auto-update(nvim): 2025-02-13 01:54:09

;

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+#import "@youwen/zen:0.1.0": *
+
+#show: zen.with(
+  title: "Homework 5",
+  author: "Youwen Wu",
+)
+
+#set heading(numbering: none)
+#show heading.where(level: 2): it => [#it.body.]
+#show heading.where(level: 3): it => [#it.body.]
+
+#set par(first-line-indent: 0pt, spacing: 1em)
+
+
+Problems:
+
+2.4: \#4bcde, 5ajq, 6ce, 7a, 9, 10, 12abc
+
+2.5: \#1abc, 3, 10
+
+#outline()
+
+= 2.4
+
+== 4
+
+=== b
+
+$3 + 11 + 19 + dots.c + (8n - 5) = 4n^2 - n$
+
+First we show the base case for $n = 1$.
+
+$ 3 = 4 (1^2) - 1 = 3 $
+
+Now we proceed by induction. Assume $sum _(k=1) ^n (8k-5) = 4(n^2) - n$.
+
+$
+  sum_(k=1)^(n+1) (8k-5) &= 4((n+1)^2) - (n+1) \
+  (sum_(k=1)^(n) (8k-5)) + (8(n+1) - 5) &= 4(n^2 + 2n + 1) - n-1 \
+  (sum_(k=1)^(n) (8k-5)) + 8n + 3 &= 4n^2 - n + 8n + 3 \
+  (sum_(k=1)^(n) (8k-5)) &= 4n^2 - n \
+$
+
+and we know that this is true by our original assumption. So by the PMI, this
+is true $forall n in NN$.
+
+=== c
+
+$sum_(i=1)^n 2^i = 2^(n+1) - 2$
+
+For the base case:
+
+$ 2^1 = 2(1 + 1) - 2 \ 2 = 2 $
+
+Assume that
+
+$
+  sum_(i=1)^n 2^i = 2^(n+1) - 2
+$
+
+Proceeding by induction,
+
+$
+  sum_(i=1)^(n+1) 2^i = 2^(n+1+1) - 2 \
+  (sum_(i=1)^(n) 2^i) + 2^(n+1) = 2^(n+2) - 2 \
+  (sum_(i=1)^(n) 2^i) = 2^(n+2) - 2^(n+1) - 2 \
+  (sum_(i=1)^(n) 2^i) = 2^(n+1) dot (2 - 1) - 2 \
+  (sum_(i=1)^(n) 2^i) = 2^(n+1) - 2 \
+$
+
+So by the PMI, it holds for all $n in NN$.
+
+=== d
+
+$1 dot 1! + 2 dot 2! + 3 dot 3! + dots.c + n dot n! = (n + 1)! - 1$
+
+For the base case: $1 dot 1! = (1 + 1)! - 1 \ 1 = 1$
+
+Assuming the inductive hypothesis,
+
+$
+  sum^n_(k=1) k dot k! = (n + 1)! - 1 \
+$
+
+Now we proceed by induction
+
+$
+  sum^(n+1)_(k=1) k dot k! = (n + 2)! - 1 \
+  sum^(n)_(k=1) k dot k! + (n + 1) dot (n+1)! = (n + 2) dot (n + 1)! - 1 \
+  sum^(n)_(k=1) k dot k! = (n + 2) dot (n + 1)! - (n + 1) dot (n+1)! - 1 \
+  sum^(n)_(k=1) k dot k! = (n + 1)! dot (n+2 - (n + 1) - 1 \
+  sum^(n)_(k=1) k dot k! = (n + 1)! - 1 \
+$
+
+So by the PMI, this is true for all $n in NN$

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

@@ -0,0 +1,790 @@
+#import "@youwen/zen:0.1.0": *
+#import "@preview/mitex:0.2.5": *
+
+#show: zen.with(
+  title: "Homework 5",
+  author: "Youwen Wu",
+)
+
+#set heading(numbering: none)
+#show heading.where(level: 2): it => [#it.body.]
+#show heading.where(level: 3): it => [#it.body.]
+
+#set par(first-line-indent: 0pt, spacing: 1em)
+
+
+Problems:
+
+2.4: \#4bcde, 5ajq, 6ce, 7a, 9, 10, 12abc
+
+2.5: \#1abc, 3, 10
+
+#outline()
+
+= 2.4
+
+== 4
+
+=== b
+
+$3 + 11 + 19 + dots.c + (8n - 5) = 4n^2 - n$
+
+#proof[
+  First we show the base case for $n = 1$.
