#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. `)