mathlib documentation

data.nat.factorization

Prime factorizations #

n.factorization is the finitely supported function ℕ →₀ ℕ mapping each prime factor of n to its multiplicity in n. For example, since 2000 = 2^4 * 5^3,

TODO #

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noncomputable def nat.factorization (n : ℕ) :
ℕ →₀ ℕ

n.factorization is the finitely supported function ℕ →₀ ℕ mapping each prime factor of n to its multiplicity in n.

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Basic facts about factorization #

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@[simp]
theorem nat.factorization_prod_pow_eq_self {n : ℕ} (hn : n ≠ 0) :
n.factorization.prod pow = n
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@[simp]
theorem nat.factors_count_eq {n p : ℕ} :
list.count p n.factors = ⇑(n.factorization) p

We can write both n.factorization p and n.factors.count p to represent the power of p in the factorization of n: we declare the former to be the simp-normal form. However, since factorization is a finsupp it's noncomputable. This theorem can also be used in reverse to compute values of factorization n p when required.

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theorem nat.eq_of_factorization_eq {a b : ℕ} (ha : a ≠ 0) (hb : b ≠ 0) (h : ∀ (p : ℕ), ⇑(a.factorization) p = ⇑(b.factorization) p) :
a = b
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theorem nat.factorization_inj  :
set.inj_on nat.factorization {x : ℕ | x ≠ 0}

Every nonzero natural number has a unique prime factorization

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@[simp]
theorem nat.factorization_zero  :
0.factorization = 0
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@[simp]
theorem nat.factorization_one  :
1.factorization = 0
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@[simp]
theorem nat.support_factorization {n : ℕ} :
n.factorization.support = n.factors.to_finset

The support of n.factorization is exactly n.factors.to_finset

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theorem nat.factor_iff_mem_factorization {n p : ℕ} :
p ∈ n.factorization.support ↔ p ∈ n.factors
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theorem nat.prime_of_mem_factorization {n p : ℕ} (hp : p ∈ n.factorization.support) :
nat.prime p
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theorem nat.pos_of_mem_factorization {n p : ℕ} (hp : p ∈ n.factorization.support) :
0 < p
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theorem nat.le_of_mem_factorization {n p : ℕ} (h : p ∈ n.factorization.support) :
p ≤ n
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theorem nat.factorization_eq_zero_of_non_prime (n p : ℕ) (hp : ¬nat.prime p) :
⇑(n.factorization) p = 0
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theorem nat.prime.factorization_pos_of_dvd {n p : ℕ} (hp : nat.prime p) (hn : n ≠ 0) (h : p ∣ n) :
0 < ⇑(n.factorization) p
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theorem nat.factorization_eq_zero_iff (n : ℕ) :
n.factorization = 0 ↔ n = 0 ∨ n = 1

The only numbers with empty prime factorization are 0 and 1

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@[simp]
theorem nat.factorization_mul {a b : ℕ} (ha : a ≠ 0) (hb : b ≠ 0) :
(a * b).factorization = a.factorization + b.factorization

For nonzero a and b, the power of p in a * b is the sum of the powers in a and b

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theorem nat.factorization_mul_support {a b : ℕ} (ha : a ≠ 0) (hb : b ≠ 0) :
(a * b).factorization.support = a.factorization.support ∪ b.factorization.support
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@[simp]
theorem nat.factorization_pow (n k : ℕ) :
(n ^ k).factorization = k • n.factorization

For any p, the power of p in n^k is k times the power in n

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@[simp]
theorem nat.prime.factorization {p : ℕ} (hp : nat.prime p) :
p.factorization = finsupp.single p 1

The only prime factor of prime p is p itself, with multiplicity 1

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theorem nat.prime.factorization_pow {p k : ℕ} (hp : nat.prime p) :
(p ^ k).factorization = finsupp.single p k

For prime p the only prime factor of p^k is p with multiplicity k

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theorem nat.eq_pow_of_factorization_eq_single {n p k : ℕ} (hn : n ≠ 0) (h : n.factorization = finsupp.single p k) :
n = p ^ k

If the factorization of n contains just one number p then n is a power of p

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theorem nat.prod_factorization_eq_prod_factors {n : ℕ} {β : Type u_1} [comm_monoid β] (f : ℕ → β) :
n.factorization.prod (λ (p k : ℕ), f p) = ∏ (p : ℕ) in n.factors.to_finset, f p

If a product over n.factorization doesn't use the multiplicities of the prime factors then it's equal to the corresponding product over n.factors.to_finset

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theorem nat.factorization_prod {α : Type u_1} {S : finset α} {g : α → ℕ} (hS : ∀ (x : α), x ∈ S → g x ≠ 0) :
(S.prod g).factorization = ∑ (x : α) in S, (g x).factorization

For any p : ℕ and any function g : α → ℕ that's non-zero on S : finset α, the power of p in S.prod g equals the sum over x ∈ S of the powers of p in g x. Generalises factorization_mul, which is the special case where S.card = 2 and g = id.

