import Mathlib.Algebra.BigOperators.Basic import Mathlib.Data.Real.Basic import Mathlib.Data.Finset.Basic import Mathlib.Tactic.LibrarySearch /-! # Apostol.Chapter_1_11 -/ namespace Apostol.Chapter_1_11 /-! ## Exercise 4 Prove that the greatest-integer function has the properties indicated. -/ /-- ### Exercise 4a `⌊x + n⌋ = ⌊x⌋ + n` for every integer `n`. -/ theorem exercise_4a (x : ℝ) (n : ℤ) : ⌊x + n⌋ = ⌊x⌋ + n := Int.floor_add_int x n /-- ### Exercise 4b.1 `⌊-x⌋ = -⌊x⌋` if `x` is an integer. -/ theorem exercise_4b_1 (x : ℤ) : ⌊-x⌋ = -⌊x⌋ := by simp only [Int.floor_int, id_eq] /-- ### Exercise 4b.2 `⌊-x⌋ = -⌊x⌋ - 1` otherwise. -/ theorem exercise_4b_2 (x : ℝ) (h : ∃ n : ℤ, x ∈ Set.Ioo ↑n (↑n + (1 : ℝ))) : ⌊-x⌋ = -⌊x⌋ - 1 := by rw [Int.floor_neg] suffices ⌈x⌉ = ⌊x⌋ + 1 by have := congrArg (HMul.hMul (-1)) this simp only [neg_mul, one_mul, neg_add_rev, add_comm] at this exact this have ⟨n, hn⟩ := h have hn' : x ∈ Set.Ico ↑n (↑n + (1 : ℝ)) := Set.mem_of_subset_of_mem Set.Ioo_subset_Ico_self hn rw [Int.ceil_eq_iff, Int.floor_eq_on_Ico n x hn'] simp only [Int.cast_add, Int.cast_one, add_sub_cancel] apply And.intro · exact (Set.mem_Ioo.mp hn).left · exact le_of_lt (Set.mem_Ico.mp hn').right /-- ### Exercise 4c `⌊x + y⌋ = ⌊x⌋ + ⌊y⌋` or `⌊x⌋ + ⌊y⌋ + 1`. -/ theorem exercise_4c (x y : ℝ) : ⌊x + y⌋ = ⌊x⌋ + ⌊y⌋ ∨ ⌊x + y⌋ = ⌊x⌋ + ⌊y⌋ + 1 := by have hx : x = Int.floor x + Int.fract x := Eq.symm (add_eq_of_eq_sub' rfl) have hy : y = Int.floor y + Int.fract y := Eq.symm (add_eq_of_eq_sub' rfl) by_cases Int.fract x + Int.fract y < 1 · refine Or.inl ?_ rw [Int.floor_eq_iff] simp only [Int.cast_add] apply And.intro · exact add_le_add (Int.floor_le x) (Int.floor_le y) · conv => lhs; rw [hx, hy, add_add_add_comm]; arg 1; rw [add_comm] rwa [add_comm, ← add_assoc, ← sub_lt_iff_lt_add', ← sub_sub, add_sub_cancel, add_sub_cancel] · refine Or.inr ?_ rw [Int.floor_eq_iff] simp only [Int.cast_add, Int.cast_one] have h := le_of_not_lt h apply And.intro · conv => lhs; rw [← add_rotate] conv => rhs; rw [hx, hy, add_add_add_comm]; arg 1; rw [add_comm] rwa [← sub_le_iff_le_add', ← sub_sub, add_sub_cancel, add_sub_cancel] · conv => lhs; rw [hx, hy, add_add_add_comm]; arg 1; rw [add_comm] conv => lhs; rw [add_comm, ← add_assoc] conv => rhs; rw [add_assoc] rw [← sub_lt_iff_lt_add', ← sub_sub, add_sub_cancel, add_sub_cancel] exact add_lt_add (Int.fract_lt_one x) (Int.fract_lt_one y) namespace Hermite /-- Constructs a partition of `[0, 1)` that looks as follows: ``` [0, 1/n), [1/n, 2/n), ..., [(n-1)/n, 1) ``` -/ def partition (n : ℕ) (i : ℕ) : Set ℝ := Set.Ico (i / n) ((i + 1) / n) /-- The indexed union of the family of sets of a `partition` is equal to `[0, 1)`. -/ theorem partition_eq_Ico_zero_one : (⋃ i ∈ Finset.range n, partition n i) = Set.Ico 0 1 := by sorry /-- The fractional portion of any real number is always in `[0, 1)`. -/ theorem fract_mem_Ico_zero_one (x : ℝ) : Int.fract x ∈ Set.Ico 0 1 := ⟨Int.fract_nonneg x, Int.fract_lt_one x⟩ /-- The fractional portion of any real number always exists in some member of the indexed family of sets formed by any `partition`. -/ theorem fract_mem_partition (r : ℝ) (hr : r ∈ Set.Ico 0 1) : ∀ n : ℕ, ∃ j : ℕ, r ∈ Set.Ico ↑(j / n) ↑((j + 1) / n) := by sorry end Hermite /-- ### Exercise 5 The formulas in Exercises 4(d) and 4(e) suggest a generalization for `⌊nx⌋`. State and prove such a generalization. -/ theorem exercise_5 (n : ℕ) (x : ℝ) : ⌊n * x⌋ = Finset.sum (Finset.range n) (fun i => ⌊x + i/n⌋) := by let r := Int.fract x have hx : x = ⌊x⌋ + r := Eq.symm (add_eq_of_eq_sub' rfl) sorry /-- ### Exercise 4d `⌊2x⌋ = ⌊x⌋ + ⌊x + 1/2⌋` -/ theorem exercise_4d (x : ℝ) : ⌊2 * x⌋ = ⌊x⌋ + ⌊x + 1/2⌋ := by suffices ⌊x⌋ + ⌊x + 1/2⌋ = Finset.sum (Finset.range 2) (fun i => ⌊x + i/2⌋) by rw [this] exact exercise_5 2 x unfold Finset.sum simp rw [add_comm] /-- ### Exercise 4e `⌊3x⌋ = ⌊x⌋ + ⌊x + 1/3⌋ + ⌊x + 2/3⌋` -/ theorem exercise_4e (x : ℝ) : ⌊3 * x⌋ = ⌊x⌋ + ⌊x + 1/3⌋ + ⌊x + 2/3⌋ := by suffices ⌊x⌋ + ⌊x + 1/3⌋ + ⌊x + 2/3⌋ = Finset.sum (Finset.range 3) (fun i => ⌊x + i/3⌋) by rw [this] exact exercise_5 3 x unfold Finset.sum simp conv => rhs; rw [← add_rotate']; arg 2; rw [add_comm] rw [← add_assoc] /-- ### Exercise 7b If `a` and `b` are positive integers with no common factor, we have the formula `Σ_{n=1}^{b-1} ⌊na / b⌋ = ((a - 1)(b - 1)) / 2`. When `b = 1`, the sum on the left is understood to be `0`. Derive the result analytically as follows: By changing the index of summation, note that `Σ_{n=1}^{b-1} ⌊na / b⌋ = Σ_{n=1}^{b-1} ⌊a(b - n) / b⌋`. Now apply Exercises 4(a) and (b) to the bracket on the right. -/ theorem exercise_7b : True := sorry end Apostol.Chapter_1_11