Continuing working on Apostol 1.11 exercises.

Bookshelf/Apostol/Chapter_1_07.tex

@@ -22,7 +22,7 @@ The properties of area in this set of exercises are to be deduced from the
 
 Prove that each of the following sets is measurable and has zero area:
 
-\subsection*{\proceeding{Exercise 1a}}%
+\subsection*{\unverified{Exercise 1a}}%
 \label{sub:exercise-1a}
 
 A set consisting of a single point.
@@ -38,7 +38,7 @@ A set consisting of a single point.
 
 \end{proof}
 
-\subsection*{\proceeding{Exercise 1b}}%
+\subsection*{\unverified{Exercise 1b}}%
 \label{sub:exercise-1b}
 
 A set consisting of a finite number of points in a plane.
@@ -97,7 +97,7 @@ A set consisting of a finite number of points in a plane.
 
 \end{proof}
 
-\subsection*{\proceeding{Exercise 1c}}%
+\subsection*{\unverified{Exercise 1c}}%
 \label{sub:exercise-1c}
 
 The union of a finite collection of line segments in a plane.
@@ -382,24 +382,19 @@ Prove that the formula is valid for right triangles and parallelograms.
   Likewise,
     \begin{equation}
       \label{sub:exercise-4b-eq2}
-      I_T = \frac{1}{2}(I_R - H_L + 2).
+      I_T = \frac{1}{2}(I_R - (H_L - 2)).
     \end{equation}
   The following shows the lattice point area formula is in agreement with
     the expected result:
     \begin{align*}
       I_T + \frac{1}{2}B_T - 1
-        & = \frac{1}{2}(I_R - H_L + 2) + \frac{1}{2}B_T - 1
+        & = \frac{1}{2}(I_R - (H_L - 2)) + \frac{1}{2}B_T - 1
           & \eqref{sub:exercise-4b-eq2} \\
-        & = \frac{1}{2}\left[ I_R - H_L + 2 + B_T - 2 \right] \\
         & = \frac{1}{2}\left[ I_R - H_L + B_T \right] \\
         & = \frac{1}{2}\left[ I_R - H_L + \frac{1}{2}B_R - 1 + H_L \right]
           & \eqref{sub:exercise-4b-eq1} \\
         & = \frac{1}{2}\left[ I_R + \frac{1}{2}B_R - 1 \right] \\
-        & = \frac{1}{2}\left[ (w - 1)(h - 1) + \frac{1}{2}(2(w + h)) - 1 \right]
+        & = \frac{1}{2}\left[ wh \right] & \text{\nameref{sub:exercise-4a}}.
-          & \text{\nameref{sub:exercise-4a}} \\
-        & = \frac{1}{2}\left[ (w - 1)(h - 1) + w + h - 1 \right] \\
-        & = \frac{1}{2}\left[ wh - w - h + 1 + w + h - 1 \right] \\
-        & = \frac{wh}{2}.
     \end{align*}
 
   We do not prove this formula is valid for parallelograms here.

Bookshelf/Apostol/Chapter_1_11.lean

@@ -1,12 +1,14 @@
-import Mathlib.Algebra.BigOperators.Basic
 import Mathlib.Data.Real.Basic
-import Mathlib.Data.Finset.Basic
 import Mathlib.Tactic.LibrarySearch
 
+import Common.Real.Floor
+
 /-! # Apostol.Chapter_1_11 -/
 
 namespace Apostol.Chapter_1_11
 
+open BigOperators
+
 /-! ## Exercise 4
 
 Prove that the greatest-integer function has the properties indicated.
@@ -76,49 +78,14 @@ theorem exercise_4c (x y : ℝ)
       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
+  : ⌊n * x⌋ = Finset.sum (Finset.range n) (fun i => ⌊x + i/n⌋) :=
-  let r := Int.fract x
+  Real.Floor.floor_mul_eq_sum_range_floor_add_index_div n x
-  have hx : x = ⌊x⌋ + r := Eq.symm (add_eq_of_eq_sub' rfl)
-  sorry
 
