auto-update(nvim): 2025-02-21 02:25:44

documents/by-course/math-8/course-notes/main.typ

@@ -127,9 +127,9 @@ middle_, these comprise the axioms of a system of propositional logic.
   + $or$ connects the smallest propositions surrounding it.
 ]
 
-= Notes on Logic and Proofs, 1.2
+= Lecture #datetime(day: 8, month: 1, year: 2025).display()
 
-_Prototypical example for this section:_ If $sin pi = 1$, then $6$ is prime.
+== More propositional forms
 
 #definition[
   For a *antedecent* $P$ and *consequent* $Q$, the *conditional sentence* $P =>
@@ -143,9 +143,6 @@ Q$ is the proposition "If $P$, then $Q$."
 
 A conditional may be true even when the antedecent and consequent are unrelated.
 
-= Lecture #datetime(day: 8, month: 1, year: 2025).display()
-
-== More propositional forms
 
 #definition[
   Let $P$ and $Q$ be propositions. The *biconditional sentence*
@@ -996,7 +993,7 @@ iota$ which is an inclusion function $iota' : A -> C$.
 
 #definition[
   Let $R$ be an equivalence relation on $A$. Recall $A\/R$ is the set of all
-  equivalence classes. The *canonical map* is
+  equivalence classes under $R$. The *canonical map* is
 
   $
     f : A -> A\/R, f(x) = overline(x), forall x in A

documents/by-course/math-8/pset-5/main.typ

@@ -10,7 +10,7 @@
 #show heading.where(level: 2): it => [#it.body.]
 #show heading.where(level: 3): it => [#it.body.]
 
-#set par(first-line-indent: 0pt, spacing: 1em)
+#set par(spacing: 1em)
 
 
 Problems:
@@ -19,8 +19,6 @@ Problems:
 
 2.5: \#1abc, 3, 10
 
-#outline()
-
 = 2.4
 
 == 4
@@ -228,43 +226,69 @@ If a set $A$ has $n$ elements, then $cal(P) (A)$ has $2^n$ elements.
 
   Ok now let's do it the annoying way.
 
-  #mitext(`
+  Consider the set with $0$ elements, $X = {}$. Then $cal(P)(X) = {emptyset}$
-Base Case ($n=0$). \\
+  so indeed it has cardinality $2^0 = 1$. Otherwise suppose that any set $X$
-If $A$ has $0$ elements, then $A = \varnothing$. Its power set is
+  with $n$ elements indeed has $|cal(P)(X)| = 2^n$.
-\[
-\mathcal{P}(A) = \{\varnothing\},
-\]
-which has exactly one element. Hence
-\[
-|\mathcal{P}(A)| = 1 = 2^0.
-\]
-So the statement holds for $n=0$.
 
-Now proceed by induction. Assume the statement is true for some integer $k \ge
+  Then, consider a set $Y$ of $n+1$ elements. For each subset of $cal(P)(Y)$,
-0$. That is, suppose that for any set $A$ with $k$ elements, we have
+  we choose an arbitrary element $k in Y$, and define $Z := Y backslash {k}$.
-\[
+  Note that $Z$ has $n$ elements so our inductive hypothesis says it has $2^n$
-|\mathcal{P}(A)| = 2^k.
+  elements. We seek to show that this implies $cal(P)(Y)$ has $2^(n+1)$
-\]
+  elements.
 
-Let $B$ be a set with $k+1$ elements. Choose one element $x \in B$, and let $A = B \setminus \{x\}$. Then $A$ has $k$ elements. By the inductive hypothesis,
+  Every subset of $Y$ in $cal(P)(Y)$ satisfies the following: either the subset
-\[
+  does not contain $k$ and so it's also a subset of $Z$, or its only difference
-|\mathcal{P}(A)| = 2^k.
+  from a subset of $Z$ is the addition of $k$.
-\]
-Now observe that any subset of $B$ is either:
-\begin{itemize}
-\item A subset of $A$ (does not contain $x$), or
-\item Of the form $S \cup \{x\}$ where $S \subseteq A$ (does contain $x$).
-\end{itemize}
-Thus every subset of $A$ gives rise to exactly one subset of $B$ that excludes $x$, and exactly one subset of $B$ that includes $x$. Therefore,
-\[
-|\mathcal{P}(B)| = |\mathcal{P}(A)| + |\mathcal{P}(A)|
-= 2^k + 2^k
-= 2 \cdot 2^k
-= 2^{k+1}.
-\]
 