+
+  $ 3 = 4 (1^2) - 1 = 3 $
+
+  Now we proceed by induction. Assume $sum _(k=1) ^n (8k-5) = 4(n^2) - n$.
+
+  $
+    sum_(k=1)^(n+1) (8k-5) &= 4((n+1)^2) - (n+1) \
+    (sum_(k=1)^(n) (8k-5)) + (8(n+1) - 5) &= 4(n^2 + 2n + 1) - n-1 \
+    (sum_(k=1)^(n) (8k-5)) + 8n + 3 &= 4n^2 - n + 8n + 3 \
+    (sum_(k=1)^(n) (8k-5)) &= 4n^2 - n \
+  $
+
+  and we know that this is true by our original assumption. So by the PMI, this
+  is true $forall n in NN$.
+]
+
+=== c
+
+#proof[
+  $sum_(i=1)^n 2^i = 2^(n+1) - 2$
+
+  For the base case:
+
+  $ 2^1 = 2(1 + 1) - 2 \ 2 = 2 $
+
+  Assume that
+
+  $
+    sum_(i=1)^n 2^i = 2^(n+1) - 2
+  $
+
+  Proceeding by induction,
+
+  $
+    sum_(i=1)^(n+1) 2^i = 2^(n+1+1) - 2 \
+    (sum_(i=1)^(n) 2^i) + 2^(n+1) = 2^(n+2) - 2 \
+    (sum_(i=1)^(n) 2^i) = 2^(n+2) - 2^(n+1) - 2 \
+    (sum_(i=1)^(n) 2^i) = 2^(n+1) dot (2 - 1) - 2 \
+    (sum_(i=1)^(n) 2^i) = 2^(n+1) - 2 \
+  $
+
+  So by the PMI, it holds for all $n in NN$.
+]
+
+=== d
+
+#proof[
+  $1 dot 1! + 2 dot 2! + 3 dot 3! + dots.c + n dot n! = (n + 1)! - 1$
+
+  For the base case: $1 dot 1! = (1 + 1)! - 1 \ 1 = 1$
+
+  Assuming the inductive hypothesis,
+
+  $
+    sum^n_(k=1) k dot k! = (n + 1)! - 1 \
+  $
+
+  Now we proceed by induction
+
+  $
+    sum^(n+1)_(k=1) k dot k! = (n + 2)! - 1 \
+    sum^(n)_(k=1) k dot k! + (n + 1) dot (n+1)! = (n + 2) dot (n + 1)! - 1 \
+    sum^(n)_(k=1) k dot k! = (n + 2) dot (n + 1)! - (n + 1) dot (n+1)! - 1 \
+    sum^(n)_(k=1) k dot k! = (n + 1)! dot (n+2 - (n + 1) - 1 \
+    sum^(n)_(k=1) k dot k! = (n + 1)! - 1 \
+  $
+
+  So by the PMI, this is true for all $n in NN$.
+]
+
+=== e
+
+#proof[
+  $1^3 + 2^3 + dots.c + n^3 = [(n(n+1)) / 2]^2$
+
+  First checking the base case:
+
+  $
+    1^3 = [(1(1 + 1)) / 2]^2 \
+    1 = 1
+  $
+
+  Now assume the inductive hypothesis
+
+  $ sum_(k=1)^n k^3 = [(n(n+1)) / 2]^2 $
+
+  Proceeding by induction,
+
+  $
+    sum_(k=1)^(n + 1) k^3 = [((n+1)(n+2)) / 2]^2 \
+    sum_(k=1)^(n) k^3 + (n+1)^3 = ((n+1)^2 (n+2)^2) / 4 \
+    sum_(k=1)^(n) k^3 = ((n+1)^2 (n+2)^2) / 4 - (n+1)^3 \
+    sum_(k=1)^(n) k^3 = ((n+1)^2 [(n+2)^2 - 4(n+1)]) / 4 \
+    sum_(k=1)^(n) k^3 = ((n+1)^2 n^2) / 4 \
+    sum_(k=1)^(n) k^3 = [(n (n+1)) / 2]^2 \
+  $
+
+  So by the PMI it is true $forall n in NN$.
+]
+
+== 5
+
+=== a
+
+$n^3 + 5n + 6$ is divisible by 3.
+
+#proof[
+  Checking the base case,
+
+  $
+    3 | 1^3 + 5(1) + 6 \
+    3 | 12
+  $
+
+  Now assume the inductive hypothesis.
