Completely multiplicative functions
DefinitionCompletely multiplicative arithmetic function#✓fully proved
Let \(R\) be a type with a multiplication and a zero. An arithmetic function \(f\) with values in \(R\) (so \(f(0) = 0\)) is completely multiplicative if \(f(mn) = f(m) f(n)\) for all \(m, n \in \mathbb{N} = \{0, 1, 2, \dots\}\).
The normalization \(f(1) = 1\) is not part of the definition.
def IsCompletelyMultiplicative [Mul R] [Zero R] (f : ArithmeticFunction R) : Prop :=
∀ m n : ℕ, f (m * n) = f m * f n
Used by 7
TheoremCompletely multiplicative functions preserve products#✓fully proved
Let \(R\) be a type with a multiplication and a zero, and let \(f\) be a completely multiplicative arithmetic function with values in \(R\). Then \(f(mn) = f(m) f(n)\) for all \(m, n \in \mathbb{N}\).
@[simp]
theorem IsCompletelyMultiplicative.map_mul [Mul R] [Zero R] {f : ArithmeticFunction R}
(hf : f.IsCompletelyMultiplicative) (m n : ℕ) : f (m * n) = f m * f n
Uses 1
TheoremNormalized completely multiplicative functions are multiplicative#✓fully proved
Let \(R\) be a monoid with zero and let \(f\) be a completely multiplicative arithmetic function with values in \(R\) such that \(f(1) = 1\). Then \(f\) is multiplicative: \(f(1) = 1\) and \(f(mn) = f(m) f(n)\) for all coprime natural numbers \(m\) and \(n\).
theorem IsCompletelyMultiplicative.isMultiplicative [MonoidWithZero R]
{f : ArithmeticFunction R} (hf : f.IsCompletelyMultiplicative) (h1 : f 1 = 1) :
f.IsMultiplicative
Uses 1 · Used by 1
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