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

#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 $S' subset.eq NN$, and it is nonempty. By
  the well ordering principle, $S'$ always has a least element. Call this least
  element $s in S'$. Now we show that $s$ is the least element in $S'$ if and
  only if $-s$ is the greatest element in $S$.

  We say that $s$ is the least element in $S'$ iff. for every $x in S'$ where
  $x != s$, we have $s < x$. Note that $-s in S$ and for every $x$, $-x$ is in
  $S$. Also, these are all of the elements in $S$, that is, it has no other
  elements besides $-x$ and $-s$ for all possible $x$. So, equivalently, for
  every $-x$ such that $-x != -s$, $-s > -x$. So if $s$ is the least element in
  $S'$, then $-s in S$ is the greatest element in $S$.

  Thus, as $s$ is guaranteed to exist by the WOP, $-s$ is also guaranteed to
  exist, and $S$ always has a greatest element.

]