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The $2$ -adic absolute value gives an ultrametric on $Q$ , in which every triangle is isosceles and every point of a ball is a centre

Example

Write $2 : = 1 + 1$ . Call an integer (The integers as equivalence classes of pairs of naturals) even if it is $2 t$ for some integer $t$ , and odd if it is $2 t + 1$ for some integer $t$ .

The $2$ -adic valuation. Every nonzero integer $m$ can be written in exactly one way as

$m = 2^{j} u with j \in N and u odd,$

and $v_{2} (m) : = j$ is the $2$ -adic valuation of $m$ (Integer powers $a^{m}$ ). For a nonzero rational $x$ (The rationals as equivalence classes of pairs of integers), written $x = a / b$ with $a$ and $b$ nonzero integers, the integer

$v_{2} (x) : = v_{2} (a) - v_{2} (b)$

does not depend on the chosen representation. The $2$ -adic absolute value is

$∣ x ∣_{2} : = 2^{- v_{2} (x)} (x \neq 0), ∣ 0 ∣_{2} : = 0,$

read inside $R$ through the embeddings $Z \to Q \to R$ (The integers embed in the rationals, The unique embedding of ℚ into an ordered field), and the $2$ -adic distance is

$d_{2} (x, y) : = ∣ x - y ∣_{2} (x, y \in Q) .$

Claims.

$d_{2}$ is an ultrametric on $Q$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric): it satisfies (M1), (M2) and the strong triangle inequality $d_{2} (x, z) \leq \max {d_{2} (x, y), d_{2} (y, z)}$ .
Every triangle is isosceles: if $d_{2} (x, y) \neq d_{2} (y, z)$ then $d_{2} (x, z) = \max {d_{2} (x, y), d_{2} (y, z)}$ .
Every point of a ball is a centre: if $y \in B (x, r)$ then $B (y, r) = B (x, r)$ (Open ball, closed ball and sphere in a metric space).

Why $p = 2$ and not a general prime. The general $p$ -adic valuation needs primality and unique factorisation in $Z$ , which are developed on Primes, Euclid's Lemma and the Fundamental Theorem of Arithmetic (The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ , Euclid's lemma: if $p$ is prime and $p ∣ a b$ then $p ∣ a$ or $p ∣ b$ , The fundamental theorem of arithmetic: every integer $n \geq 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i < r} p_{i} = \prod_{j < s} q_{j}$ with every $p_{i}$ and $q_{j}$ prime, then $r = s$ and $q_{i} = p_{π (i)}$ for some $π \in Sym (r)$ , The $p$ -adic valuation extends to the nonzero rationals by $v_{p} (a / b) : = v_{p} (a) - v_{p} (b) \in Z$ , independently of the representation; it satisfies $v_{p} (x y) = v_{p} (x) + v_{p} (y)$ , and $v_{p} (x + y) \geq \min {v_{p} (x), v_{p} (y)}$ whenever $x$ , $y$ and $x + y$ are nonzero) and are therefore available here; this item nevertheless develops the case $p = 2$ from parity alone, so that the ultrametric geometry below rests on nothing but the discreteness of $Z$ . At $p = 2$ everything reduces to parity, which is available: The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ partitions $N$ into the ranges of its two index maps, and that is what claim 2 of the verification turns into "even or odd, never both".

Claims 2 and 3 use nothing about $d_{2}$ beyond the strong triangle inequality, so they hold in every ultrametric space.

Facts & Assumptions

Given: The integers $Z$ and rationals $Q$ with their arithmetic; the successor $σ$ on $N$ ; the element $2 = 1 + 1$ ; the index maps $e, o$ of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ; and integers $m, a, c$ and rationals $x, y, z$ as introduced in the steps.

[L1]

Ring and field arithmetic: $Z$ is a commutative ring (The integers form a commutative ring) and $Q$ a field (The rationals form a field, The rationals as equivalence classes of pairs of integers); $Z$ is totally ordered and its order is compatible with addition and with multiplication by positives (The integers form a totally ordered ring); nonzero integers have nonzero product and cancel (The integers have no zero divisors; multiplicative cancellation).

[L2]

Induction (The principle of mathematical induction) and strong induction (Strong (complete) induction) on $N$ (The natural numbers $N$ (von Neumann)).

[L3]

The index maps: $e_{0} = 0$ , $e_{σ (j)} = σ (σ (e_{j}))$ , $o_{0} = σ (0)$ , $o_{σ (j)} = σ (σ (o_{j}))$ , and $N$ is the disjoint union of the ranges of $e$ and $o$ , each natural lying in exactly one range and being hit exactly once (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ).

