#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(spacing: 1em) Problems: 2.4: \#4bcde, 5ajq, 6ce, 7a, 9, 10, 12abc 2.5: \#1abc, 3, 10 = 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. Consider the set with $0$ elements, $X = {}$. Then $cal(P)(X) = {emptyset}$ so indeed it has cardinality $2^0 = 1$. Otherwise suppose that any set $X$ with $n$ elements indeed has $|cal(P)(X)| = 2^n$. Then, consider a set $Y$ of $n+1$ elements. For each subset of $cal(P)(Y)$, 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 $2^n$ elements. We seek to show that this implies $cal(P)(Y)$ has $2^(n+1)$ elements. 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 from a subset of $Z$ is the addition of $k$. 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. // `) ] == 6 === c $(n+1)! > 2^(n+3)$ for all $n >= 5$. Consider $n = 5$. Then $6! > 2^(8)$. Now proceed by induction. Suppose that $(n+1)! > 2^(n + 3)$. 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. If we consider $n = 1$, then $2! < 2^(4)$ so the PMI does not hold for all $NN$. // #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 $(n! > 3n)$ for all $n >= 4$. Consider $n = 4$. Then $4! > 3(4)$. Suppose $(n! > 3n)$. Then $ (n+1)! > 3(n+1) \ (n+1)n! > 3n + 3 $ Assume $n >= 4$. Then the above statement holds true for any $n$. So we conclude the statement is always true for $n >= 4$. // #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 $ (sect.big^n_(i=1) A_i)^c = union.big^n_(i=1) A_i^c $ 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 $A_i$. 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 $y in.not A_i$. 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 $z in.not A_i$. Now note this is actually precisely the definition of $Y$. Hence $Z = Y$. // #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 Suppose there was one point. Then there would be $(1^2 - 1)/2 = 0$ line 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 $(n^2 - n)/2$. 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. // #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 First, note that with 1 disk, it takes $2^1 - 1 = 1$ moves to solve the puzzle. 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 stack, such that our situation is equivalent to when there are $n$ disks (because any disk can be stacked on top of the largest disk, our set of 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 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 $2^n - 1$ moves. So it took us a total of $2 dot (2^n - 1) + 1$ moves to move the entire stack to another peg, or $2^(n+1) - 1$ 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 Every natural number greater than or equal to 11 can be written in the form $2s + 5t$ for some naturals $s$ and $t$. First, we see that $11 = 2(3) + 5(1)$, $12 = 2(2) + 5(2)$, $13 = 2(4) + 5(1)$, $14 = 2(2) + 5(2)$. Now suppose that we can write all naturals up to $11 <= k <= n$ as $2s + 5t$ for naturals $s,t$. Note that we have shown naturals from 11 to 15 can be written in the $2s + 5t$ form. Then we note $n >= 15$ iff. $n - 4 >= 11$, and $n - 4$ is covered 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 $s, t$. === 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 \( t \ge 2 \). `) #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(` 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 \( t \ge 2 \). `) #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(` 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 \( n \), \[ a_n = 2^n. \] `) #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. `) #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. ]