-Since $B$ was any set with $k+1$ elements, the statement holds for $k+1$. By the principle of mathematical induction, the proof is complete.
+  Therefore, for each subset of $Z$, we can introduce $k$, and it's still a
-`)
+  subset of $Y$, and in fact completes $cal(P)(Y)$ from $cal(P)(Z)$, due to our
+  previous assertion. So for each element in $cal(P)(Z)$, there is one
+  additional element in $cal(P)(Y)$. In other words,
+  $
+    |cal(P)(Z)| dot 2 = |cal(P)(Y)|
+  $
+  Recall our inductive hypothesis states $|cal(P)(Z)| = 2^n$ because $Z$ has
+  $n$ elements, so $|cal(P)(X)| = 2 dot |cal(P)(Z)| = 2^(n+1)$, completing our
+  induction. Therefore for any arbitrary set $X$ with $n$ elements, its power
+  set $cal(P)(X)$ has $2^n$ elements.
+
+  //   #mitext(`
+  // Base Case ($n=0$). \\
+  // If $A$ has $0$ elements, then $A = \varnothing$. Its power set is
+  // \[
+  // \mathcal{P}(A) = \{\varnothing\},
+  // \]
+  // which has exactly one element. Hence
+  // \[
+  // |\mathcal{P}(A)| = 1 = 2^0.
+  // \]
+  // So the statement holds for $n=0$.
+  //
+  // Now proceed by induction. Assume the statement is true for some integer $k \ge
+  // 0$. That is, suppose that for any set $A$ with $k$ elements, we have
+  // \[
+  // |\mathcal{P}(A)| = 2^k.
+  // \]
+  //
+  // Let $B$ be a set with $k+1$ elements. Choose one element $x \in B$, and let $A = B \setminus \{x\}$. Then $A$ has $k$ elements. By the inductive hypothesis,
+  // \[
+  // |\mathcal{P}(A)| = 2^k.
+  // \]
+  // Now observe that any subset of $B$ is either:
+  // \begin{itemize}
+  // \item A subset of $A$ (does not contain $x$), or
+  // \item Of the form $S \cup \{x\}$ where $S \subseteq A$ (does contain $x$).
+  // \end{itemize}
+  // Thus every subset of $A$ gives rise to exactly one subset of $B$ that excludes $x$, and exactly one subset of $B$ that includes $x$. Therefore,
+  // \[
+  // |\mathcal{P}(B)| = |\mathcal{P}(A)| + |\mathcal{P}(A)|
+  // = 2^k + 2^k
+  // = 2 \cdot 2^k
+  // = 2^{k+1}.
+  // \]
+  //
+  // Since $B$ was any set with $k+1$ elements, the statement holds for $k+1$. By the principle of mathematical induction, the proof is complete.
+  // `)
 
 ]
 
@@ -272,224 +296,261 @@ Since $B$ was any set with $k+1$ elements, the statement holds for $k+1$. By the
 
 === c
 
-#mitext(`
+$(n+1)! > 2^(n+3)$ for all $n >= 5$.
-Let \( n = 5 \). Then:
-\[
-(5+1)! = 6! = 720 \quad\text{and}\quad 2^{5+3} = 2^8 = 256.
-\]
-Since \( 720 > 256 \), the inequality holds for \( n = 5 \).
 
-Assume that for some natural number \( n \ge 5 \) the inequality holds for all integers \( m \) with \( 5 \le m \le n \). In particular, assume
+Consider $n = 5$. Then $6! > 2^(8)$. Now proceed by induction. Suppose that
-\[
+$(n+1)! > 2^(n + 3)$.
-(n+1)! > 2^{n+3}.
-\]
-We must show that
-\[
-(n+2)! > 2^{(n+1)+3} = 2^{n+4}.
-\]
 
-Starting with the left-hand side, we have:
+Then $(n + 1 + 1)! > 2^(n+3 + 1)$ so $(n + 2)(n+1)! > 2 dot 2^(n + 3)$. We note
-\[
+that $n+2$ is always at least $2$, so this is true by our inductive hypothesis.
-(n+2)! = (n+2)(n+1)!.
-\]
-Using the inductive hypothesis,
-\[
-(n+2)! > (n+2) \cdot 2^{n+3}.
-\]
-Since \( n \ge 5 \), it follows that \( n+2 \ge 7 \). Therefore,
-\[
-(n+2) \cdot 2^{n+3} \ge 7 \cdot 2^{n+3}.
-\]
-But clearly,
-\[
-7 \cdot 2^{n+3} > 2 \cdot 2^{n+3} = 2^{n+4}.
-\]
-Thus,
-\[
-(n+2)! > 2^{n+4},
-\]
-which completes the inductive step.
 
-By the generalized principle of mathematical induction, the inequality
+If we consider $n = 1$, then $2! < 2^(4)$ so the PMI does not hold for all $NN$.
-\[
-(n+1)! > 2^{n+3}
-\]
-holds for all natural numbers \( n \ge 5 \).
 
-The inequality is stated to hold for all \( n \ge 5 \). However, for \( n < 5 \) the inequality fails. For instance, when \( n = 4 \):
+// #mitext(`
-\[
+// Let \( n = 5 \). Then:
-(4+1)! = 5! = 120 \quad\text{and}\quad 2^{4+3} = 2^7 = 128.
+// \[
-\]
+// (5+1)! = 6! = 720 \quad\text{and}\quad 2^{5+3} = 2^8 = 256.
-Since \( 120 \) is not greater than \( 128 \), the inequality is false for \( n = 4 \).
+// \]
-`)
+// Since \( 720 > 256 \), the inequality holds for \( n = 5 \).
+//
+// Assume that for some natural number \( n \ge 5 \) the inequality holds for all integers \( m \) with \( 5 \le m \le n \). In particular, assume
+// \[
+// (n+1)! > 2^{n+3}.
+// \]
+// We must show that
+// \[
+// (n+2)! > 2^{(n+1)+3} = 2^{n+4}.
+// \]
+//
+// Starting with the left-hand side, we have:
+// \[
+// (n+2)! = (n+2)(n+1)!.
+// \]
+// Using the inductive hypothesis,
+// \[
+// (n+2)! > (n+2) \cdot 2^{n+3}.
+// \]
+// Since \( n \ge 5 \), it follows that \( n+2 \ge 7 \). Therefore,
+// \[
+// (n+2) \cdot 2^{n+3} \ge 7 \cdot 2^{n+3}.
+// \]
+// But clearly,
+// \[
+// 7 \cdot 2^{n+3} > 2 \cdot 2^{n+3} = 2^{n+4}.
+// \]
+// Thus,
+// \[
+// (n+2)! > 2^{n+4},
+// \]
+// which completes the inductive step.
+//
+// By the generalized principle of mathematical induction, the inequality
+// \[
+// (n+1)! > 2^{n+3}
+// \]
+// holds for all natural numbers \( n \ge 5 \).
+//
+// The inequality is stated to hold for all \( n \ge 5 \). However, for \( n < 5 \) the inequality fails. For instance, when \( n = 4 \):
+// \[
+// (4+1)! = 5! = 120 \quad\text{and}\quad 2^{4+3} = 2^7 = 128.
+// \]
+// Since \( 120 \) is not greater than \( 128 \), the inequality is false for \( n = 4 \).
+// `)
 
 === e
 
-#mitext(`
+$(n! > 3n)$ for all $n >= 4$.
-Let \( n = 4 \). Then:
-\[
-4! = 24 \quad\text{and}\quad 3 \times 4 = 12.
-\]
-Since \( 24 > 12 \), the inequality holds for \( n = 4 \).
 