 /-- ### Exercise 4d
 
@@ -157,6 +124,9 @@ 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
+theorem exercise_7b (ha : a > 0) (hb : b > 0) (hp : Nat.coprime a b)
+  : ∑ n in (Finset.range b).filter (· > 0), ⌊n * ((a : ℕ) : ℝ) / b⌋ =
+      ((a - 1) * (b - 1)) / 2 := by
+  sorry
 
 end Apostol.Chapter_1_11

Bookshelf/Apostol/Chapter_1_11.tex

@@ -2,6 +2,8 @@
 
 \input{../../preamble}
 
+\externaldocument[C:1:07:]{Chapter_1_07}[Chapter_1_07.pdf]
+
 \newcommand{\lean}[1]{\leanref
   {./Chapter\_1\_11.html\#Apostol.Chapter\_1\_11.#1}
   {Apostol.Chapter\_1\_11.#1}}
@@ -15,7 +17,7 @@
 
 Prove that the greatest-integer function has the properties indicated:
 
-\subsection*{\proceeding{Exercise 4a}}%
+\subsection*{\verified{Exercise 4a}}%
 \label{sub:exercise-4a}
 
 $\floor{x + n} = \floor{x} + n$ for every integer $n$.
@@ -24,9 +26,19 @@ $\floor{x + n} = \floor{x} + n$ for every integer $n$.
 
   \lean{exercise\_4a}
 
+  \divider
+
+  Let $x$ be a real number and $n$ an integer.
+  Let $m = \floor{x + n}$.
+  By definition of the floor function, $m$ is the unique integer such that
+    $m \leq x + n < m + 1$.
+  Then $m - n \leq x < (m - n) + 1$.
+  That is, $m - n = \floor{x}$.
+  Rearranging terms we see that $m = \floor{x} + n$ as expected.
+
 \end{proof}
 
-\subsection*{\proceeding{Exercise 4b}}%
+\subsection*{\verified{Exercise 4b}}%
 \label{sub:exercise-4b}
 
 $\floor{-x} =
@@ -37,16 +49,45 @@ $\floor{-x} =
 
 \begin{proof}
 
-  \  % Force space prior to *Proof.*
+  \ \vspace{6pt}
 
-  \begin{enumerate}[(a)]
+  \lean{exercise\_4b\_1}
-    \item \lean{exercise\_4b\_1}
+
-    \item \lean{exercise\_4b\_2}
+  \lean{exercise\_4b\_2}
-  \end{enumerate}
+
+  \divider
+
+  There are two cases to consider:
+
+  \paragraph{Case 1}%
+
+    Suppose $x$ is an integer.
+    Then $x = \floor{x}$ and $-x = \floor{-x}$.
+    It immediately follows that $$\floor{-x} = -x = -\floor{x}.$$
+
+  \paragraph{Case 2}%
+
+    Suppose $x$ is not an integer.
+    Let $m = \floor{-x}$.
+    By definition of the floor function, $m$ is the unique integer such that
+      $m \leq -x < m + 1$.
+    Equivalently, $-m - 1 < x \leq -m$.
+    Since $x$ is not an integer, it follows $-m - 1 \leq x < -m$.
+    Then, by definition of the floor function, $\floor{x} = -m - 1$.
+    Solving for $m$ yields $$\floor{-x} = m = -\floor{x} - 1.$$
+
+  \paragraph{Conclusion}%
+
+    The above two cases are exhaustive. Thus
+      $$\floor{-x} =
+        \begin{cases}
+          -\floor{x} & \text{if } x \text{ is an integer}, \\
+          -\floor{x} - 1 & \text{otherwise}.
+        \end{cases}$$
 
 \end{proof}
 
-\subsection*{\proceeding{Exercise 4c}}%
+\subsection*{\verified{Exercise 4c}}%
 \label{sub:exercise-4c}
 
 $\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$.
@@ -55,6 +96,40 @@ $\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$.
 
   \lean{exercise\_4c}
 
+  \divider
+
+  Rewrite $x$ and $y$ as the sum of their floor and fractional components:
+    $x = \floor{x} + \{x\}$ and $y = \floor{y} + \{y\}$.
+  Now
+    \begin{align}
+      \floor{x + y}
+        & = \floor{\floor{x} + \{x\} + \floor{y} + \{y\}} \nonumber \\
+        & = \floor{\floor{x} + \floor{y} + \{x\} + \{y\}} \nonumber \\
+        & = \floor{x} + \floor{y} + \floor{\{x\} + \{y\}}
+        & \text{\nameref{sub:exercise-4a}} \label{sub:exercise-4c-eq1}
+    \end{align}
+  There are two cases to consider:
+
+  \paragraph{Case 1}%
+
+    Suppose $\{x\} + \{y\} < 1$.
+    Then $\floor{\{x\} + \{y\}} = 0$.
+    Substituting this value into \eqref{sub:exercise-4c-eq1} yields
+      $$\floor{x + y} = \floor{x} + \floor{y}.$$
+
+  \paragraph{Case 2}%
+
+    Suppose $\{x\} + \{y\} \geq 1$.
+    Because $\{x\}$ and $\{y\}$ are both less than $1$, $\{x\} + \{y\} < 2$.
+    Thus $\floor{\{x\} + \{y\}} = 1$.
+    Substituting this value into \eqref{sub:exercise-4c-eq1} yields
+      $$\floor{x + y} = \floor{x} + \floor{y} + 1.$$
+
+  \paragraph{Conclusion}%
+
+    Since the above two cases are exhaustive, it follows
+      $\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$.
+
 \end{proof}
 
