TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31

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Ordered coprime pairs in a box have asymptotic density 6 over pi squared

Statement

Let $C (x)$ be the number of ordered pairs $(a, b)$ of positive integers with $a, b \leq x$ and $\gcd (a, b) = 1$ (Coprime integers: $\gcd (a, b) = 1$ ). Then

$C (x) = \frac{6}{π^{2}} x^{2} + O (x \log x)$

for real $x \geq 1$ .

Facts & Assumptions

Given: A real $x \geq 1$ and $N : = ⌊ x ⌋$ .

Proof

technique · direct

1.1givenalgebra

Since $a, b \leq x$ is equivalent to $a, b \leq N$ , the count is unchanged if $x$ is replaced by $N$ . For $m = 1$ there is exactly one coprime pair with $\max (a, b) = 1$ , namely $(1, 1)$ . For $m \geq 2$ , the coprime pairs with $\max (a, b) = m$ are exactly $(m, b)$ with $1 \leq b \leq m$ and $\gcd (m, b) = 1$ , together with $(a, m)$ with $1 \leq a \leq m$ and $\gcd (a, m) = 1$ ; these two families are disjoint because $(m, m)$ is not coprime. Hence $C (x) = 1 + 2 \sum_{m = 2}^{N} φ (m) = 2 \sum_{m = 1}^{N} φ (m) - 1.$

2.1step 1.1givenalgebra∎

Applying The summatory totient function is 3 over pi squared times x squared plus O(x log x) at $x = N$ gives $C (x) = \frac{6}{π^{2}} N^{2} + O (N \log N) .$ Since $N = x + O (1)$ , this is $C (x) = \frac{6}{π^{2}} x^{2} + O (x \log x) .$

Depends on

Used by

The proportion of pairs in 1,...,n² that are coprime tends to 6 over pi squared Corollary

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Karl-Dieter Crisman, Number Theory: In Context and Interactive (standard reference, not scraped)