716 lines
20 KiB
Text
716 lines
20 KiB
Text
#import "@youwen/zen:0.1.0": *
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#import "@preview/mitex:0.2.5": *
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#show: zen.with(
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title: "Homework 5",
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author: "Youwen Wu",
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)
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#set heading(numbering: none)
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#show heading.where(level: 2): it => [#it.body.]
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#show heading.where(level: 3): it => [#it.body.]
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#set par(spacing: 1em)
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Problems:
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2.4: \#4bcde, 5ajq, 6ce, 7a, 9, 10, 12abc
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2.5: \#1abc, 3, 10
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= 2.4
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== 4
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=== b
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$3 + 11 + 19 + dots.c + (8n - 5) = 4n^2 - n$
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#proof[
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First we show the base case for $n = 1$.
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$ 3 = 4 (1^2) - 1 = 3 $
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Now we proceed by induction. Assume $sum _(k=1) ^n (8k-5) = 4(n^2) - n$.
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$
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sum_(k=1)^(n+1) (8k-5) &= 4((n+1)^2) - (n+1) \
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(sum_(k=1)^(n) (8k-5)) + (8(n+1) - 5) &= 4(n^2 + 2n + 1) - n-1 \
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(sum_(k=1)^(n) (8k-5)) + 8n + 3 &= 4n^2 - n + 8n + 3 \
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(sum_(k=1)^(n) (8k-5)) &= 4n^2 - n \
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$
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and we know that this is true by our original assumption. So by the PMI, this
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is true $forall n in NN$.
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]
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=== c
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#proof[
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$sum_(i=1)^n 2^i = 2^(n+1) - 2$
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For the base case:
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$ 2^1 = 2(1 + 1) - 2 \ 2 = 2 $
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Assume that
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$
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sum_(i=1)^n 2^i = 2^(n+1) - 2
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$
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Proceeding by induction,
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$
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sum_(i=1)^(n+1) 2^i = 2^(n+1+1) - 2 \
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(sum_(i=1)^(n) 2^i) + 2^(n+1) = 2^(n+2) - 2 \
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(sum_(i=1)^(n) 2^i) = 2^(n+2) - 2^(n+1) - 2 \
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(sum_(i=1)^(n) 2^i) = 2^(n+1) dot (2 - 1) - 2 \
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(sum_(i=1)^(n) 2^i) = 2^(n+1) - 2 \
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$
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So by the PMI, it holds for all $n in NN$.
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]
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=== d
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#proof[
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$1 dot 1! + 2 dot 2! + 3 dot 3! + dots.c + n dot n! = (n + 1)! - 1$
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For the base case: $1 dot 1! = (1 + 1)! - 1 \ 1 = 1$
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Assuming the inductive hypothesis,
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$
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sum^n_(k=1) k dot k! = (n + 1)! - 1 \
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$
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Now we proceed by induction
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$
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sum^(n+1)_(k=1) k dot k! = (n + 2)! - 1 \
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sum^(n)_(k=1) k dot k! + (n + 1) dot (n+1)! = (n + 2) dot (n + 1)! - 1 \
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sum^(n)_(k=1) k dot k! = (n + 2) dot (n + 1)! - (n + 1) dot (n+1)! - 1 \
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sum^(n)_(k=1) k dot k! = (n + 1)! dot (n+2 - (n + 1) - 1 \
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sum^(n)_(k=1) k dot k! = (n + 1)! - 1 \
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$
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So by the PMI, this is true for all $n in NN$.
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]
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=== e
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#proof[
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$1^3 + 2^3 + dots.c + n^3 = [(n(n+1)) / 2]^2$
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First checking the base case:
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$
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1^3 = [(1(1 + 1)) / 2]^2 \
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1 = 1
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$
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Now assume the inductive hypothesis
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$ sum_(k=1)^n k^3 = [(n(n+1)) / 2]^2 $
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Proceeding by induction,
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$
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sum_(k=1)^(n + 1) k^3 = [((n+1)(n+2)) / 2]^2 \
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sum_(k=1)^(n) k^3 + (n+1)^3 = ((n+1)^2 (n+2)^2) / 4 \
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sum_(k=1)^(n) k^3 = ((n+1)^2 (n+2)^2) / 4 - (n+1)^3 \
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sum_(k=1)^(n) k^3 = ((n+1)^2 [(n+2)^2 - 4(n+1)]) / 4 \
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sum_(k=1)^(n) k^3 = ((n+1)^2 n^2) / 4 \
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sum_(k=1)^(n) k^3 = [(n (n+1)) / 2]^2 \
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$
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So by the PMI it is true $forall n in NN$.