-Assume that for some natural number \( n \ge 4 \) the inequality holds for all integers \( m \) with \( 4 \le m \le n \). In particular, assume that
+Consider $n = 4$. Then $4! > 3(4)$. Suppose $(n! > 3n)$. Then
-\[
+$
-n! > 3n.
+  (n+1)! > 3(n+1) \
-\]
+  (n+1)n! > 3n + 3
-We must show that
+$
-\[
+Assume $n >= 4$. Then the above statement holds true for any $n$. So we
-(n+1)! > 3(n+1).
+conclude the statement is always true for $n >= 4$.
-\]
 
-Starting with the left-hand side:
+// #mitext(`
-\[
+// Let \( n = 4 \). Then:
-(n+1)! = (n+1) \cdot n!.
+// \[
-\]
+// 4! = 24 \quad\text{and}\quad 3 \times 4 = 12.
-By the inductive hypothesis, we have:
+// \]
-\[
+// Since \( 24 > 12 \), the inequality holds for \( n = 4 \).
-(n+1)! > (n+1) \cdot 3n.
+//
-\]
+// Assume that for some natural number \( n \ge 4 \) the inequality holds for all integers \( m \) with \( 4 \le m \le n \). In particular, assume that
-Since \( n \ge 4 \) (so \( n \ge 2 \)), it follows that:
+// \[
-\[
+// n! > 3n.
-(n+1) \cdot 3n \ge 3(n+1).
+// \]
-\]
+// We must show that
-To see this, note that for \( n \ge 2 \) we have:
+// \[
-\[
+// (n+1)! > 3(n+1).
-3n(n+1) = 3(n+1)n > 3(n+1),
+// \]
-\]
+//
-because \( n > 1 \). Therefore,
+// Starting with the left-hand side:
-\[
+// \[
-(n+1)! > 3(n+1),
+// (n+1)! = (n+1) \cdot n!.
-\]
+// \]
-which completes the inductive step.
+// By the inductive hypothesis, we have:
-
+// \[
-By the generalized principle of mathematical induction, the inequality
+// (n+1)! > (n+1) \cdot 3n.
-\[
+// \]
-n! > 3n
+// Since \( n \ge 4 \) (so \( n \ge 2 \)), it follows that:
-\]
+// \[
-holds for all natural numbers \( n \ge 4 \).
+// (n+1) \cdot 3n \ge 3(n+1).
-
+// \]
-The claim is that \( n! > 3n \) for all \( n \ge 4 \). However, for some smaller natural numbers the inequality is false. For instance, when \( n = 3 \):
+// To see this, note that for \( n \ge 2 \) we have:
-\[
+// \[
-3! = 6 \quad\text{and}\quad 3 \times 3 = 9.
+// 3n(n+1) = 3(n+1)n > 3(n+1),
-\]
+// \]
-`)
+// because \( n > 1 \). Therefore,
+// \[
+// (n+1)! > 3(n+1),
+// \]
+// which completes the inductive step.
+//
+// By the generalized principle of mathematical induction, the inequality
+// \[
+// n! > 3n
+// \]
+// holds for all natural numbers \( n \ge 4 \).
+//
+// The claim is that \( n! > 3n \) for all \( n \ge 4 \). However, for some smaller natural numbers the inequality is false. For instance, when \( n = 3 \):
+// \[
+// 3! = 6 \quad\text{and}\quad 3 \times 3 = 9.
+// \]
+// `)
 
 == 7
 
 === a
 
-#mitext(`
+$
-Let \(\{A_i : i \in \mathbb{N}\}\) be an indexed family of sets. Then for every natural number \( n\ge1 \),
+  (sect.big^n_(i=1) A_i)^c = union.big^n_(i=1) A_i^c
-\[
+$
-\left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c
-\]
-and
-\[
-\left(\bigcap_{i=1}^{n} A_i\right)^c = \bigcup_{i=1}^{n} A_i^c.
-\]
 
-For \( n=1 \), we have
+First consider $sect.big^n_(i=1) A_i$. Let $X := sect.big^n_(i=1) A_i$.
-\[
+Then $x in X$ if and only if $x in A_i$ for every $A_i$.
-\left(\bigcup_{i=1}^{1} A_i\right)^c = A_1^c \quad \text{and} \quad \bigcap_{i=1}^{1} A_i^c = A_1^c.
-\]
-Thus, the identity holds for \( n=1 \).
 
-Assume that for some \( n \ge 1 \) the statement holds; that is, assume
+Now consider $Y := (sect.big^n_(i=1) A_i)^c$. Then $y in Y$ if and only if $y
-\[
+in.not X$, that is, if and only there exists some $A_i$ such that $y in.not
-\left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c.
+A_i$. Now let $Z := union.big^n_(i=1) A_i^c$. Every element $z in Z$ is in some
-\]
+$A^c _i$, that is, $z in Z$ if and only if there exists some $A_i$ such that $z
-We need to show that
+in.not A_i$. Now note this is actually precisely the definition of $Y$. Hence
-\[
+$Z = Y$.
-\left(\bigcup_{i=1}^{n+1} A_i\right)^c = \bigcap_{i=1}^{n+1} A_i^c.
-\]
 
