, . . .
2101
ω 2012
201
ω 1222
ω (02)2112
abs. values
all
| |3
all
| |3
| |3
all
| |2 , | |3 , | |5
| |2 , | |3 , | |5
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p
-representation
q
MD algorithm – properties
For the sequence (si )i≥1 from the MD algorithm we have:
(i) if s > 0 and t = 1, (si )i≥1 is eventually zero,
(ii) if s > 0 and t > 1, (si )i≥1 is either eventually zero or eventually
negative,
(iii) if s < 0, (si )i≥1 is negative,
p−1
t, then si < si+1 ,
(iv) if si < − p−q
p−1
p−1
(v) if − p−q
t ≤ si < 0, then − p−q
t ≤ si+1 < 0.
− p−1
p−q t
-1 0
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p
-representation
q
MD algorithm – properties
For the sequence (si )i≥1 from the MD algorithm we have:
(i) if s > 0 and t = 1, (si )i≥1 is eventually zero,
(ii) if s > 0 and t > 1, (si )i≥1 is either eventually zero or eventually
negative,
(iii) if s < 0, (si )i≥1 is negative,
p−1
(iv) if si < − p−q
t, then si < si+1 ,
p−1
p−1
(v) if − p−q
t ≤ si < 0, then − p−q
t ≤ si+1 < 0.
Proposition
Let x ∈ Q. Then < x > p is eventually periodic with period <
q
p−1
p−q t.
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
Finite qp -expansion
Let x have finite qp -expansion, then
n
X
s
ai p i
= n+1 ,
x=
q q
q
s > 0.
i=0
Define for all positive integers m the set
)
(
D s E
INF(m) := s s > 0, m 1 p is infinite .
q
q q
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
Finite qp -expansion
Let x have finite qp -expansion, then
n
X
s
ai p i
= n+1 ,
x=
q q
q
s > 0.
i=0
Define for all positive integers m the set
)
(
D s E
INF(m) := s s > 0, m 1 p is infinite .
q
q q
Example (p = 3, q = 2)
INF(1) = ∅, INF(2) = {1}, INF(3) = {1, 2, 3, 5, 7, 11},
INF(4) = {1, 2, 3, 4, 5, . . . , 22, 23, 25, 29, 31, 33, 37, 41, 49},
#INF(5) = 277, #INF(6) = 1101
p
-representation
q
p-adic numbers
Positive integers
Negative integers
Rationals
p
-representation
q
Finite case
Finite qp -expansion – II
2
0
m=0
0
−1
1
2
3
4
m=1
− 12
1
2
3
2
5
2
7
2
m=2
− 34
− 14
1
4
3
4
5
4
7
4
9
4
11
4
13
4
15
4
m=3
1
8
3
8
5
8
7
8
9
8
11
8
13
8
15
8
17
8
19
8
21
8
23
8
25
8
m=4
29
16
33
16
37
16
41
16
49
16
27
8
29
8
31
8
p-adic numbers
Positive integers
Negative integers
Rationals
p
-representation
q
Finite case
Finite qp -expansion – II
1
m=0
2
0
2
0
−1
1
1
0
2
3
4
m=1
− 12
1
2
3
2
5
2
7
2
m=2
− 34
− 14
1
4
3
4
5
4
7
4
9
4
11
4
13
4
15
4
m=3
1
8
3
8
5
8
7
8
9
8
11
8
13
8
15
8
17
8
19
8
21
8
23
8
25
8
m=4
29
16
33
16
37
16
41
16
49
16
27
8
29
8
31
8
p-adic numbers
Positive integers
Negative integers
Rationals
p
-representation
q
Finite case
Finite qp -expansion – II
1
m=0
2
0
2
0
−1
1
1
2
3
4
1
2
0
1
0
m=1
− 12
1
2
3
2
0
1
5
2
0
2
7
2
1
0
2
m=2
− 34
− 14
1
4
3
4
5
4
7
4
9
4
11
4
13
4
15
4
m=3
1
8
3
8
5
8
7
8
9
8
11
8
13
8
15
8
17
8
19
8
21
8
23
8
25
8
m=4
29
16
33
16
37
16
41
16
49
16
27
8
29
8
31
8
p-adic numbers
Positive integers
Negative integers
Rationals
p
-representation
q
Finite case
Finite qp -expansion – II
1
m=0
2
0
2
0
−1
1
1
2
3
4
1
2
0
1
2
0
m=1
− 12
1
2
1
3
2
0
2
1
5
2
0
2
7
2
1
0
2
m=2
− 34
− 14
1
4
3
4
5
4
7
4
9
4
11
4
13
4
15
4
m=3
1
8
3
8
5
8
7
8
9
8
11
8
13
8
15
8
17
8
19
8
21
8
23
8
25
8
m=4
29
16
33
16
37
16
41
16
49
16
27
8
29
8
31
8
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p
-representation
q
Alternative algorithm
Algorithm
Let x = qsm , s, m positive integers. Put h0 = s and hi+1 and bi ∈ Ap
define as follows: for i = 0, 1, . . . , m − 1 by
hi
m−(i+1)
q
=p
hi+1
m−(i+1)
q
for i = m, m + 1, . . . by
qhi = phi+1 + bi .
