92 lines
2.4 KiB
Plaintext
92 lines
2.4 KiB
Plaintext
import Mathlib.Data.Real.Basic
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/-! # Apostol.Chapter_1_11 -/
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namespace Apostol.Chapter_1_11
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/-! ## Exercise 4
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Prove that the greatest-integer function has the properties indicated.
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-/
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/-- ### Exercise 4a
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`⌊x + n⌋ = ⌊x⌋ + n` for every integer `n`.
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-/
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theorem exercise_4a (x : ℝ) (n : ℤ) : ⌊x + n⌋ = ⌊x⌋ + n :=
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Int.floor_add_int x n
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/-- ### Exercise 4b.1
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`⌊-x⌋ = -⌊x⌋` if `x` is an integer.
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-/
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theorem exercise_4b_1 (x : ℤ) : ⌊-x⌋ = -⌊x⌋ := by
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simp only [Int.floor_int, id_eq]
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/-- ### Exercise 4b.2
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`⌊-x⌋ = -⌊x⌋ - 1` otherwise.
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-/
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theorem exercise_4b_2 (x : ℝ) (h : ∃ n : ℤ, x ∈ Set.Ioo ↑n (↑n + (1 : ℝ)))
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: ⌊-x⌋ = -⌊x⌋ - 1 := by
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rw [Int.floor_neg]
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suffices ⌈x⌉ = ⌊x⌋ + 1 by
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have := congrArg (HMul.hMul (-1)) this
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simp only [neg_mul, one_mul, neg_add_rev, add_comm] at this
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exact this
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have ⟨n, hn⟩ := h
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have hn' : x ∈ Set.Ico ↑n (↑n + (1 : ℝ)) :=
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Set.mem_of_subset_of_mem Set.Ioo_subset_Ico_self hn
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rw [Int.ceil_eq_iff, Int.floor_eq_on_Ico n x hn']
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simp only [Int.cast_add, Int.cast_one, add_sub_cancel]
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apply And.intro
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· exact (Set.mem_Ioo.mp hn).left
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· exact le_of_lt (Set.mem_Ico.mp hn').right
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/-- ### Exercise 4c
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`⌊x + y⌋ = ⌊x⌋ + ⌊y⌋` or `⌊x⌋ + ⌊y⌋ + 1`.
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-/
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theorem exercise_4c (x y : ℝ)
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: ⌊x + y⌋ = ⌊x⌋ + ⌊y⌋ ∨ ⌊x + y⌋ = ⌊x⌋ + ⌊y⌋ + 1 := by
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sorry
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/-- ### Exercise 4d
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`⌊2x⌋ = ⌊x⌋ + ⌊x + 1/2⌋`
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-/
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theorem exercise_4d (x : ℝ)
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: ⌊2 * x⌋ = ⌊x⌋ + ⌊x + 1/2⌋ := by
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sorry
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/-- ### Exercise 4e
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`⌊3x⌋ = ⌊x⌋ + ⌊x + 1/3⌋ + ⌊x + 2/3⌋`
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-/
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theorem exercise_4e (x : ℝ)
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: ⌊3 * x⌋ = ⌊x⌋ + ⌊x + 1/3⌋ + ⌊x + 2/3⌋ := by
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sorry
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/-- ### Exercise 5
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The formulas in Exercises 4(d) and 4(e) suggest a generalization for `⌊nx⌋`.
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State and prove such a generalization.
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-/
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theorem exercise_5 (n : ℕ) (x : ℝ)
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: ⌊n * x⌋ = 10 := by
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sorry
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/-- ### Exercise 7b
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If `a` and `b` are positive integers with no common factor, we have the formula
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`Σ_{n=1}^{b-1} ⌊na / b⌋ = ((a - 1)(b - 1)) / 2`. When `b = 1`, the sum on the
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left is understood to be `0`.
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Derive the result analytically as follows: By changing the index of summation,
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note that `Σ_{n=1}^{b-1} ⌊na / b⌋ = Σ_{n=1}^{b-1} ⌊a(b - n) / b⌋`. Now apply
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Exercises 4(a) and (b) to the bracket on the right.
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-/
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theorem exercise_7b : True := sorry
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end Apostol.Chapter_1_11
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