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]
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== 5
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=== a
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$n^3 + 5n + 6$ is divisible by 3.
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#proof[
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Checking the base case,
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$
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3 | 1^3 + 5(1) + 6 \
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3 | 12
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$
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Now assume the inductive hypothesis.
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$
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3 | n^3 + 5n + 6
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$
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Proceeding by induction,
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$
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3 | (n+1)^3 + 5(n+1) + 6 \
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3 | (n+1)^3 + 5(n+1) + 6 \
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3 | n^3 + 3n^2 + 3n + 1 + 5n + 5 + 6 \
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3 | n^3 + 5n + 6 + 3n^2 + 3n + 6 \
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$
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By our inductive hypothesis, we know that $3 | n^3 + 5n + 6$. Additionally we
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clearly see that $3 | 3n^2 + 3n + 6$ as it can be factored out. So the
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$(n+1)^"th"$ case is true and by the PMI, it is true for all $n in NN$.
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]
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=== j
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$3^n >= 1 + 2^n$
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#proof[
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Base case: $ 3^1 >= 1 + 2^1 \ 3 >= 3 $
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Assume the inductive hypothesis
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$
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3^n >= 1 + 2^n
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$
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Now proceed by induction
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$
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3^(n+1) >= 1 + 2^(n+1)
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$
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We want to show that this statement is always true. Rewrite the left side as
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$3 dot 3^n$, then we can derive the following inequality from our inductive
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hypothesis:
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$
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3 dot 3^n >= 3(1 + 2^n) \
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3 dot 3^n >= (2 + 1)(1 + 2^n) \
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3 dot 3^n >= 2 + 2^(n+1) + 1 + 2^n \
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$
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And clearly we have
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$
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2 + 2^(n+1) + 1 + 2^n >= 1 + 2^(n+1)
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$
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So,
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$
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3^(n+1) >= 1 + 2^(n+1) + 2 + 2^n >= 1 + 2^(n+1)
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$
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We have shown that it applies to the $(n+1)^"th"$ case when the $n^"th"$ is
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true. Therefore by the PMI it is true $forall n in NN$.
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]
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=== q
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If a set $A$ has $n$ elements, then $cal(P) (A)$ has $2^n$ elements.
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#proof[
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Just for fun, consider an $n$-tuple that encodes a subset of $A$. Each entry
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corresponds to a different element in $A$, and is 1 if that element is in the
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subset, and 0 if it is not. Now to count the cardinality of $cal(P)(A)$, we
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simply need to count each possible combination of entries in our tuple, as
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each tuple corresponds to a unique subset of $A$. Since there are 2
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alternatives for each entry and $n$ entries, by the general multiplication
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principle, there are $2^n$ variants of this tuple so the cardinality of
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$cal(P)(A)$ is $2^n$.
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---
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Ok now let's do it the annoying way.
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Consider the set with $0$ elements, $X = {}$. Then $cal(P)(X) = {emptyset}$
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so indeed it has cardinality $2^0 = 1$. Otherwise suppose that any set $X$
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with $n$ elements indeed has $|cal(P)(X)| = 2^n$.
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Then, consider a set $Y$ of $n+1$ elements. For each subset of $cal(P)(Y)$,
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we choose an arbitrary element $k in Y$, and define $Z := Y backslash {k}$.
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Note that $Z$ has $n$ elements so our inductive hypothesis says it has $2^n$
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elements. We seek to show that this implies $cal(P)(Y)$ has $2^(n+1)$
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elements.
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Every subset of $Y$ in $cal(P)(Y)$ satisfies the following: either the subset
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does not contain $k$ and so it's also a subset of $Z$, or its only difference
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from a subset of $Z$ is the addition of $k$.