-Notice that
+// #mitext(`
-\[
+// Let \(\{A_i : i \in \mathbb{N}\}\) be an indexed family of sets. Then for every natural number \( n\ge1 \),
-\bigcup_{i=1}^{n+1} A_i = \left(\bigcup_{i=1}^{n} A_i\right) \cup A_{n+1}.
+// \[
-\]
+// \left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c
-Taking the complement of both sides, and using De Morgan's Law for two sets, we obtain:
+// \]
-\[
+// and
-\left(\bigcup_{i=1}^{n+1} A_i\right)^c = \left(\left(\bigcup_{i=1}^{n} A_i\right) \cup A_{n+1}\right)^c = \left(\bigcup_{i=1}^{n} A_i\right)^c \cap A_{n+1}^c.
+// \[
-\]
+// \left(\bigcap_{i=1}^{n} A_i\right)^c = \bigcup_{i=1}^{n} A_i^c.
-By the induction hypothesis,
+// \]
-\[
+//
-\left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c.
+// For \( n=1 \), we have
-\]
+// \[
-Thus, we have:
+// \left(\bigcup_{i=1}^{1} A_i\right)^c = A_1^c \quad \text{and} \quad \bigcap_{i=1}^{1} A_i^c = A_1^c.
-\[
+// \]
-\left(\bigcup_{i=1}^{n+1} A_i\right)^c = \left(\bigcap_{i=1}^{n} A_i^c\right) \cap A_{n+1}^c = \bigcap_{i=1}^{n+1} A_i^c.
+// Thus, the identity holds for \( n=1 \).
-\]
+//
-
+// Assume that for some \( n \ge 1 \) the statement holds; that is, assume
-This completes the inductive step.
+// \[
-
+// \left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c.
-By the PMI, we conclude that for every natural number \( n\ge1 \),
+// \]
-\[
+// We need to show that
-\left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c.
+// \[
-\]
+// \left(\bigcup_{i=1}^{n+1} A_i\right)^c = \bigcap_{i=1}^{n+1} A_i^c.
-
+// \]
-Thus, De Morgan's Laws hold for any indexed family \(\{A_i : i \in \mathbb{N}\}\).
+//
-`)
+// Notice that
+// \[
+// \bigcup_{i=1}^{n+1} A_i = \left(\bigcup_{i=1}^{n} A_i\right) \cup A_{n+1}.
+// \]
+// Taking the complement of both sides, and using De Morgan's Law for two sets, we obtain:
+// \[
+// \left(\bigcup_{i=1}^{n+1} A_i\right)^c = \left(\left(\bigcup_{i=1}^{n} A_i\right) \cup A_{n+1}\right)^c = \left(\bigcup_{i=1}^{n} A_i\right)^c \cap A_{n+1}^c.
+// \]
+// By the induction hypothesis,
+// \[
+// \left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c.
+// \]
+// Thus, we have:
+// \[
+// \left(\bigcup_{i=1}^{n+1} A_i\right)^c = \left(\bigcap_{i=1}^{n} A_i^c\right) \cap A_{n+1}^c = \bigcap_{i=1}^{n+1} A_i^c.
+// \]
+//
+// This completes the inductive step.
+//
+// By the PMI, we conclude that for every natural number \( n\ge1 \),
+// \[
+// \left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c.
+// \]
+//
+// Thus, De Morgan's Laws hold for any indexed family \(\{A_i : i \in \mathbb{N}\}\).
+// `)
 
 == 9
 
-#mitext(`
+Suppose there was one point. Then there would be $(1^2 - 1)/2 = 0$ line
-Given \(n\) points \(P_1, P_2, \dots, P_n\) in a plane with no three collinear, the number of line segments joining every pair of points is
+segments which is correct. Now we proceed by induction. Suppose that for
-\[
+$n$ points, the number of line segments joining all pairs of points are $(n^2 -
-\frac{n^2 - n}{2}.
+n)/2$. Now introduce an additional point such that there are $n+1$ points. For
-\]
+each of the previous $n$ points, we draw a line connecting it to the newly
+added point. So we introduce $n$ additional lines. Therefore we have
+$
+  (n^2 - n) / 2 + n
+$
+lines, which we can rewrite
+$
+  (n^2 + n) / 2 = ((n+1)^2 - (n+1)) / 2
+$
+So we conclude that our hypothesis holds for $n+1$ points and therefore holds
+for all $n in NN$ points.
 
-For \(n=2\), there is exactly 1 line segment joining the two points. The formula gives
+// #mitext(`
-\[
+// Given \(n\) points \(P_1, P_2, \dots, P_n\) in a plane with no three collinear, the number of line segments joining every pair of points is
-\frac{2^2 - 2}{2} = \frac{4-2}{2} = 1.
+// \[
-\]
+// \frac{n^2 - n}{2}.
-So the statement holds for \(n=2\).
+// \]
-
+//
-Assume that for some \(n \ge 2\), the number of line segments joining \(n\) points is
+// For \(n=2\), there is exactly 1 line segment joining the two points. The formula gives
-\[
+// \[
-\frac{n^2 - n}{2}.
+// \frac{2^2 - 2}{2} = \frac{4-2}{2} = 1.
-\]
+// \]
-Now consider \(n+1\) points. When we add a new point \(P_{n+1}\), this new point can be connected to each of the \(n\) existing points, thereby adding \(n\) new segments. Hence, the total number of segments becomes
+// So the statement holds for \(n=2\).
-\[
+//
-\frac{n^2 - n}{2} + n.
+// Assume that for some \(n \ge 2\), the number of line segments joining \(n\) points is
-\]
+// \[
-Simplify the expression:
+// \frac{n^2 - n}{2}.
-\[
+// \]
-\frac{n^2 - n}{2} + n = \frac{n^2 - n + 2n}{2} = \frac{n^2 + n}{2} = \frac{n(n+1)}{2}.
+// Now consider \(n+1\) points. When we add a new point \(P_{n+1}\), this new point can be connected to each of the \(n\) existing points, thereby adding \(n\) new segments. Hence, the total number of segments becomes
-\]
+// \[
-Therefore, by the PMI, the number of line segments joining all pairs of \(n\) points is
+// \frac{n^2 - n}{2} + n.
-\[
+// \]
-\frac{n^2 - n}{2}
+// Simplify the expression:
-\]
+// \[
-for all \(n \ge 2\).
+// \frac{n^2 - n}{2} + n = \frac{n^2 - n + 2n}{2} = \frac{n^2 + n}{2} = \frac{n(n+1)}{2}.
-`)
+// \]
+// Therefore, by the PMI, the number of line segments joining all pairs of \(n\) points is
+// \[
+// \frac{n^2 - n}{2}
+// \]
+// for all \(n \ge 2\).
+// `)
 
