2 Numerators, \(p\)-adic congruences, and von Staudt–Clausen
Throughout, congruences between rational numbers are read \(p\)-adically: we write \(x \equiv y \pmod{p}\) for rationals \(x,y\) when \(x - y \in p\mathbb {Z}_p\).
If a rational number \(q\) has \(|q| {\gt} 1\), then some prime divides \(\operatorname {num}(q)\).
Since \(|q| {\gt} 1\), the numerator of \(|q|\) strictly exceeds its denominator; in particular \(|\operatorname {num}(q)| \ge 2\), so the numerator has a prime factor.
Let \(q, r\) be rationals with \(q \equiv r \pmod{p}\). If \(p\mid \operatorname {num}(q)\), then \(p\mid \operatorname {num}(r)\).
Divisibility of the numerator by \(p\) is equivalent to \(q \in p\mathbb {Z}_p\), which is invariant under perturbation by elements of \(p\mathbb {Z}_p\).
Let \(q\) be a rational and \(n\) a positive natural number. If \(p\mid \operatorname {num}(q/n)\), then \(p\mid \operatorname {num}(q)\).
Dividing by \(n\) can only cancel factors of the numerator.
Let \(p\) be a prime and \(n\) a positive even index with \((p-1)\mid n\). Then \(p\nmid \operatorname {num}(B_n/n)\).
By the von Staudt–Clausen theorem,
the sum running over primes \(q\) with \((q-1)\mid n\). Since \((p-1)\mid n\), the prime \(p\) occurs in the correction sum, so \(B_n + 1/p\) has denominator prime to \(p\). If \(p\) divided \(\operatorname {num}(B_n)\), then adding \(1/p\) would produce a denominator divisible by \(p\), a contradiction; by lemma 2.3 the same holds for the divided numerator.
Let \(p\) be a prime and \(n\) an even index. Then \(pB_n \in \mathbb {Z}_p\).
By von Staudt–Clausen, \(B_n\) differs from an integer by the correction sum \(\sum _{(q-1)\mid n} 1/q\). Each summand has prime denominator, so multiplying by \(p\) clears the single possible factor of \(p\) in the denominator and leaves a \(p\)-adic integer.