 \subsection*{\proceeding{Exercise 4d}}%
@@ -66,6 +141,11 @@ $\floor{2x} = \floor{x} + \floor{x + \frac{1}{2}}.$
 
   \lean{exercise\_4d}
 
+  \divider
+
+  This is immediately proven by applying Hermite's Identity as shown in
+    \nameref{sec:exercise-5}.
+
 \end{proof}
 
 \subsection*{\proceeding{Exercise 4e}}%
@@ -77,6 +157,11 @@ $\floor{3x} = \floor{x} + \floor{x + \frac{1}{3}} + \floor{x + \frac{2}{3}}.$
 
   \lean{exercise\_4e}
 
+  \divider
+
+  This is immediately proven by applying Hermite's Identity as shown in
+    \nameref{sec:exercise-5}.
+
 \end{proof}
 
 \section*{\proceeding{Exercise 5}}%
@@ -116,7 +201,7 @@ State and prove such a generalization.
 
   \paragraph{Left-Hand Side}%
 
-    Consider the left-hande side of identity \eqref{sec:exercise-5-eq1}
+    Consider the left-hand side of identity \eqref{sec:exercise-5-eq1}.
     By \eqref{sec:exercise-5-eq2}, $nr \in \ico{j}{j + 1}$.
     Therefore $\floor{nr} = j$.
     Thus
@@ -182,13 +267,32 @@ Prove that the number of lattice points in $S$ is equal to the sum
 
 \begin{proof}
 
-  Define $S_i = \mathbb{Z} \cap \ioc{0}{f(i)}$ for all $i \in \mathbb{Z}$.
+  Let $i = a, \ldots, b$ and define $S_i = \mathbb{N} \cap \ioc{0}{f(i)}$.
-  By definition, the set of lattice points of $S$ is given by
+  By construction, the number of lattice points in $S$ is
-    $$L = \{ (i, j) : i = a, \ldots, b \land j \in S_i \}.$$
+    \begin{equation}
-  By construction, it follows $$\sum_{j \in S_i} 1 = \floor{f(i)}.$$
+      \label{sec:exercise-6-eq1}
-  Therefore $$\abs{L}
+      \sum_{n = a}^b \abs{S_n}.
-    = \sum_{i=a}^b \sum_{j \in S_i} 1
+    \end{equation}
-    = \sum_{i=1}^b \floor{f(i)}.$$
+  All that remains is to show $\abs{S_i} = \floor{f(i)}$.
+  There are two cases to consider:
+
+  \paragraph{Case 1}%
+
+    Suppose $f(i)$ is an integer.
+    Then the number of integers in $\ioc{0}{f(i)}$ is $f(i) = \floor{f(i)}$.
+
+  \paragraph{Case 2}%
+
+    Suppose $f(i)$ is not an integer.
+    Then the number of integers in $\ioc{0}{f(i)}$ is the same as that of
+      $\ioc{0}{\floor{f(i)}}$.
+    Once again, that number is $\floor{f(i)}$.
+
+  \paragraph{Conclusion}%
+
+    By cases 1 and 2, $\abs{S_i} = \floor{f(i)}$.
+    Substituting this identity into \eqref{sec:exercise-6-eq1} finishes the
+      proof.
 
 \end{proof}
 
@@ -199,6 +303,9 @@ If $a$ and $b$ are positive integers with no common factor, we have the formula
   $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}} = \frac{(a - 1)(b - 1)}{2}.$$
 When $b = 1$, the sum on the left is understood to be $0$.
 
