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AlexKontorovich/PrimeNumberTheoremAnd
Source indexedtheorem · leanprover/lean4:v4.32.0

Lcm.Criterion.not_highlyAbundant_1

PrimeNumberTheoremAnd.IEANTN.Lcm · PrimeNumberTheoremAnd/IEANTN/Lcm.lean:664 to 723

Source documentation

\subsection{A sufficient condition}

We give a sufficient condition for σ(M)σ(Ln)\sigma(M) \geq \sigma(L_n).

Exact Lean statement

@[blueprint
  "lem:criterion-sufficient"
  (title := "A sufficient inequality")
  (statement := /--
  Suppose
  \[
    \frac{\sigma(M)}{M}
    \Bigl(1 - \frac{4 p_1 p_2 p_3}{q_1 q_2 q_3}\Bigr)
    \;\ge\; \frac{\sigma(L_n)}{L_n}.
  \]
  Then \(\sigma(M) \ge \sigma(L_n)\), and so \(L_n\) is not highly abundant.
  -/)
  (proof := /--
  By Lemma~\ref{lem:M-basic},
  \[
    \frac{L_n}{M} < \Bigl(1 - \frac{4 p_1 p_2 p_3}{q_1 q_2 q_3}\Bigr)^{-1}.
  \]
  Hence
  \[
    \frac{\sigma(M)}{M} \ge \frac{\sigma(L_n)}{L_n}
    \Bigl(1 - \frac{4 p_1 p_2 p_3}{q_1 q_2 q_3}\Bigr)^{-1}
    > \frac{\sigma(L_n)}{L_n} \cdot \frac{M}{L_n}.
  \]
  Multiplying both sides by \(M\) gives
  \[
    \sigma(M) > \sigma(L_n) \cdot \frac{M}{L_n}
  \]
  and hence
  \[
    \sigma(M) \ge \sigma(L_n),
  \]
  since \(M/L_n<1\) and both sides are integers.
  -/)
  (proofUses := ["lem:M-basic"])
  (latexEnv := "lemma")]
theorem Criterion.not_highlyAbundant_1 (c : Criterion)
    (h : σnorm c.M * (1 - (4 : ℝ) * (∏ i, c.p i) / (∏ i, c.q i)) ≥ σnorm (L c.n)) :
    ¬HighlyAbundant (L c.n)

Complete declaration

Lean source

Canonical source
Full Lean sourceLean 4
@[blueprint  "lem:criterion-sufficient"  (title := "A sufficient inequality")  (statement := /--  Suppose  \[    \frac{\sigma(M)}{M}    \Bigl(1 - \frac{4 p_1 p_2 p_3}{q_1 q_2 q_3}\Bigr)    \;\ge\; \frac{\sigma(L_n)}{L_n}.  \]  Then \(\sigma(M) \ge \sigma(L_n)\), and so \(L_n\) is not highly abundant.  -/)  (proof := /--  By Lemma~\ref{lem:M-basic},  \[    \frac{L_n}{M} < \Bigl(1 - \frac{4 p_1 p_2 p_3}{q_1 q_2 q_3}\Bigr)^{-1}.  \]  Hence  \[    \frac{\sigma(M)}{M} \ge \frac{\sigma(L_n)}{L_n}    \Bigl(1 - \frac{4 p_1 p_2 p_3}{q_1 q_2 q_3}\Bigr)^{-1}    > \frac{\sigma(L_n)}{L_n} \cdot \frac{M}{L_n}.  \]  Multiplying both sides by \(M\) gives  \[    \sigma(M) > \sigma(L_n) \cdot \frac{M}{L_n}  \]  and hence  \[    \sigma(M) \ge \sigma(L_n),  \]  since \(M/L_n<1\) and both sides are integers.  -/)  (proofUses := ["lem:M-basic"])  (latexEnv := "lemma")]theorem Criterion.not_highlyAbundant_1 (c : Criterion)    (h : σnorm c.M * (1 - (4 : ) * (∏ i, c.p i) / (∏ i, c.q i))  σnorm (L c.n)) :    ¬HighlyAbundant (L c.n) := by  intro hHA  have hM_pos : (0 : ) < c.M := cast_pos.mpr c.M_pos  have hLn_pos : (0 : ) < L c.n := cast_pos.mpr c.L_pos  have hσnorm_Ln_pos : 0 < σnorm (L c.n) := by    rw [σnorm]; exact div_pos (cast_pos.mpr <| by rw [σ, sigma_pos_iff]; exact c.L_pos) hLn_pos  have hprod_q_pos : (0 : ) < (∏ i, c.q i) := cast_pos.mpr (prod_pos fun i _  (c.hq i).pos)  have h1_sub_pos : (0 : ) < 1 - (4 : ) * (∏ i, c.p i) / (∏ i, c.q i) := by    rw [sub_pos, div_lt_one hprod_q_pos]; exact_mod_cast c.prod_p_le_prod_q  have h1_sub_lt : 1 - (4 : ) * (∏ i, c.p i) / (∏ i, c.q i) < c.M / L c.n := by    have hinv_lt := c.Ln_div_M_lt    rw [lt_inv_comm₀ (div_pos hLn_pos hM_pos) h1_sub_pos, inv_div] at hinv_lt    exact hinv_lt  have hσM_gt : (σ c.M : ) > σ (L c.n) := by    have hσnorm_gt : σnorm c.M > σnorm (L c.n) * (L c.n / c.M) :=      calc σnorm c.M  σnorm (L c.n) / (1 - (4 : ) * (∏ i, c.p i) / (∏ i, c.q i)) := by            rw [ge_iff_le, div_le_iff₀ h1_sub_pos]; exact h        _ > σnorm (L c.n) / (c.M / L c.n) := by gcongr        _ = σnorm (L c.n) * (L c.n / c.M) := by rw [div_div_eq_mul_div, mul_div_assoc]    calcc.M : ) = σnorm c.M * c.M := by field_simp [σnorm]      _ > σnorm (L c.n) * (L c.n / c.M) * c.M := by nlinarith      _ = σ (L c.n) := by field_simp [σnorm, c.M_pos.ne']  exact not_lt.mpr (cast_lt.mp hσM_gt).le (hHA c.M c.M_lt)