Equivalence between ℕ+ and ℕ →₀ ℕ with support in the primes. #

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theorem nat.prod_pow_factorization_eq_self {f : ℕ →₀ ℕ} (hf : ∀ (p : ℕ), p ∈ f.support → nat.prime p) :
(f.prod pow).factorization = f

Any finsupp f : ℕ →₀ ℕ whose support is in the primes is equal to the factorization of the product ∏ (a : ℕ) in f.support, a ^ f a.

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theorem nat.eq_factorization_iff {n : ℕ} {f : ℕ →₀ ℕ} (hn : n ≠ 0) (hf : ∀ (p : ℕ), p ∈ f.support → nat.prime p) :
f = n.factorization ↔ f.prod pow = n
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noncomputable def nat.factorization_equiv  :
ℕ+ ≃ ↥{f : ℕ →₀ ℕ | ∀ (p : ℕ), p ∈ f.support → nat.prime p}

The equiv between ℕ+ and ℕ →₀ ℕ with support in the primes.

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theorem nat.factorization_equiv_apply (n : ℕ+) :
(⇑nat.factorization_equiv n).val = n.val.factorization
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theorem nat.factorization_equiv_inv_apply {f : ℕ →₀ ℕ} (hf : ∀ (p : ℕ), p ∈ f.support → nat.prime p) :
(⇑(nat.factorization_equiv.symm) ⟨f, hf⟩).val = f.prod pow

Factorization and divisibility #

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theorem nat.dvd_of_mem_factorization {n p : ℕ} (h : p ∈ n.factorization.support) :
p ∣ n
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theorem nat.pow_factorization_dvd (n p : ℕ) :
p ^ ⇑(n.factorization) p ∣ n
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theorem nat.pow_succ_factorization_not_dvd {n p : ℕ} (hn : n ≠ 0) (hp : nat.prime p) :
¬p ^ (⇑(n.factorization) p + 1) ∣ n
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theorem nat.factorization_le_iff_dvd {d n : ℕ} (hd : d ≠ 0) (hn : n ≠ 0) :
d.factorization ≤ n.factorization ↔ d ∣ n
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theorem nat.factorization_le_factorization_mul_left {a b : ℕ} (hb : b ≠ 0) :
a.factorization ≤ (a * b).factorization
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theorem nat.factorization_le_factorization_mul_right {a b : ℕ} (ha : a ≠ 0) :
b.factorization ≤ (a * b).factorization
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theorem nat.prime.pow_dvd_iff_le_factorization {p k n : ℕ} (pp : nat.prime p) (hn : n ≠ 0) :
p ^ k ∣ n ↔ k ≤ ⇑(n.factorization) p
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theorem nat.prime.pow_dvd_iff_dvd_pow_factorization {p k n : ℕ} (pp : nat.prime p) (hn : n ≠ 0) :
p ^ k ∣ n ↔ p ^ k ∣ p ^ ⇑(n.factorization) p
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theorem nat.exists_factorization_lt_of_lt {a b : ℕ} (ha : a ≠ 0) (hab : a < b) :
∃ (p : ℕ), ⇑(a.factorization) p < ⇑(b.factorization) p
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@[simp]
theorem nat.factorization_div {d n : ℕ} (h : d ∣ n) :
(n / d).factorization = n.factorization - d.factorization
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theorem nat.dvd_iff_div_factorization_eq_tsub (d n : ℕ) (hd : d ≠ 0) (hdn : d ≤ n) :
d ∣ n ↔ (n / d).factorization = n.factorization - d.factorization
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theorem nat.dvd_iff_prime_pow_dvd_dvd (n d : ℕ) :
d ∣ n ↔ ∀ (p k : ℕ), nat.prime p → p ^ k ∣ d → p ^ k ∣ n
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theorem nat.prod_prime_factors_dvd (n : ℕ) :
∏ (p : ℕ) in n.factors.to_finset, p ∣ n
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theorem nat.factorization_gcd {a b : ℕ} (ha_pos : a ≠ 0) (hb_pos : b ≠ 0) :
(a.gcd b).factorization = a.factorization ⊓ b.factorization

Factorization and coprimes #

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theorem nat.factorization_mul_apply_of_coprime {p a b : ℕ} (hab : a.coprime b) :
⇑((a * b).factorization) p = ⇑(a.factorization) p + ⇑(b.factorization) p

For coprime a and b, the power of p in a * b is the sum of the powers in a and b

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theorem nat.factorization_mul_of_coprime {a b : ℕ} (hab : a.coprime b) :
(a * b).factorization = a.factorization + b.factorization

For coprime a and b, the power of p in a * b is the sum of the powers in a and b

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theorem nat.factorization_eq_of_coprime_left {p a b : ℕ} (hab : a.coprime b) (hpa : p ∈ a.factors) :
⇑((a * b).factorization) p = ⇑(a.factorization) p

If p is a prime factor of a then the power of p in a is the same that in a * b, for any b coprime to a.

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theorem nat.factorization_eq_of_coprime_right {p a b : ℕ} (hab : a.coprime b) (hpb : p ∈ b.factors) :
⇑((a * b).factorization) p = ⇑(b.factorization) p

If p is a prime factor of b then the power of p in b is the same that in a * b, for any a coprime to b.