+\note{When $b = 1$, the proofs of (a) and (b) are trivial. We continue under the
+  assumption $b > 1$.}
+
 \subsection*{\unverified{Exercise 7a}}%
 \label{sub:exercise-7a}
 
@@ -207,7 +314,78 @@ Derive this result by a geometric argument, counting lattice points in a right
 
 \begin{proof}
 
-  TODO
+  Let $f \colon [1, b - 1] \rightarrow \mathbb{R}$ be given by $f(x) = ax / b$.
+  Let $S$ denote the set of points $(x, y)$ satisfying $1 \leq x \leq b - 1$,
+    $0 < y \leq f(x)$.
+  By \nameref{sec:exercise-6}, the number of lattice points of $S$ is equal to
+    the sum
+    \begin{equation}
+      \label{sub:exercise-7a-eq1}
+      \sum_{n=1}^{b-1} \floor{f(n)} = \sum_{n=1}^{b-1} \floor{\frac{na}{b}}.
+    \end{equation}
+  Define $T$ to be the triangle of width $w = b$ and height $h = f(b) = a$
+    as $$T = \{ (x, y) : 0 < x < b, 0 < y \leq f(x) \}.$$
+  By construction, $T$ does not introduce any additional lattice points.
+  Thus $S$ and $T$ have the same number of lattice points.
+  Let $H_L$ denote the number of boundary points on $T$'s hypotenuse.
+  We prove that (i) $H_L = 2$ and (ii) that $T$ has $\frac{(a - 1)(b - 1)}{2}$
+    lattice points.
+
+  \paragraph{(i)}%
+  \label{par:exercise-7a-i}
+
+    Consider the line $L$ overlapping the hypotenuse of $T$.
+    By construction, $T$'s hypotenuse has endpoints $(0, 0)$ and $(b, a)$.
+    By hypothesis, $a$ and $b$ are positive, excluding the possibility of $L$
+      being vertical.
+    Define the slope of $L$ as $$m = \frac{a}{b}.$$
+    $H_L$ coincides with the number of indices $i = 0, \ldots, b$ such that
+      $(i, i * m)$ is a lattice point.
+    But $a$ and $b$ are coprime by hypothesis and $i \leq b$.
+    Thus $i * m$ is an integer if and only if $i = 0$ or $i = b$.
+    Thus $H_L = 2$.
+
+  \paragraph{(ii)}%
+
+    Next we count the number of lattice points in $T$.
+    Let $R$ be the overlapping retangle of width $w$ and height $h$, situated
+      with bottom-left corner at $(0, 0)$.
+    Let $I_R$ denote the number of interior lattice points of $R$.
+    Let $I_T$ and $B_T$ denote the interior and boundary lattice points of $T$
+      respectively.
+    By \nameref{C:1:07:sub:exercise-4b-eq2},
+      \begin{align}
+        I_T
+          & = \frac{1}{2}(I_R - (H_L - 2)) \nonumber \\
+          & = \frac{1}{2}(I_R - (2 - 2))
+            & \text{\nameref{par:exercise-7a-i}} \nonumber \\
+            & = \frac{1}{2}I_R. & \label{sub:exercise-7a-eq2}
+      \end{align}
+    Furthermore, since both the adjacent and opposite side of $T$ are not
+      included in $T$ and there exist no lattice points on $T$'s hypotenuse
+      besides the endpoints, it follows
+      \begin{equation}
+        \label{sub:exercise-7a-eq3}
+        B_T = 0.
+      \end{equation}
+    Thus the number of lattice points of $T$ equals
+      \begin{align}
+        I_T + B_T
+          & = I_T & \eqref{sub:exercise-7a-eq3} \nonumber \\
+          & = \frac{1}{2}I_R & \eqref{sub:exercise-7a-eq2} \nonumber \\
+          & = \frac{(b - 1)(a - 1)}{2}.
+            & \text{\nameref{C:1:07:sub:exercise-4a}}
+              \label{sub:exercise-7a-eq4}
+      \end{align}
+
+  \paragraph{Conclusion}%
+
+    By \eqref{sub:exercise-7a-eq1} the number of lattice points of $S$ is equal
+      to the sum $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}}.$$
+    But the number of lattice points of $S$ is the same as that of $T$.
+    By \eqref{sub:exercise-7a-eq4}, the number of lattice points in $T$ is equal
+      to $$\frac{(b - 1)(a - 1)}{2}.$$
+    Thus $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}} = \frac{(a - 1)(b - 1)}{2}.$$
 
 \end{proof}
 
@@ -223,6 +401,35 @@ Now apply Exercises 4(a) and (b) to the bracket on the right.
 