 == 10
 
-#mitext(`
+First, note that with 1 disk, it takes $2^1 - 1 = 1$ moves to solve the puzzle.
-The Tower of Hanoi problem with \( n \) disks can be solved in exactly \( 2^n - 1 \) moves.
+Now suppose that for $n$ disks, it takes $2^n - 1$ moves to solve. We proceed
-
+by induction on $n+1$ disks. First, ignore the disk at the very bottom of the
-For \( n = 1 \):
+stack, such that our situation is equivalent to when there are $n$ disks
-There is only one disk, and it can be moved directly from the source peg to the destination peg in one move.
+(because any disk can be stacked on top of the largest disk, our set of
-Since \( 2^1 - 1 = 1 \), the statement holds for \( n = 1 \).
+possible moves is unchanged from $n$ disks). Then we can move each of these
-
+disks to another peg in $2^n - 1$ moves. Now move the largest disk, which now
-Assume that for some \( k \ge 1 \) the Tower of Hanoi with \( k \) disks can be solved in \( 2^k - 1 \) moves (this is the inductive hypothesis). We now show that a Tower of Hanoi with \( k+1 \) disks can be solved in \( 2^{k+1} - 1 \) moves.
+no more disks above it, to the other free peg. Now we again can ignore the
-
+largest disk and move our $n$ disks on top of the largest disk in $2^n - 1$
-1. Move the top \( k \) disks from the source peg to the auxiliary peg.
+moves. So it took us a total of $2 dot (2^n - 1) + 1$ moves to move the entire
-   By the inductive hypothesis, this takes \( 2^k - 1 \) moves.
+stack to another peg, or $2^(n+1) - 1$ moves.
-2. Move the largest disk (the \((k+1)^\text{th}\) disk) from the source peg to the destination peg.
-   This requires 1 move.
-3. Move the \( k \) disks from the auxiliary peg to the destination peg.
-   Again by the inductive hypothesis, this requires \( 2^k - 1 \) moves.
-
-\[
-\text{Total Moves} = (2^k - 1) + 1 + (2^k - 1) = 2 \cdot 2^k - 1 = 2^{k+1} - 1.
-\]
-
-This completes the inductive step.
-
-By the PMI, the Tower of Hanoi with \( n \) disks can be solved in \( 2^n - 1 \) moves.
-`)
 
 == 12
 
@@ -522,48 +583,20 @@ is no reason why this should hold true for all $NN$.
 
 === a
 
-#mitext(`
+Every natural number greater than or equal to 11 can be written in the form $2s
-Every natural number \( n \ge 11 \) can be written in the form
++ 5t$ for some naturals $s$ and $t$.
-\[
-n = 2s + 5t,
-\]
-for some nonnegative integers \( s \) and \( t \).
 
-We verify the statement for the initial numbers:
+First, we see that $11 = 2(3) + 5(1)$, $12 = 2(2) + 5(2)$, $13 = 2(4) + 5(1)$,
-- \( n = 11 \): \( 11 = 2\cdot 3 + 5\cdot 1 \)
+$14 = 2(2) + 5(2)$. Now suppose that we can write all naturals up to $11 <= k
-- \( n = 12 \): \( 12 = 2\cdot 1 + 5\cdot 2 \)
+<= n$ as $2s + 5t$ for naturals $s,t$. Note that we have shown naturals from 11
-- \( n = 13 \): \( 13 = 2\cdot 4 + 5\cdot 1 \)
+to 15 can be written in the $2s + 5t$ form. Then we note $n >= 15$ iff. $n - 4 >= 11$,
-- \( n = 14 \): \( 14 = 2\cdot 2 + 5\cdot 2 \)
+and $n - 4$ is covered
-- \( n = 15 \): \( 15 = 2\cdot 5 + 5\cdot 1 \)
+by our inductive hypothesis, so we can write
+$
+  n + 1 = (n - 4) + 4 + 1 = 2s + 5t + 5 = 2s + 5(t + 1)
+$
+So indeed $n+1$ can be written as $2s + 5t$ for some naturals $s, t$.
 
-Thus, the claim holds for all \( 11 \le n \le 15 \).
-
-Assume as the induction hypothesis that for every integer \( m \) with \( 11 \le m < n \) (where \( n \ge 16 \)) there exist nonnegative integers \( s \) and \( t \) such that
-\[
-m = 2s + 5t.
-\]
-
-Since \( n \ge 16 \), notice that:
-\[
-n - 2 \ge 14.
-\]
-Because \( n-2 \) is at least 11 (indeed, \( n-2 \ge 14 \)), the induction hypothesis applies. Therefore, there exist nonnegative integers \( s \) and \( t \) such that:
-\[
-n - 2 = 2s + 5t.
-\]
-
-Then,
-\[
-n = (n - 2) + 2 = 2s + 5t + 2 = 2(s + 1) + 5t.
-\]
-If we set \( s' = s + 1 \) (which is clearly a nonnegative integer), we obtain:
-\[
-n = 2s' + 5t.
-\]
-Thus, \( n \) can be written in the desired form.
-
-By the PCI, every natural number \( n \ge 11 \) can be written as \( 2s + 5t \) for some nonnegative integers \( s \) and \( t \).
-`)
 