+
+  $
+    3 | n^3 + 5n + 6
+  $
+
+  Proceeding by induction,
+
+  $
+    3 | (n+1)^3 + 5(n+1) + 6 \
+    3 | (n+1)^3 + 5(n+1) + 6 \
+    3 | n^3 + 3n^2 + 3n + 1 + 5n + 5 + 6 \
+    3 | n^3 + 5n + 6 + 3n^2 + 3n + 6 \
+  $
+
+  By our inductive hypothesis, we know that $3 | n^3 + 5n + 6$. Additionally we
+  clearly see that $3 | 3n^2 + 3n + 6$ as it can be factored out. So the
+  $(n+1)^"th"$ case is true and by the PMI, it is true for all $n in NN$.
+]
+
+=== j
+
+$3^n >= 1 + 2^n$
+
+#proof[
+  Base case: $ 3^1 >= 1 + 2^1 \ 3 >= 3 $
+
+  Assume the inductive hypothesis
+
+  $
+    3^n >= 1 + 2^n
+  $
+
+  Now proceed by induction
+
+  $
+    3^(n+1) >= 1 + 2^(n+1)
+  $
+
+  We want to show that this statement is always true. Rewrite the left side as
+  $3 dot 3^n$, then we can derive the following inequality from our inductive
+  hypothesis:
+
+  $
+    3 dot 3^n >= 3(1 + 2^n) \
+    3 dot 3^n >= (2 + 1)(1 + 2^n) \
+    3 dot 3^n >= 2 + 2^(n+1) + 1 + 2^n \
+  $
+
+  And clearly we have
+
+  $
+    2 + 2^(n+1) + 1 + 2^n >= 1 + 2^(n+1)
+  $
+
+  So,
+
+  $
+    3^(n+1) >= 1 + 2^(n+1) + 2 + 2^n >= 1 + 2^(n+1)
+  $
+
+  We have shown that it applies to the $(n+1)^"th"$ case when the $n^"th"$ is
+  true. Therefore by the PMI it is true $forall n in NN$.
+]
+
+=== q
+
+If a set $A$ has $n$ elements, then $cal(P) (A)$ has $2^n$ elements.
+
+#proof[
+  Just for fun, consider an $n$-tuple that encodes a subset of $A$. Each entry
+  corresponds to a different element in $A$, and is 1 if that element is in the
+  subset, and 0 if it is not. Now to count the cardinality of $cal(P)(A)$, we
+  simply need to count each possible combination of entries in our tuple, as
+  each tuple corresponds to a unique subset of $A$. Since there are 2
+  alternatives for each entry and $n$ entries, by the general multiplication
+  principle, there are $2^n$ variants of this tuple so the cardinality of
+  $cal(P)(A)$ is $2^n$.
+
+  ---
+
+  Ok now let's do it the annoying way.
+
+  #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.
+`)
+
+]
+
+== 6
+
+=== c
+
+#mitext(`
+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
+\[
+(n+1)! > 2^{n+3}.
+\]
+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(`
+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
+\[
+n! > 3n.
+\]
+We must show that
+\[
+(n+1)! > 3(n+1).
+\]
+
+Starting with the left-hand side:
+\[
+(n+1)! = (n+1) \cdot n!.
+\]
+By the inductive hypothesis, we have:
+\[
+(n+1)! > (n+1) \cdot 3n.
+\]
+Since \( n \ge 4 \) (so \( n \ge 2 \)), it follows that:
+\[
+(n+1) \cdot 3n \ge 3(n+1).
+\]
+To see this, note that for \( n \ge 2 \) we have:
+\[
+3n(n+1) = 3(n+1)n > 3(n+1),
+\]
+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 \( n\ge1 \),
+\[
+\left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c
+\]
+and
+\[
+\left(\bigcap_{i=1}^{n} A_i\right)^c = \bigcup_{i=1}^{n} A_i^c.
+\]
+
+For \( n=1 \), we have
+\[
+\left(\bigcup_{i=1}^{1} A_i\right)^c = A_1^c \quad \text{and} \quad \bigcap_{i=1}^{1} A_i^c = A_1^c.
+\]
+Thus, the identity holds for \( n=1 \).
+
+Assume that for some \( n \ge 1 \) the statement holds; that is, assume
+\[
+\left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c.
+\]
+We need to show that
+\[
+\left(\bigcup_{i=1}^{n+1} A_i\right)^c = \bigcap_{i=1}^{n+1} A_i^c.