[L4]

Addition on $N$ : $m + 0 = m$ , $m + σ (k) = σ (m + k)$ (Addition of natural numbers) and $σ (m) + n = σ (m + n)$ (Left successor law for addition); the order is total and transitive (Order on the natural numbers, $\leq$ is a linear order on $N$ ), and $n \neq 0$ gives $n \geq 1$ (Discreteness: $σ (n)$ is the immediate successor).

[L5]

The embedding $N \to Z$ is injective, preserves addition, multiplication and order, and its image is exactly the set of integers $\geq 0$ (The naturals embed in the integers); the embeddings $Z \to Q \to R$ are injective and order preserving (The integers embed in the rationals, The unique embedding of ℚ into an ordered field).

[L6]

Powers: $a^{0} = 1$ , $a^{n + 1} = a^{n} a$ , $a^{m + n} = a^{m} a^{n}$ and $a^{- m} = (a^{m})^{- 1}$ , valid for integer exponents when $a \neq 0$ , and $a \neq 0$ gives $a^{n} \neq 0$ (Integer powers $a^{m}$ , Laws of integer exponents); for $a > 1$ and $k \geq 1$ one has $a^{k} \geq a > 1$ , and $a > 0$ gives $a^{n} > 0$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L7]

Order in $R$ : $0 < 1$ hence $0 < 1 < 2$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities); products and inverses of positives are positive and scaling preserves inequalities (Sign rules for products and monotonicity of multiplication, Inverses of positives are positive, and reciprocation reverses order, Field, Ordered field, Complete ordered field (least-upper-bound property)).

[L8]

Metric notions: the axioms (M1), (M2), (M3), the strong form (M3'), and the fact that a function satisfying (M1), (M2), (M3') is nonnegative and hence satisfies (M3), the maximum of two nonnegative reals being at most their sum (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, Nonnegativity of a metric is a consequence of the other axioms, not an axiom, Maximum and minimum of a set, Every nonempty finite set of reals has a maximum and a minimum); balls are as in Open ball, closed ball and sphere in a metric space.

Verification

technique · direct

1.1L2L3L4

For every $j \in N$ one has $e_{j} = j + j$ and $o_{j} = σ (j + j)$ , by induction on $j$ : at $j = 0$ , $e_{0} = 0 = 0 + 0$ and $o_{0} = σ (0) = σ (0 + 0)$ ; and if $e_{j} = j + j$ and $o_{j} = σ (j + j)$ , then $e_{σ (j)} = σ (σ (e_{j})) = σ (σ (j + j)) = σ (j) + σ (j)$ and $o_{σ (j)} = σ (σ (o_{j})) = σ (σ (σ (j + j))) = σ (σ (j) + σ (j))$ , using $σ (p) + q = σ (p + q)$ and $p + σ (q) = σ (p + q)$ .

1.2L1L4L5

No integer $k$ satisfies $0 < k < 1$ : such a $k$ would be positive, hence the image of a natural $n \neq 0$ , so $n \geq 1$ and, the embedding being order preserving, $k \geq 1$ , contradicting $k < 1$ .

1.3L1

A product of two odd integers is odd: $(2 a + 1) (2 c + 1) = 2 (2 a c + a + c) + 1$ by ring arithmetic.

2.1step 1.1L3

Every natural number is $j + j$ for exactly one $j$ , or $σ (j + j)$ for exactly one $j$ , and never both: this is the disjoint-union statement for the ranges of $e$ and $o$ , rewritten through step 1.1.

2.2step 1.2L1L7

No integer is both even and odd: if $2 s = 2 t + 1$ then the integer $k : = s - t$ satisfies $2 k = 1$ , and $k \leq 0$ would give $2 k \leq 0 < 1$ while $k \geq 1$ would give $2 k \geq 2 > 1$ , so $0 < k < 1$ , which step 1.2 forbids.

3.1step 2.1L1L5

Every integer is even or odd: an integer $m \geq 0$ is the image of a natural $n$ , which by step 2.1 is $j + j$ or $σ (j + j)$ , so $m$ is $2 t$ or $2 t + 1$ with $t$ the image of $j$ , the embedding preserving addition and successors; and if $m < 0$ then $- m > 0$ is $2 t$ or $2 t + 1$ , whence $m = 2 (- t)$ or $m = 2 (- t - 1) + 1$ .

3.2step 2.2L1L4L6

The representation is unique: if $2^{j} u = 2^{j^{'}} u^{'}$ with $u, u^{'}$ odd and, without loss of generality, $j \leq j^{'}$ , then $2^{j^{'}} = 2^{j} 2^{j^{'} - j}$ and cancelling the nonzero factor $2^{j}$ gives $u = 2^{j^{'} - j} u^{'}$ ; if $j^{'} > j$ then $j^{'} - j \geq 1$ and $u = 2 (2^{j^{'} - j - 1} u^{'})$ is even as well as odd, which step 2.2 forbids; so $j = j^{'}$ and then $u = u^{'}$ .