Return b = · · · b2 b1 b0 .
+ bi ,
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p
-representation
q
Alternative algorithm
Algorithm
Let x = qsm , s, m positive integers. Put h0 = s and hi+1 and bi ∈ Ap
define as follows: for i = 0, 1, . . . , m − 1 by
hi
m−(i+1)
q
=p
hi+1
m−(i+1)
q
+ bi ,
for i = m, m + 1, . . . by
qhi = phi+1 + bi .
Return b = · · · b2 b1 b0 .
It holds that b =< x > p and
q
(
hi q i
si =
hi q m
i = 0, 1, . . . , m − 1
i = m, m + 1, . . . .
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p
-representation
q
Finite qp -expansions III
Proposition
INF(1) = ∅ and
INF(m) = A(m) ∪ B(m), m = 2, 3, . . . , where
n
o
A(m) =
−kp + aq m−1 | k ≥ 1, a ∈ Ap ∩ N
n
o
B(m) =
pk + aq m−1 | k ∈ F (m − 1), a ∈ Ap .
Moreover, for m ≥ 2 we get
p−1
max INF(m)
=
m
q
p−q
p
q -expansion
m
m
p
p
−1 −
,
q
q
of this number is ω (p − q)(p − 1)m .
p-adic numbers
Positive integers
Negative integers
Rationals
Finite case
p
-representation
q
Finite qp -expansions – example
Example (p = 12, q = 5)
INF(1) = ∅
INF(2) = {1, 2, 3, 4, 6, 7, 8, 9, 11, 13, 14, . . . , 26, 28, 31, 33, 38, 43}
#INF(3) = 1480,
#INF(4) = 26100,
#INF(5) = 403549.
p-adic numbers
Positive integers
Negative integers
Rationals
p
q -representation
Finite case
p
-representation
q
I
Definition
A left infinite word · · · a−`0 +1 a−`0 , `0 ∈ N, over Ap is a
p
q -representation of x ∈ Qr if a−`0 > 0 or `0 = 0 and
∞
X
ak p k
x=
q q
k =−`0
with respect to | |r .
If the qp -representation is not finite (i.e., does not begin in infinitely
many zeros), then r is a prime factor of p.
p-adic numbers
Positive integers
Negative integers
Rationals
p
q -representation
Let p =
r1j1 r2j2
Finite case
p
-representation
q
II
· · · rkjk .
Theorem
Let x ∈ Qri for some i ∈ {1, . . . , k }.
(i) If k = 1, there exists a unique qp -representation of x in Qri .
(ii) If k > 1, there exist uncountably many qp -representations of x
in Qri .
p-adic numbers
Positive integers
Negative integers
Rationals
p
q -representation
Let p =
r1j1 r2j2
Finite case
p
-representation
q
II
· · · rkjk .
Theorem
Let x ∈ Qri for some i ∈ {1, . . . , k }.
(i) If k = 1, there exists a unique qp -representation of x in Qri .
(ii) If k > 1, there exist uncountably many qp -representations of x
in Qri .
Let x ∈ Q and let i1 , . . . , i` ∈ {1, . . . , k }, ` < k , be distinct.
(i) There exist uncountably many qp -representations which work in
all fields Qri1 , . . . , Qri` .
(ii) There exists a unique qp -representation which works in all
fields Qr1 , . . . , Qrk ; namely, the qp -expansion < x > p .
q
(ii) The qp -expansion < x > p is the only qp -representation which is
q
eventually periodic.