+\]
+
+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(`
+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
+\[
+\frac{n^2 - n}{2}.
+\]
+
+For \(n=2\), there is exactly 1 line segment joining the two points. The formula gives
+\[
+\frac{2^2 - 2}{2} = \frac{4-2}{2} = 1.
+\]
+So the statement holds for \(n=2\).
+
+Assume that for some \(n \ge 2\), the number of line segments joining \(n\) points is
+\[
+\frac{n^2 - n}{2}.
+\]
+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
+\[
+\frac{n^2 - n}{2} + n.
+\]
+Simplify the expression:
+\[
+\frac{n^2 - n}{2} + n = \frac{n^2 - n + 2n}{2} = \frac{n^2 + n}{2} = \frac{n(n+1)}{2}.
+\]
+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(`
+The Tower of Hanoi problem with \( n \) disks can be solved in exactly \( 2^n - 1 \) moves.
+
+For \( n = 1 \):
+There is only one disk, and it can be moved directly from the source peg to the destination peg in one move.
+Since \( 2^1 - 1 = 1 \), the statement holds for \( n = 1 \).
+
+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.
+
+1. Move the top \( k \) disks from the source peg to the auxiliary peg.
+   By the inductive hypothesis, this takes \( 2^k - 1 \) 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
+
+=== a
+
+F.
+
+The inductive step assumes that two overlapping sets of $n$ horses share at
+least $n−1$ horses in common, so that they must all be the same color. But
+when $n=1$, two different 1‐horse sets do not overlap at all, so the argument
+that “both sets share a horse of the same color” no longer applies.
+
+=== b
+
+F.
+
+The inductive step is not shown, so there is no justification for the claim
+that it is divisible.
+
+=== c
+
+C.
+
+Though the inductive reasoning is right, the base case is not shown, so there
+is no reason why this should hold true for all $NN$.
+
+= 2.5
+
+== 1
+
+=== a
+
+#mitext(`
+Every natural number \( n \ge 11 \) can be written in the form
+\[
+n = 2s + 5t,
+\]
+for some nonnegative integers \( s \) and \( t \).
+
+We verify the statement for the initial numbers:
+- \( n = 11 \): \( 11 = 2\cdot 3 + 5\cdot 1 \)
+- \( n = 12 \): \( 12 = 2\cdot 1 + 5\cdot 2 \)
+- \( n = 13 \): \( 13 = 2\cdot 4 + 5\cdot 1 \)
+- \( n = 14 \): \( 14 = 2\cdot 2 + 5\cdot 2 \)
+- \( n = 15 \): \( 15 = 2\cdot 5 + 5\cdot 1 \)
+
+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
+
+#mitext(`
+Every natural number \( n > 22 \) (i.e. every \( n \ge 23 \)) can be written in the form
+\[
+n = 3s + 4t,
+\]
+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.
+`)
+
+=== c
+
+#mitext(`
+Every natural number \( n > 33 \) (i.e. every \( n \ge 34 \)) can be written in the form
+\[
+n = 4s + 5t,
+\]
+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.
+`)
+
+== 3
+
+#mitext(`
+Let the sequence \(\{a_n\}\) be defined by
+\[
+a_1 = 2,\quad a_2 = 4,\quad a_{n+2} = 5a_{n+1} - 6a_n \quad \text{for all } n \ge 1.
+\]
+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.
+`)
+
+== 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.
+`)

documents/by-course/math-8/pset-5/package.nix

@@ -0,0 +1,37 @@
+{
+  pkgs,
+  typstPackagesCache,
+  typixLib,
+  cleanTypstSource,
+  flakeSelf,
+  ...
+}:
+let
+  src = cleanTypstSource ./.;
+  commonArgs = {
+    typstSource = "main.typ";
+
+    fontPaths = [
+      # Add paths to fonts here
+      # "${pkgs.roboto}/share/fonts/truetype"
+    ];
+
+    virtualPaths = [
+      # Add paths that must be locally accessible to typst here
+      # {
+      #   dest = "icons";
+      #   src = "${inputs.font-awesome}/svgs/regular";
+      # }
+    ];
+
+    XDG_CACHE_HOME = typstPackagesCache;
+    SOURCE_DATE_EPOCH = builtins.toString flakeSelf.lastModified;
+  };
+
+in
+typixLib.buildTypstProject (
+  commonArgs
+  // {
+    inherit src;
+  }
+)