   \lean{exercise\_7b}
 
+  \divider
+
+  Let $n = 1, \ldots, b - 1$.
+  By hypothesis, $a$ and $b$ are coprime.
+  Furthermore, $n < b$ for all values of $n$.
+  Thus $an / b$ is not an integer.
+  By \nameref{sub:exercise-4b},
+    \begin{equation}
+      \label{sub:exercise-7b-eq1}
+      \floor{-\frac{an}{b}} = -\floor{\frac{an}{b}} - 1.
+    \end{equation}
+  Consider the following:
+    \begin{align*}
+      \sum_{n=1}^{b-1} \floor{\frac{na}{b}}
+        & = \sum_{n=1}^{b-1} \floor{\frac{a(b - n)}{b}} \\
+        & = \sum_{n=1}^{b-1} \floor{\frac{ab - an}{b}} \\
+        & = \sum_{n=1}^{b-1} \floor{-\frac{an}{b} + a} \\
+        & = \sum_{n=1}^{b-1} \floor{-\frac{an}{b}} + a.
+          & \text{\nameref{sub:exercise-4a}} \\
+        & = \sum_{n=1}^{b-1} -\floor{\frac{an}{b}} - 1 + a
+          & \eqref{sub:exercise-7b-eq1} \\
+        & = -\sum_{n=1}^{b-1} \floor{\frac{an}{b}} - \sum_{n=1}^{b-1} 1 +
+          \sum_{n=1}^{b-1} a \\
+        & = -\sum_{n=1}^{b-1} \floor{\frac{an}{b}} - (b - 1) + a(b - 1).
+    \end{align*}
+  Rearranging the above yields
+    $$2\sum_{n=1}^{b-1} \floor{\frac{an}{b}} = (a - 1)(b - 1).$$
+  Dividing both sides of the above identity concludes the proof.
+
 \end{proof}
 