 === b
 
@@ -573,55 +606,32 @@ Every natural number \( n > 22 \) (i.e. every \( n \ge 23 \)) can be written in
 n = 3s + 4t,
 \]
 with integers \( s \ge 3 \) and \( t \ge 2 \).
-
-We prove the statement by complete (strong) induction.
-
-We explicitly verify the claim for a few numbers:
-- For \( n = 23 \):
-  \( 23 = 3 \cdot 5 + 4 \cdot 2 \) (here, \( s = 5 \ge 3 \) and \( t = 2 \ge 2 \)).
-- For \( n = 24 \):
-  \( 24 = 3 \cdot 4 + 4 \cdot 3 \) (here, \( s = 4 \ge 3 \) and \( t = 3 \ge 2 \)).
-- For \( n = 25 \):
-  \( 25 = 3 \cdot 3 + 4 \cdot 4 \) (here, \( s = 3 \ge 3 \) and \( t = 4 \ge 2 \)).
-
-Thus, the statement holds for \( n = 23, 24, 25 \).
-
-Assume that for every integer \( m \) with \( 23 \le m \le n \) (where \( n \ge 25 \)), there exist integers \( s \ge 3 \) and \( t \ge 2 \) such that
-\[
-m = 3s + 4t.
-\]
-
-The number \( n+1 \) can also be written in the form
-\[
-n+1 = 3s' + 4t',
-\]
-with \( s' \ge 3 \) and \( t' \ge 2 \).
-
-Since \( n \ge 25 \), observe that
-\[
-n+1 - 3 = n-2 \ge 23.
-\]
-By the inductive hypothesis, there exist integers \( s \ge 3 \) and \( t \ge 2 \) such that
-\[
-n-2 = 3s + 4t.
-\]
-Then
-\[
-n+1 = (n-2) + 3 = 3s + 4t + 3 = 3(s+1) + 4t.
-\]
-Define \( s' = s+1 \) (so that \( s' \ge 3+1 = 4 \ge 3 \)) and let \( t' = t \) (which satisfies \( t' \ge 2 \)). This shows that
-\[
-n+1 = 3s' + 4t',
-\]
-with the required conditions.
-
-By the PCI, every natural number \( n > 22 \) (i.e. \( n \ge 23 \)) can be written in the form
-\[
-n = 3s + 4t,
-\]
-where \( s \ge 3 \) and \( t \ge 2 \) are integers.
 `)
 
+#proof[
+  First, let's consider a few base cases.
+  $
+    23 &= 3(5) + 4(2) \
+    24 &= 3(4) + 4(3) \
+    25 &= 3(3) + 4(4) \
+  $
+  So for 23, 24, and 25, we can write them in the form $3s + 4t$ such that $s
+  >= 3$ and $t >= 2$. Now we proceed by strong
+  induction. Suppose that any natural $23 <= k <= n$ can be written as $k = 3s
+  + 4t$ for $t >= 2$ and $s >= 3$. Note that when $n >= 26$, $n - 3 >= 23$
+  (otherwise, it's already covered in our base case and we are done). Then we
+  can assume $n-3$ is always covered by the inductive hypothesis and can be
+  written in the desired form $3s + 4t$. Now we seek to show that this implies
+  $n+1$ can also be written in this form.
+  $
+    (n+1) = (n-3) + 4 = 3s + 4t + 4 = 3s + 4(t+1)
+  $
+  And $s >= 3$, $t + 1 >= 2$. So indeed any natural number greater than 22 can
+  be written as $3s + 4t$ for some naturals $t >= 2$ and $s >= 3$.
+]
+
+
+
 === c
 
 #mitext(`
@@ -630,66 +640,32 @@ Every natural number \( n > 33 \) (i.e. every \( n \ge 34 \)) can be written in
 n = 4s + 5t,
 \]
 where \( s \) and \( t \) are integers with \( s \ge 3 \) and \( t \ge 2 \).
-
-We prove the statement by complete induction.
-
-We verify the claim for the first four numbers:
-
-- \( n = 34 \):
-  \( 34 = 4\cdot 6 + 5\cdot 2 \)
-  (Here, \( s = 6 \ge 3 \) and \( t = 2 \ge 2 \).)
-
-- \( n = 35 \):
-  \( 35 = 4\cdot 5 + 5\cdot 3 \)
-  (Here, \( s = 5 \ge 3 \) and \( t = 3 \ge 2 \).)
-
-- \( n = 36 \):
-  \( 36 = 4\cdot 4 + 5\cdot 4 \)
-  (Here, \( s = 4 \ge 3 \) and \( t = 4 \ge 2 \).)
-
-- \( n = 37 \):
-  \( 37 = 4\cdot 3 + 5\cdot 5 \)
-  (Here, \( s = 3 \ge 3 \) and \( t = 5 \ge 2 \).)
-
-Thus, the statement holds for \( n = 34, 35, 36, \) and \( 37 \).
-
-Inductive Hypothesis:
-Assume that for every integer \( m \) with \( 34 \le m \le n \) (where \( n \ge 37 \)) there exist integers \( s \ge 3 \) and \( t \ge 2 \) such that
-\[
-m = 4s + 5t.
-\]
-
-The number \( n+1 \) can also be written in the form
-\[
-n+1 = 4s' + 5t',
-\]
-with \( s' \ge 3 \) and \( t' \ge 2 \).
-
-Since \( n \ge 37 \), we have:
-\[
-n+1 - 4 = n - 3 \ge 37 - 3 = 34.
-\]
-Thus, \( n-3 \) is at least 34, and by the inductive hypothesis, there exist integers \( s \ge 3 \) and \( t \ge 2 \) such that
-\[
-n - 3 = 4s + 5t.
-\]
-Then,
-\[
-n+1 = (n-3) + 4 = 4s + 5t + 4 = 4(s+1) + 5t.
-\]
-Define \( s' = s+1 \) and \( t' = t \). Since \( s \ge 3 \), it follows that \( s' \ge 4 \ge 3 \), and \( t' = t \ge 2 \). Thus, we obtain the required representation:
-\[
-n+1 = 4s' + 5t',
-\]
-with \( s' \ge 3 \) and \( t' \ge 2 \).
-
-By the PCI, every natural number \( n > 33 \) (i.e. \( n \ge 34 \)) can be written in the form
-\[
-n = 4s + 5t,
-\]
-where \( s \ge 3 \) and \( t \ge 2 \) are integers.
 `)
 