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theorem nat.factorization_disjoint_of_coprime {a b : ℕ} (hab : a.coprime b) :
disjoint a.factorization.support b.factorization.support

The prime factorizations of coprime a and b are disjoint

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theorem nat.factorization_mul_support_of_coprime {a b : ℕ} (hab : a.coprime b) :
(a * b).factorization.support = a.factorization.support ∪ b.factorization.support

For coprime a and b the prime factorization a * b is the union of those of a and b

Induction principles involving factorizations #

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def nat.rec_on_prime_pow {P : ℕ → Sort u_1} (h0 : P 0) (h1 : P 1) (h : Π (a p n : ℕ), nat.prime p → ¬p ∣ a → 0 < n → P a → P ((p ^ n) * a)) (a : ℕ) :
P a

Given P 0, P 1 and a way to extend P a to P (p ^ n * a) for prime p not dividing a, we can define P for all natural numbers.

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  • nat.rec_on_prime_pow h0 h1 h = λ (a : ℕ), a.strong_rec_on (λ (n : ℕ), nat.rec_on_prime_pow._match_1 h0 h1 h n)
  • nat.rec_on_prime_pow._match_1 h0 h1 h (k + 2) = λ (hk : Π (m : ℕ), m < k + 2 → P m), let p : ℕ := (k + 2).min_fac, t : ℕ := list.count p (k + 2).factors in _.mpr (h ((k + 2) / p ^ t) p t _ _ _ (hk ((k + 2) / p ^ t) _))
  • nat.rec_on_prime_pow._match_1 h0 h1 h 1 = λ (_x : Π (m : ℕ), m < 1 → P m), h1
  • nat.rec_on_prime_pow._match_1 h0 h1 h 0 = λ (_x : Π (m : ℕ), m < 0 → P m), h0
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def nat.rec_on_pos_prime_pos_coprime {P : ℕ → Sort u_1} (hp : Π (p n : ℕ), nat.prime p → 0 < n → P (p ^ n)) (h0 : P 0) (h1 : P 1) (h : Π (a b : ℕ), 1 < a → 1 < b → a.coprime b → P a → P b → P (a * b)) (a : ℕ) :
P a

Given P 0, P 1, and P (p ^ n) for positive prime powers, and a way to extend P a and P b to P (a * b) when a, b are positive coprime, we can define P for all natural numbers.

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def nat.rec_on_prime_coprime {P : ℕ → Sort u_1} (h0 : P 0) (hp : Π (p n : ℕ), nat.prime p → P (p ^ n)) (h : Π (a b : ℕ), 1 < a → 1 < b → a.coprime b → P a → P b → P (a * b)) (a : ℕ) :
P a

Given P 0, P (p ^ n) for all prime powers, and a way to extend P a and P b to P (a * b) when a, b are positive coprime, we can define P for all natural numbers.

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def nat.rec_on_mul {P : ℕ → Sort u_1} (h0 : P 0) (h1 : P 1) (hp : Π (p : ℕ), nat.prime p → P p) (h : Π (a b : ℕ), P a → P b → P (a * b)) (a : ℕ) :
P a

Given P 0, P 1, P p for all primes, and a way to extend P a and P b to P (a * b), we can define P for all natural numbers.

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  • nat.rec_on_mul h0 h1 hp h = let hp : Π (p n : ℕ), nat.prime p → P (p ^ n) := λ (p n : ℕ) (hp' : nat.prime p), nat.rec_on_mul._match_1 h1 hp h p hp' n in nat.rec_on_prime_coprime h0 hp (λ (a b : ℕ) (_x : 1 < a) (_x : 1 < b) (_x : a.coprime b), h a b)
  • nat.rec_on_mul._match_1 h1 hp h p hp' (n + 1) = h p (p ^ n.add 0) (hp p hp') (nat.rec_on_mul._match_1 h1 hp h p hp' n)
  • nat.rec_on_mul._match_1 h1 hp h p hp' 0 = h1
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theorem nat.multiplicative_factorization {β : Type u_1} [comm_monoid β] (f : ℕ → β) (h_mult : ∀ (x y : ℕ), x.coprime y → f (x * y) = (f x) * f y) (hf : f 1 = 1) {n : ℕ} :
n ≠ 0 → f n = n.factorization.prod (λ (p k : ℕ), f (p ^ k))

For any multiplicative function f with f 1 = 1 and any n ≠ 0, we can evaluate f n by evaluating f at p ^ k over the factorization of n

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theorem nat.multiplicative_factorization' {β : Type u_1} [comm_monoid β] (f : ℕ → β) (h_mult : ∀ (x y : ℕ), x.coprime y → f (x * y) = (f x) * f y) (hf0 : f 0 = 1) (hf1 : f 1 = 1) {n : ℕ} :
f n = n.factorization.prod (λ (p k : ℕ), f (p ^ k))

For any multiplicative function f with f 1 = 1 and f 0 = 1, we can evaluate f n by evaluating f at p ^ k over the factorization of n