 \section*{\unverified{Exercise 8}}%

Common/Real/Floor.lean

@@ -0,0 +1,149 @@
+import Mathlib.Algebra.BigOperators.Basic
+import Mathlib.Data.Real.Basic
+import Mathlib.Data.Finset.Basic
+import Mathlib.Tactic.LibrarySearch
+
+/-! # Common.Real.Floor
+
+A collection of useful definitions and theorems around the floor function.
+-/
+
+namespace Real.Floor
+
+/--
+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⟩
+
+/-! ## Hermite's Identity
+
+Definitions and theorems in support of proving Hermite's identity.
+-/
+
+namespace Hermite
+
+/--
+A partition of `[0, 1)` that looks as follows:
+
+```
+[0, 1/n), [1/n, 2/n), ..., [(n-1)/n, 1)
+```
+
+This is expected to be used as an indexing function of a union of sets, e.g.
+`⋃ i ∈ Finset.range n, partition n i`.
+-/
+def partition (n : ℕ) (i : ℕ) : Set ℝ := Set.Ico (↑i / n) ((↑i + 1) / n)
+
+/--
+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,
+      j < n ∧ r ∈ Set.Ico (((j : ℕ) : ℝ) / n) ((↑j + 1) / n) := by
+  sorry
+
+/--
+The indexed union of the family of sets of a `partition` is a subset of `[0, 1)`.
+-/
+theorem partition_subset_Ico_zero_one
+  : (⋃ i ∈ Finset.range n, partition n i) ⊆ Set.Ico 0 1 := by
+  simp only [
+    Finset.mem_range,
+    gt_iff_lt,
+    zero_lt_one,
+    not_true,
+    ge_iff_le,
+    Set.unionᵢ_subset_iff
+  ]
+  intro i hi x hx
+  have hn : (0 : ℝ) < n := calc (0 : ℝ)
+    _ ≤ i := Nat.cast_nonneg i
+    _ < n := Nat.cast_lt.mpr hi
+  apply And.intro
+  · have h_zero_le_i_div_n : (0 : ℝ) ≤ i / n := by
+      rw [← mul_le_mul_right hn, zero_mul, div_mul, div_self, div_one]
+      · exact Nat.cast_nonneg i
+      · exact ne_iff_lt_or_gt.mpr (Or.inr hn)
+    calc (0 : ℝ)
+      _ ≤ i / n := h_zero_le_i_div_n
+      _ ≤ x := hx.left
+  · have h_succ_div_n_le_one : (i + 1) / n ≤ (1 : ℝ) := by
+      rw [div_le_one_iff]
+      refine Or.inl ?_
+      exact ⟨hn, by norm_cast⟩
+    calc x
+      _ < (i + 1) / n := hx.right
+      _ ≤ 1 := h_succ_div_n_le_one
+
+/--
+`[0, 1)` is a subset of the indexed union of the family of sets of a `partition`.
+-/
+theorem Ico_zero_one_subset_partition
+  : Set.Ico 0 1 ⊆ (⋃ i ∈ Finset.range n, partition n i) := by
+  intro x hx
+  simp only [Finset.mem_range, Set.mem_unionᵢ, exists_prop]
+  unfold partition
+  exact fract_mem_partition x hx 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 :=
+  Set.Subset.antisymm_iff.mpr
+    ⟨partition_subset_Ico_zero_one, Ico_zero_one_subset_partition⟩
+
+end Hermite
+
+open BigOperators
+
+/-- ### Hermite's Identity
+
+The following decomposes the floor of a multiplication into a sum of floors.
+-/
+theorem floor_mul_eq_sum_range_floor_add_index_div (n : ℕ) (x : ℝ)
+  : ⌊n * x⌋ = ∑ i in Finset.range n, ⌊x + i / n⌋ := by
+  let r := Int.fract x
+
+  -- Here we see there exists some `j` such that `r ∈ [j / n, (j + 1) / n]`.
+  have hx : x = ⌊x⌋ + r := Eq.symm (add_eq_of_eq_sub' rfl)
+  have ⟨j, ⟨hj, hr⟩⟩ :=
+    Hermite.fract_mem_partition r (fract_mem_Ico_zero_one x) n
+
+  -- With the above definitions established, we now show the left- and
+  -- right-hand sides of the goal evaluate to the same number.
+
+  have hlhs : ⌊n * x⌋ = n * ⌊x⌋ + j := by
+    have hn : (0 : ℝ) < n := calc (0 : ℝ)
+      _ ≤ j := Nat.cast_nonneg j
+      _ < n := Nat.cast_lt.mpr hj
+    -- We prove that `nr ∈ [j, j + 1)`. It must then follow `⌊nr⌋ = j`.
+    have hnr : n * r ∈ Set.Ico ((j : ℕ) : ℝ) (j + 1) := by
+      apply And.intro
+      · have := hr.left
+        rw [← mul_le_mul_right hn, div_mul, div_self, div_one] at this
+        · rwa [mul_comm]
+        · exact ne_of_gt hn
+      · have := hr.right
+        rw [← mul_lt_mul_right hn, div_mul, div_self, div_one] at this
+        · rwa [mul_comm]
+        · exact ne_of_gt hn
+    have hnr_eq_j : ⌊n * r⌋ = j := by
+      have := Int.floor_eq_on_Ico' j (n * r) hnr
+      norm_cast at this
+    conv =>
+      lhs
+      rw [hx, mul_add, add_comm]
+      norm_cast
+      rw [Int.floor_add_int, hnr_eq_j, add_comm]
+
+  have hrhs : ∑ i in Finset.range n, ⌊x + i / n⌋ = n * ⌊x⌋ + j := by
+    sorry
+
+  -- Close out goal by showing left- and right-hand side equal a common value.
+  rw [hlhs, hrhs]
+
+end Real.Floor

DocGen4/Output/Index.lean

@@ -37,9 +37,9 @@ def index : BaseHtmlM Html := do templateExtends (baseHtml "Index") <|
         status:
         <ul>
           <li>
-            <span style="color:cyan">Cyan statements </span> indicate axioms and
+            <span style="color:darkgray">Dark gray statements </span> indicate
-            definitions. There must exist a corresponding <code>axiom</code> or
+            axioms and definitions. There must exist a corresponding
-            <code>def</code> in Lean.
+            <code>axiom</code> or <code>def</code> in Lean.
           </li>
           <li>
             <span style="color:teal">Teal statements </span> indicate those

preamble.tex

@@ -50,7 +50,7 @@
 % Indicates a statement corresponds to an axiom or definition. There must exist
 % a corresponding `axiom` or `def` in Lean.
 \DeclareRobustCommand{\defined}[1]{%
-  \texorpdfstring{\color{cyan}\faParagraph\ #1}{#1}}
+  \texorpdfstring{\color{darkgray}\faParagraph\ #1}{#1}}
 
 % Indicates a statement has a complete proof in both LaTeX *and* Lean.
 \DeclareRobustCommand{\verified}[1]{%