+#proof[
+  We seek to show the proposition. We first show a few base cases, for $n = 34,
+35, 36$.
+  $
+    34 &= 4(6) + 5(2) \
+    35 &= 4(5) + 5(3) \
+    36 &= 4(4) + 5(4)
+  $
+  Now we state our inductive hypothesis. Suppose that for some natural $n$, $34
+  <= k <= n$ can be written in the form $k = 4s + 5t$, for some integral $s >=
+  3$ and $t >= 2$. Then we proceed by strong induction. Note that if $n >= 37$,
+  then $n - 3 >= 34$ so $n - 3$ is covered by our inductive hypothesis
+  (otherwise $n$ would be covered by one of our base cases and we are done). So
+  $n-3$ can be written in the desired $4s + 5t$ form. Now we seek to show that
+  this implies $n+1$ can also be written in the desired form.
+
+  $
+    n + 1 = (n - 3) + 4 = 4s + 5t + 4 = 4(s+1) + 5t
+  $
+  $s+1 >= 3$ because $s >= 3$ and $t >= 2$ is still true. So indeed $n+1$ can
+  be written in the desired form. By strong induction, every natural $n >= 34$
+  can be written as $n = 4s + 5t$ for some integral $s >= 3$ and $t >= 2$.
+]
+
 == 3
 
 #mitext(`
@@ -701,90 +677,40 @@ Then for all natural numbers \( n \),
 \[
 a_n = 2^n.
 \]
-
-We will prove by complete (strong) induction that for every natural number \( n \), the equality
-\[
-a_n = 2^n
-\]
-holds.
-
-- For \( n = 1 \):
-  \[
-  a_1 = 2 = 2^1.
-  \]
-- For \( n = 2 \):
-  \[
-  a_2 = 4 = 2^2.
-  \]
-
-Thus, the statement is true for \( n = 1 \) and \( n = 2 \).
-
-Inductive Hypothesis:
-Assume that for all natural numbers \( j \) with \( 1 \le j \le n \) (for some \( n \ge 2 \)), we have
-\[
-a_j = 2^j.
-\]
-
-We need to show that \( a_{n+1} = 2^{n+1} \).
-
-Notice that if \( n \ge 2 \), we can apply the recurrence relation with \( j = n-1 \) (since \( n-1 \ge 1 \)):
-\[
-a_{(n-1)+2} = a_{n+1} = 5a_n - 6a_{n-1}.
-\]
-By the inductive hypothesis, we know:
-\[
-a_n = 2^n \quad \text{and} \quad a_{n-1} = 2^{n-1}.
-\]
-Thus,
-\[
-a_{n+1} = 5(2^n) - 6(2^{n-1}).
-\]
-Factor \(2^{n-1}\) from the right-hand side:
-\[
-a_{n+1} = 2^{n-1}\Bigl(5\cdot 2 - 6\Bigr) = 2^{n-1}(10-6) = 2^{n-1} \cdot 4 = 2^{n+1}.
-\]
-
-By the PCI, the equality
-\[
-a_n = 2^n
-\]
-holds for all natural numbers \( n \), completing the proof.
 `)
 
+#proof[
+
+  We already have bases cases for $n = 1$ and $n = 2$, so we just need to show
+  assume the inductive hypothesis for $n > 2$. Note that when $n > 2$,
+  $a_(n+1)$
+  is
+  $
+    a_(n+1) = 5a_(n) - 6a_(n-1)
+  $
+
+  Suppose that all $a_k$ such that $3 <= k <= n$ are given by $a_k = 2^k$. We
+  seek to show that $n+1$ is also given by this equation.
+
+  Because $a_n$ and $a_(n-1)$ are included in our inductive hypothesis,
+  we have
+  $
+    a_(n+1) = 5(2^n) - 6(2^(n-1) = 5(2^n) - 3(2^n) = 2^(n+1)
+  $
+  Therefore indeed $a_(n+1)$ is given by $2^(n+1)$ which satisfies our
+  proposition and so it is true for all natural numbers.
+]
+
 == 10
 
