A p-Adic Style Metric from Divisibility

Distance Measured by Divisibility

Ordinary distance on the integers records how large the difference between two numbers is. A different metric can instead record how many times a fixed prime divides that difference. Differences divisible by a large power of the prime are assigned small distances. The resulting rule is a genuine metric, and its triangle inequality is stronger than the usual one: the distance between two endpoints cannot exceed the larger of the two intervening distances.

Fix a prime number $p$. For a nonzero integer $n$, let $v_p(n)$ be the largest nonnegative integer $k$ such that $p^k$ divides $n$. Divisibility here applies to negative integers as well: $p^k$ divides $n$ when $n=p^kq$ for some integer $q$. The exponent $v_p(n)$ exists. Indeed, $p^0=1$ divides $n$, and if $p^k$ divides nonzero $n$, then $p^k\leq |n|$; only finitely many powers of $p\geq2$ can satisfy this bound.

Definition: For integers $x,y$, define $$ d_p(x,y)= \begin{cases} 0, & x=y,\\ p^{-v_p(x-y)}, & x\ne y. \end{cases} $$ This is the p-adic style metric on $\mathbb{Z}$ determined by $p$. The integer set is being equipped with this metric, not with the usual absolute-value distance.

For a nonzero difference, the exponent is always at least zero, so $d_p(x,y)\leq1$. A difference divisible by $p^k$ but not by $p^{k+1}$ has distance $p^{-k}$. Thus, for this metric, divisibility by a higher power means closeness.

Why the Rule Is a Metric

A metric on a set $X$ is a function $d:X\times X\to[0,\infty)$ satisfying, for all $x,y,z\in X$: $d(x,y)=0$ exactly when $x=y$; $d(x,y)=d(y,x)$; and $d(x,z)\leq d(x,y)+d(y,z)$. We will prove these properties and, more strongly, establish the strong triangle inequality.

Theorem: For every prime $p$, the function $d_p$ is a metric on $\mathbb{Z}$. Moreover, for all $x,y,z\in\mathbb{Z}$, $$ d_p(x,z)\leq \max\{d_p(x,y),d_p(y,z)\}. $$

Proof. The value $d_p(x,y)$ is nonnegative: it is either zero or a positive power $p^{-k}$. If $x=y$, its value is zero by definition. If $x\ne y$, then $x-y\ne0$, so $v_p(x-y)$ is a finite integer and $p^{-v_p(x-y)}>0$. Therefore $d_p(x,y)=0$ exactly when $x=y$.

For symmetry, $y-x=-(x-y)$. An integer divides $x-y$ exactly when it divides its negative, so $v_p(y-x)=v_p(x-y)$ when $x\ne y$. When $x=y$, both distances are zero. Hence $d_p(x,y)=d_p(y,x)$ in all cases.

To prove the stronger inequality while including zero differences, set $v_p(0)=\infty$ for this proof only and use $p^{-\infty}=0$. Let $m=\min\{v_p(x-y),v_p(y-z)\}$. If $m$ is finite, both differences are divisible by $p^m$. Their sum $(x-y)+(y-z)=x-z$ is therefore divisible by $p^m$. If $x-z\ne0$, this gives $v_p(x-z)\geq m$, and consequently $d_p(x,z)=p^{-v_p(x-z)}\leq p^{-m}$. If $x-z=0$, then $d_p(x,z)=0\leq p^{-m}$. If $m=\infty$, both $x-y$ and $y-z$ are zero, so $x=z$ and $d_p(x,z)=0$. In every case, $$ d_p(x,z)\leq p^{-m} =\max\{p^{-v_p(x-y)},p^{-v_p(y-z)}\} =\max\{d_p(x,y),d_p(y,z)\}. $$ Since the maximum of two nonnegative numbers is at most their sum, the usual triangle inequality follows as well. Thus all metric axioms hold. $\square$

A metric satisfying $d(x,z)\leq\max\{d(x,y),d(y,z)\}$ is called an ultrametric. The argument shows exactly how divisibility supplies this stronger inequality: if both successive differences have a factor $p^m$, their sum has that factor too.

Computing with the Metric

Worked Example: Distances for the Prime 2

Use $p=2$ to compute the three distances between $14$, $2$, and $6$, then check the strong triangle inequality with endpoints $14$ and $6$.

For the pair $14,2$, the difference is $14-2=12=2^2\cdot3$. The factor $3$ is not divisible by $2$, so $v_2(12)=2$, and $d_2(14,2)=2^{-2}=1/4$.

For $2,6$, the difference is $2-6=-4=-(2^2)$. Its absolute value is divisible by $2^2$, but not by $2^3=8$, so $v_2(-4)=2$ and $d_2(2,6)=2^{-2}=1/4$.