4.1step 2.1step 3.1L1L2L4L5L6

Every nonzero integer $m$ is $2^{j} u$ with $j \in N$ and $u$ odd: apply strong induction to the property $P (n)$ that every nonzero integer whose absolute value is the image of $n$ , that is every nonzero $m$ equal to the image of $n$ or to its negative, has such a representation. Given $n$ and $P (k)$ for all $k < n$ , note $n \neq 0$ since $m \neq 0$ ; by step 2.1 either $n = σ (j + j)$ , in which case $m$ is $\pm (2 t + 1)$ and hence odd by the computation of step 3.1, so $m = 2^{0} m$ works, or $n = j + j$ with $j \neq 0$ , in which case $m = 2 m^{'}$ for the nonzero integer $m^{'}$ that is the image of $j$ or its negative, and $j < n$ because $j \geq 1$ gives $n = j + j \geq j + 1 > j$ , so $P (j)$ supplies $m^{'} = 2^{i} u$ and $m = 2^{i + 1} u$ . Every nonzero integer is the image of some natural or its negative, so the conclusion holds for all of them.

5.1step 3.2step 4.1L1L6

For nonzero integers $a, c$ with $a + c \neq 0$ : $v_{2} (a + c) \geq \min {v_{2} (a), v_{2} (c)}$ . Indeed write $a = 2^{s} u$ and $c = 2^{s^{'}} u^{'}$ with $u, u^{'}$ odd and, without loss of generality, $s \leq s^{'}$ ; then $a + c = 2^{s} (u + 2^{s^{'} - s} u^{'})$ , the bracket is a nonzero integer, so by step 4.1 it equals $2^{i} q$ with $q$ odd, whence $a + c = 2^{s + i} q$ and, by the uniqueness of step 3.2, $v_{2} (a + c) = s + i \geq s = \min {s, s^{'}}$ .

5.2step 1.3step 3.2step 4.1L1L6

The valuation of a nonzero rational is well defined: if $a / b = c / d$ with $a, b, c, d$ nonzero integers then $a d = c b$ , and writing $a = 2^{s} u$ , $b = 2^{t} w$ , $c = 2^{s^{'}} u^{'}$ , $d = 2^{t^{'}} w^{'}$ with all four of $u, w, u^{'}, w^{'}$ odd gives $a d = 2^{s + t^{'}} (u w^{'})$ and $c b = 2^{s^{'} + t} (u^{'} w)$ with $u w^{'}$ and $u^{'} w$ odd by step 1.3; the uniqueness of step 3.2 applied to the nonzero integer $a d$ forces $s + t^{'} = s^{'} + t$ , that is $s - t = s^{'} - t^{'}$ .

6.1step 5.2L1L6L7

Basic properties of $∣ \cdot ∣_{2}$ : for $x \neq 0$ the value $2^{- v_{2} (x)}$ is a positive real, since $2 > 0$ and powers and inverses of positives are positive; so $∣ x ∣_{2} > 0$ for $x \neq 0$ and $∣ x ∣_{2} = 0$ exactly when $x = 0$ . Moreover $∣ - x ∣_{2} = ∣ x ∣_{2}$ , because $- x = (- a) / b$ and $- (2^{s} u) = 2^{s} (- u)$ with $- u = 2 (- t - 1) + 1$ odd when $u = 2 t + 1$ , so $v_{2} (- x) = v_{2} (x)$ .

7.1step 5.1step 5.2L1L5L6L7

Strong triangle inequality for $∣ \cdot ∣_{2}$ : $∣ x + y ∣_{2} \leq \max {∣ x ∣_{2}, ∣ y ∣_{2}}$ . If $x = 0$ , or $y = 0$ , or $x + y = 0$ , this is immediate from step 6.1. Otherwise write $x = a / b$ and $y = c / b$ over a common nonzero denominator $b$ , with $a, c$ nonzero integers, so that $x + y = (a + c) / b$ with $a + c \neq 0$ ; then $v_{2} (x) = v_{2} (a) - v_{2} (b)$ , $v_{2} (y) = v_{2} (c) - v_{2} (b)$ and $v_{2} (x + y) = v_{2} (a + c) - v_{2} (b)$ , so step 5.1 gives $v_{2} (x + y) \geq \min {v_{2} (x), v_{2} (y)}$ ; finally $t \mapsto 2^{t}$ is strictly increasing on $Z$ , since for $t < t^{'}$ one has $2^{t^{'}} = 2^{t} 2^{t^{'} - t}$ with $2^{t} > 0$ and $2^{t^{'} - t} > 1$ , so $2^{- v_{2} (x + y)} \leq 2^{- \min {v_{2} (x), v_{2} (y)}} = \max {2^{- v_{2} (x)}, 2^{- v_{2} (y)}}$ .