 #mitext(`
 Every nonempty subset \( S \) of \(\mathbb{Z}^-\) (the set of negative integers) has a largest element.
-
-Let \( S \subseteq \mathbb{Z}^- \) be nonempty. Define the set
-\[
-T = \{ -s : s \in S \}.
-\]
-Since every element \( s \) in \( S \) is negative, each \( -s \) is a positive integer. Hence, \( T \) is a nonempty subset of the positive integers \(\mathbb{N}\).
-
-By the well-ordering principle of \(\mathbb{N}\), the set \( T \) has a least element, say \( m \). Thus,
-\[
-m \in T \quad \text{and} \quad m \le t \quad \text{for all } t \in T.
-\]
-
-Since \( m \in T \), there exists an element \( s_0 \in S \) such that
-\[
-m = -s_0.
-\]
-
-We now claim that \( s_0 \) is the largest element of \( S \). To see this, let \( s \) be any element of \( S \). Then \(-s \in T\), and by the minimality of \( m \) we have
-\[
-m \le -s.
-\]
-Substituting \( m = -s_0 \), we obtain
-\[
--s_0 \le -s.
-\]
-Multiplying both sides by \(-1\) (which reverses the inequality) gives
-\[
-s_0 \ge s.
-\]
-Since \( s \) was an arbitrary element of \( S \), it follows that \( s_0 \) is an upper bound of \( S \) and, being an element of \( S \), is the largest element of \( S \).
-
-Thus, every nonempty subset of \(\mathbb{Z}^-\) has a largest element.
 `)
+
+#proof[
+  We construct a new set, $S' = {-x | x in S}$. Then we note that $S'$ is a set
+  of positive integers, and in fact $S' subset.eq NN$. By the well ordering
+  principle, $S'$ always has a least element. Call this least element $k in
+  S'$. $k$ is the least element in $S'$ if and only if $-k$ is the greatest
+  element in $S$. Thus, $S$ always has a largest element.
+]

documents/flake.lock

@@ -5,11 +5,11 @@
         "nixpkgs-lib": "nixpkgs-lib"
       },
       "locked": {
-        "lastModified": 1736143030,
+        "lastModified": 1738453229,
-        "narHash": "sha256-+hu54pAoLDEZT9pjHlqL9DNzWz0NbUn8NEAHP7PQPzU=",
+        "narHash": "sha256-7H9XgNiGLKN1G1CgRh0vUL4AheZSYzPm+zmZ7vxbJdo=",
         "owner": "hercules-ci",
         "repo": "flake-parts",
-        "rev": "b905f6fc23a9051a6e1b741e1438dbfc0634c6de",
+        "rev": "32ea77a06711b758da0ad9bd6a844c5740a87abd",
         "type": "github"
       },
       "original": {
@@ -20,11 +20,11 @@
     },
     "nixpkgs": {
       "locked": {
-        "lastModified": 1736012469,
+        "lastModified": 1739866667,
-        "narHash": "sha256-/qlNWm/IEVVH7GfgAIyP6EsVZI6zjAx1cV5zNyrs+rI=",
+        "narHash": "sha256-EO1ygNKZlsAC9avfcwHkKGMsmipUk1Uc0TbrEZpkn64=",
         "owner": "NixOS",
         "repo": "nixpkgs",
-        "rev": "8f3e1f807051e32d8c95cd12b9b421623850a34d",
+        "rev": "73cf49b8ad837ade2de76f87eb53fc85ed5d4680",
         "type": "github"
       },
       "original": {
@@ -36,14 +36,14 @@
     },
     "nixpkgs-lib": {
       "locked": {
-        "lastModified": 1735774519,
+        "lastModified": 1738452942,
-        "narHash": "sha256-CewEm1o2eVAnoqb6Ml+Qi9Gg/EfNAxbRx1lANGVyoLI=",
+        "narHash": "sha256-vJzFZGaCpnmo7I6i416HaBLpC+hvcURh/BQwROcGIp8=",
         "type": "tarball",
-        "url": "https://github.com/NixOS/nixpkgs/archive/e9b51731911566bbf7e4895475a87fe06961de0b.tar.gz"
+        "url": "https://github.com/NixOS/nixpkgs/archive/072a6db25e947df2f31aab9eccd0ab75d5b2da11.tar.gz"
       },
       "original": {
         "type": "tarball",
-        "url": "https://github.com/NixOS/nixpkgs/archive/e9b51731911566bbf7e4895475a87fe06961de0b.tar.gz"
+        "url": "https://github.com/NixOS/nixpkgs/archive/072a6db25e947df2f31aab9eccd0ab75d5b2da11.tar.gz"
       }
     },
     "root": {
@@ -62,11 +62,11 @@
         ]
       },
       "locked": {
-        "lastModified": 1735930054,
+        "lastModified": 1738982361,
-        "narHash": "sha256-30Q6QmUHT/RBoPrrGfSNE9mjub3qLHJ0IDPUxuyHDAQ=",
+        "narHash": "sha256-QWDOo/+9pGu63knSlrhPiESSC+Ij/QYckC3yH8QPK4k=",
         "owner": "loqusion",
         "repo": "typix",
-        "rev": "29144f9e4131628afb800cc54806da1a702f7c80",
+        "rev": "bdb42d3e9a8722768e2168e31077129207870f92",
         "type": "github"
       },
       "original": {
@@ -78,11 +78,11 @@
     "typst-packages": {
       "flake": false,
       "locked": {
-        "lastModified": 1736182943,
+        "lastModified": 1740119901,
-        "narHash": "sha256-w8acKWK2aKkCsaOJcR9GLtkrsPJxDrrPN/77h1OkhuM=",
+        "narHash": "sha256-VzjqNki0yam7wrKbNkgMdyKyChEYBo0vjK0gwJZCFs4=",
         "owner": "typst",
         "repo": "packages",
-        "rev": "820818337a0f2a27dfc880f5ed96634914ed592f",
+        "rev": "92ae895b0076ac5d12ca81665b2af56f628b09f0",
         "type": "github"
       },
       "original": {
@@ -93,11 +93,11 @@
     },
     "zen-typ": {
       "locked": {
-        "lastModified": 1737198395,
+        "lastModified": 1740095364,
-        "narHash": "sha256-V3runX3cITanGPVF/Sfjak2/FNs6DuAwayQ44Iedal8=",
+        "narHash": "sha256-czzheiWIwJeze1aAzLmOgh+y9RoG8Qyzeicn1fd0ESY=",
         "owner": "youwen5",
         "repo": "zen.typ",
-        "rev": "3564895682f394f3c9be291332cc19c6ab695a60",
+        "rev": "d761990a004ad8dee8fc642a03fd4f11348a5fe2",
         "type": "github"
       },
       "original": {