From 3d4a68f1f3f6989e1fe0ccef4a033890d41e1ee0 Mon Sep 17 00:00:00 2001 From: Youwen Wu Date: Thu, 13 Feb 2025 01:54:09 -0800 Subject: [PATCH] auto-update(nvim): 2025-02-13 01:54:09 --- ; | 95 +++ documents/by-course/math-8/pset-5/main.typ | 790 ++++++++++++++++++ documents/by-course/math-8/pset-5/package.nix | 37 + 3 files changed, 922 insertions(+) create mode 100644 ; create mode 100644 documents/by-course/math-8/pset-5/main.typ create mode 100644 documents/by-course/math-8/pset-5/package.nix diff --git a/; b/; new file mode 100644 index 0000000..08a1ea7 --- /dev/null +++ b/; @@ -0,0 +1,95 @@ +#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 $n in NN$ diff --git a/documents/by-course/math-8/pset-5/main.typ b/documents/by-course/math-8/pset-5/main.typ new file mode 100644 index 0000000..318844b --- /dev/null +++ b/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. +`) diff --git a/documents/by-course/math-8/pset-5/package.nix b/documents/by-course/math-8/pset-5/package.nix new file mode 100644 index 0000000..7879e7e --- /dev/null +++ b/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; + } +)