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Therefore, for each subset of $Z$, we can introduce $k$, and it's still a
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subset of $Y$, and in fact completes $cal(P)(Y)$ from $cal(P)(Z)$, due to our
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previous assertion. So for each element in $cal(P)(Z)$, there is one
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additional element in $cal(P)(Y)$. In other words,
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$
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|cal(P)(Z)| dot 2 = |cal(P)(Y)|
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$
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Recall our inductive hypothesis states $|cal(P)(Z)| = 2^n$ because $Z$ has
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$n$ elements, so $|cal(P)(X)| = 2 dot |cal(P)(Z)| = 2^(n+1)$, completing our
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induction. Therefore for any arbitrary set $X$ with $n$ elements, its power
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set $cal(P)(X)$ has $2^n$ elements.
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// #mitext(`
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// Base Case ($n=0$). \\
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// If $A$ has $0$ elements, then $A = \varnothing$. Its power set is
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// \[
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// \mathcal{P}(A) = \{\varnothing\},
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// \]
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// which has exactly one element. Hence
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// \[
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// |\mathcal{P}(A)| = 1 = 2^0.
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// \]
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// So the statement holds for $n=0$.
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//
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// Now proceed by induction. Assume the statement is true for some integer $k \ge
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// 0$. That is, suppose that for any set $A$ with $k$ elements, we have
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// \[
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// |\mathcal{P}(A)| = 2^k.
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// \]
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//
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// 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,
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// \[
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// |\mathcal{P}(A)| = 2^k.
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// \]
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// Now observe that any subset of $B$ is either:
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// \begin{itemize}
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// \item A subset of $A$ (does not contain $x$), or
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// \item Of the form $S \cup \{x\}$ where $S \subseteq A$ (does contain $x$).
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// \end{itemize}
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// 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,
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// \[
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// |\mathcal{P}(B)| = |\mathcal{P}(A)| + |\mathcal{P}(A)|
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// = 2^k + 2^k
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// = 2 \cdot 2^k
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// = 2^{k+1}.
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// \]
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//
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// 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.
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// `)
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]
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== 6
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=== c
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$(n+1)! > 2^(n+3)$ for all $n >= 5$.
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Consider $n = 5$. Then $6! > 2^(8)$. Now proceed by induction. Suppose that
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$(n+1)! > 2^(n + 3)$.
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Then $(n + 1 + 1)! > 2^(n+3 + 1)$ so $(n + 2)(n+1)! > 2 dot 2^(n + 3)$. We note
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that $n+2$ is always at least $2$, so this is true by our inductive hypothesis.
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If we consider $n = 1$, then $2! < 2^(4)$ so the PMI does not hold for all $NN$.
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// #mitext(`
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// Let \( n = 5 \). Then:
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// \[
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// (5+1)! = 6! = 720 \quad\text{and}\quad 2^{5+3} = 2^8 = 256.
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// \]
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// Since \( 720 > 256 \), the inequality holds for \( n = 5 \).
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//
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// 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
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// \[
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// (n+1)! > 2^{n+3}.
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// \]
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// We must show that
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// \[
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// (n+2)! > 2^{(n+1)+3} = 2^{n+4}.
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// \]
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//
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// Starting with the left-hand side, we have:
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// \[
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// (n+2)! = (n+2)(n+1)!.
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// \]
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// Using the inductive hypothesis,
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// \[
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// (n+2)! > (n+2) \cdot 2^{n+3}.
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// \]
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// Since \( n \ge 5 \), it follows that \( n+2 \ge 7 \). Therefore,
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// \[
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// (n+2) \cdot 2^{n+3} \ge 7 \cdot 2^{n+3}.
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// \]
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// But clearly,
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// \[
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// 7 \cdot 2^{n+3} > 2 \cdot 2^{n+3} = 2^{n+4}.
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// \]
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// Thus,
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// \[
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// (n+2)! > 2^{n+4},
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// \]
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// which completes the inductive step.
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//
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// By the generalized principle of mathematical induction, the inequality
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// \[
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// (n+1)! > 2^{n+3}
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// \]
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// holds for all natural numbers \( n \ge 5 \).
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//
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// The inequality is stated to hold for all \( n \ge 5 \). However, for \( n < 5 \) the inequality fails. For instance, when \( n = 4 \):
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// \[
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// (4+1)! = 5! = 120 \quad\text{and}\quad 2^{4+3} = 2^7 = 128.
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// \]
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// Since \( 120 \) is not greater than \( 128 \), the inequality is false for \( n = 4 \).
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// `)
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=== e
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$(n! > 3n)$ for all $n >= 4$.