For $14,6$, the difference is $14-6=8=2^3$. It is divisible by $2^3$, but not by $2^4=16$, so $v_2(8)=3$ and $d_2(14,6)=2^{-3}=1/8$.

Check: the strong triangle inequality gives $d_2(14,6)\leq\max\{d_2(14,2),d_2(2,6)\}$. Substituting the computed distances gives $1/8\leq\max\{1/4,1/4\}=1/4$, which is true. The ordinary triangle inequality also holds: $1/8\leq1/4+1/4=1/2$.

Worked Example: A Zero Difference and Negative Differences

Use $p=3$ to compute the distances among $1$, $10$, and $19$. This example includes differences that are negative and a pair of equal distances.

The distance from $1$ to $10$ uses $1-10=-9=-(3^2)$. Since $9$ is divisible by $3^2=9$ and not by $3^3=27$, $v_3(-9)=2$, so $d_3(1,10)=3^{-2}=1/9$.

The distance from $10$ to $19$ uses $10-19=-9$ again, giving $v_3(-9)=2$ and $d_3(10,19)=3^{-2}=1/9$.

The distance from $1$ to $19$ uses $1-19=-18=-(3^2\cdot2)$. Since $2$ is not divisible by $3$, the highest power of $3$ dividing $-18$ is $3^2$. Thus $v_3(-18)=2$ and $d_3(1,19)=3^{-2}=1/9$.

Check: with endpoints $1$ and $19$ and intermediate point $10$, the inequality reads $1/9\leq\max\{1/9,1/9\}=1/9$, so equality holds. Also $d_3(10,10)=0$ by the equal-points clause in the definition; no exponent for a nonzero difference is needed.

Worked Example: The Prime Changes the Distance

Compare the distance between $0$ and $12$ for $p=2$, $p=3$, and $p=5$.

For $p=2$, $12=2^2\cdot3$, with $3$ not divisible by $2$. Hence $v_2(12)=2$ and $d_2(0,12)=2^{-2}=1/4$.

For $p=3$, $12=3\cdot4$, with $4$ not divisible by $3$. Hence $v_3(12)=1$ and $d_3(0,12)=3^{-1}=1/3$.

For $p=5$, $12$ is not divisible by $5$, while $5^0=1$ divides every integer. Thus $v_5(12)=0$ and $d_5(0,12)=5^0=1$.

Check: in each case the computed distance is the defining power of its chosen prime: $2^{-2}=1/4$, $3^{-1}=1/3$, and $5^0=1$. The same integer difference therefore has different distances for different choices of prime.

What the Strong Inequality Says

The proof depends on a basic divisibility fact, not on the size of the integer differences. If $p^m$ divides each of $x-y$ and $y-z$, then it divides their sum $x-z$. In distance language, if both steps have distance at most $p^{-m}$, the direct step has distance at most $p^{-m}$ too. This is why the maximum, rather than the sum, controls the endpoint distance.

For example, if two successive differences are both divisible by $p^4$, then their sum is divisible by $p^4$, even if the sum is zero or has additional factors of $p$. The resulting endpoint distance is therefore either zero or at most $p^{-4}$. No assumption that the three integers are distinct is needed; zero differences were included in the proof.

The prime $p$ must be fixed when defining $d_p$, and it must be prime as specified. For the proof of the strong inequality, the key requirement is that divisibility by $p^m$ is preserved under addition; the metric formula and its values are then tied to the chosen prime. Changing the prime changes the function, as the calculation for $12$ demonstrates.

A Common Mistake: Reversing the Scale

Common pitfall. The exponent is not itself the distance. For $p=2$, the difference $8$ has $v_2(8)=3$. A mistaken calculation might write $d_2(0,8)=3$, treating the exponent as the distance. The definition instead gives $d_2(0,8)=2^{-v_2(8)}=2^{-3}=1/8$. Substitution into the formula corrects the error: a higher exponent produces a smaller positive distance, not a larger one.

There is a second useful check on the formula. If $x\ne y$ and $p$ does not divide $x-y$, then $v_p(x-y)=0$, so $d_p(x,y)=p^0=1$. If $p^k$ divides $x-y$ for some $k\geq1$, the distance is at most $p^{-1}<1$. These conclusions follow directly from the exponent and can help detect a misplaced sign in the power.

Check Your Understanding

For each calculation, identify the prime and use the highest power dividing the stated difference.

For $p=2$, compute $d_2(0,20)$, $d_2(0,24)$, and $d_2(20,24)$.
For $p=3$, find $d_3(5,14)$ and state the exact exponent used.
Suppose $p^3$ divides $x-y$ and $p^3$ divides $y-z$. What bound does the strong triangle inequality proof give for $d_p(x,z)$?
Why does the definition assign distance zero when the two integers are equal rather than applying the exponent formula to their difference?
Give a nonzero integer difference whose distance is $1$ for a chosen prime $p$, and verify the exponent.