8.1step 6.1step 7.1L8

Claim 1: $d_{2} (x, y) = ∣ x - y ∣_{2}$ vanishes exactly when $x = y$ by step 6.1, is symmetric because $∣ y - x ∣_{2} = ∣ - (x - y) ∣_{2} = ∣ x - y ∣_{2}$ , and satisfies $d_{2} (x, z) = ∣ (x - y) + (y - z) ∣_{2} \leq \max {∣ x - y ∣_{2}, ∣ y - z ∣_{2}} = \max {d_{2} (x, y), d_{2} (y, z)}$ by step 7.1; being nonnegative, it also satisfies the ordinary triangle inequality, the maximum of two nonnegative reals being at most their sum. So $d_{2}$ is an ultrametric on $Q$ .

9.1step 8.1L7L8

Claim 2: suppose $d_{2} (x, y) \neq d_{2} (y, z)$ and, without loss of generality, $d_{2} (x, y) < d_{2} (y, z)$ . Then $d_{2} (x, z) \leq \max {d_{2} (x, y), d_{2} (y, z)} = d_{2} (y, z)$ ; and $d_{2} (y, z) \leq \max {d_{2} (y, x), d_{2} (x, z)} = \max {d_{2} (x, y), d_{2} (x, z)}$ , where the maximum cannot be $d_{2} (x, y)$ , since that would give $d_{2} (y, z) \leq d_{2} (x, y) < d_{2} (y, z)$ ; so $d_{2} (y, z) \leq d_{2} (x, z)$ and the two are equal, that is $d_{2} (x, z) = \max {d_{2} (x, y), d_{2} (y, z)}$ .

9.2step 8.1L8

Claim 3: let $y \in B (x, r)$ , so $d_{2} (x, y) < r$ . For $z \in B (y, r)$ the strong triangle inequality gives $d_{2} (x, z) \leq \max {d_{2} (x, y), d_{2} (y, z)} < r$ , so $z \in B (x, r)$ ; and for $z \in B (x, r)$ it gives $d_{2} (y, z) \leq \max {d_{2} (y, x), d_{2} (x, z)} < r$ , so $z \in B (y, r)$ . Hence $B (y, r) = B (x, r)$ .

10.1step 8.1step 9.1step 9.2∎

Claims 1, 2 and 3 are established by steps 8.1, 9.1 and 9.2, so the $2$ -adic distance is an ultrametric on $Q$ in which every triangle is isosceles and every point of a ball is a centre.

Remarks

Where parity is spent, and where it is not. The partition of $N$ into the two ranges of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ is used exactly twice: in step 2.1, to know that a natural is even or odd, and through it in step 3.1 for the integers. Uniqueness of the valuation (step 3.2) uses only that no integer is both even and odd, which step 2.2 derives from the discreteness of $Z$ rather than from the partition.
What an ultrametric costs and what it buys. Claims 2 and 3 are formal consequences of (M3') and hold in every ultrametric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric); they are recorded here because they are the two facts that make ultrametric geometry look unlike the real line, where a triangle need not be isosceles and a ball has exactly one centre.
The general $p$ -adic absolute value, and what this item does instead. Its well-definedness needs the primality of $p$ in the form of Euclid's lemma, that $p$ dividing a product divides one of the factors. That is Euclid's lemma: if $p$ is prime and $p ∣ a b$ then $p ∣ a$ or $p ∣ b$ , and the resulting valuation on $Q^{\times}$ is The $p$ -adic valuation extends to the nonzero rationals by $v_{p} (a / b) : = v_{p} (a) - v_{p} (b) \in Z$ , independently of the representation; it satisfies $v_{p} (x y) = v_{p} (x) + v_{p} (y)$ , and $v_{p} (x + y) \geq \min {v_{p} (x), v_{p} (y)}$ whenever $x$ , $y$ and $x + y$ are nonzero, both on the primes page and both available here. This item deliberately does not use them: at $p = 2$ the statement doing the same work is that a product of odd integers is odd, proved in step 1.3 above by a one-line ring computation, so the whole development below is self-contained from parity.

The $2$ -adic absolute value gives an ultrametric on $Q$ , in which every triangle is isosceles and every point of a ball is a centre

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

The 22-adic absolute value gives an ultrametric on QQ, in which every triangle is isosceles and every point of a ball is a centre

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

The $2$ -adic absolute value gives an ultrametric on $Q$ , in which every triangle is isosceles and every point of a ball is a centre