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Consider $n = 4$. Then $4! > 3(4)$. Suppose $(n! > 3n)$. Then
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$
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(n+1)! > 3(n+1) \
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(n+1)n! > 3n + 3
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$
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Assume $n >= 4$. Then the above statement holds true for any $n$. So we
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conclude the statement is always true for $n >= 4$.
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// #mitext(`
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// Let \( n = 4 \). Then:
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// \[
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// 4! = 24 \quad\text{and}\quad 3 \times 4 = 12.
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// \]
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// Since \( 24 > 12 \), the inequality holds for \( n = 4 \).
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//
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// 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
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// \[
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// n! > 3n.
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// \]
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// We must show that
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// \[
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// (n+1)! > 3(n+1).
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// \]
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//
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// Starting with the left-hand side:
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// \[
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// (n+1)! = (n+1) \cdot n!.
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// \]
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// By the inductive hypothesis, we have:
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// \[
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// (n+1)! > (n+1) \cdot 3n.
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// \]
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// Since \( n \ge 4 \) (so \( n \ge 2 \)), it follows that:
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// \[
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// (n+1) \cdot 3n \ge 3(n+1).
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// \]
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// To see this, note that for \( n \ge 2 \) we have:
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// \[
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// 3n(n+1) = 3(n+1)n > 3(n+1),
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// \]
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// because \( n > 1 \). Therefore,
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// \[
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// (n+1)! > 3(n+1),
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// \]
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// which completes the inductive step.
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//
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// By the generalized principle of mathematical induction, the inequality
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// \[
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// n! > 3n
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// \]
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// holds for all natural numbers \( n \ge 4 \).
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//
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// The claim is that \( n! > 3n \) for all \( n \ge 4 \). However, for some smaller natural numbers the inequality is false. For instance, when \( n = 3 \):
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// \[
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// 3! = 6 \quad\text{and}\quad 3 \times 3 = 9.
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// \]
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// `)
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== 7
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=== a
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$
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(sect.big^n_(i=1) A_i)^c = union.big^n_(i=1) A_i^c
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$
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First consider $sect.big^n_(i=1) A_i$. Let $X := sect.big^n_(i=1) A_i$.
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Then $x in X$ if and only if $x in A_i$ for every $A_i$.
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Now consider $Y := (sect.big^n_(i=1) A_i)^c$. Then $y in Y$ if and only if $y
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in.not X$, that is, if and only there exists some $A_i$ such that $y in.not
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A_i$. Now let $Z := union.big^n_(i=1) A_i^c$. Every element $z in Z$ is in some
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$A^c _i$, that is, $z in Z$ if and only if there exists some $A_i$ such that $z
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in.not A_i$. Now note this is actually precisely the definition of $Y$. Hence
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$Z = Y$.
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// #mitext(`
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// Let \(\{A_i : i \in \mathbb{N}\}\) be an indexed family of sets. Then for every natural number \( n\ge1 \),
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// \[
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// \left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c
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// \]
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// and
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// \[
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// \left(\bigcap_{i=1}^{n} A_i\right)^c = \bigcup_{i=1}^{n} A_i^c.
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// \]
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//
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// For \( n=1 \), we have
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// \[
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// \left(\bigcup_{i=1}^{1} A_i\right)^c = A_1^c \quad \text{and} \quad \bigcap_{i=1}^{1} A_i^c = A_1^c.
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// \]
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// Thus, the identity holds for \( n=1 \).
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//
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// Assume that for some \( n \ge 1 \) the statement holds; that is, assume
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// \[
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// \left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c.
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// \]
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// We need to show that
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// \[
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// \left(\bigcup_{i=1}^{n+1} A_i\right)^c = \bigcap_{i=1}^{n+1} A_i^c.
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// \]
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//
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// Notice that
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// \[
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// \bigcup_{i=1}^{n+1} A_i = \left(\bigcup_{i=1}^{n} A_i\right) \cup A_{n+1}.
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// \]
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// Taking the complement of both sides, and using De Morgan's Law for two sets, we obtain:
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// \[
|
||
// \left(\bigcup_{i=1}^{n+1} A_i\right)^c = \left(\left(\bigcup_{i=1}^{n} A_i\right) \cup A_{n+1}\right)^c = \left(\bigcup_{i=1}^{n} A_i\right)^c \cap A_{n+1}^c.
|
||
// \]
|
||
// 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.
|
||
]
|