790 lines
19 KiB
Text
790 lines
19 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(first-line-indent: 0pt, 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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#outline()
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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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#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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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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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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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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#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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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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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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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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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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#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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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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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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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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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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#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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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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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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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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\[
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\left(\bigcup_{i=1}^{n+1} A_i\right)^c = \left(\left(\bigcup_{i=1}^{n} A_i\right) \cup A_{n+1}\right)^c = \left(\bigcup_{i=1}^{n} A_i\right)^c \cap A_{n+1}^c.
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\]
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By the induction hypothesis,
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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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Thus, we have:
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\[
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\left(\bigcup_{i=1}^{n+1} A_i\right)^c = \left(\bigcap_{i=1}^{n} A_i^c\right) \cap A_{n+1}^c = \bigcap_{i=1}^{n+1} A_i^c.
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\]
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This completes the inductive step.
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By the PMI, we conclude that 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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Thus, De Morgan's Laws hold for any indexed family \(\{A_i : i \in \mathbb{N}\}\).
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`)
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== 9
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#mitext(`
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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
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\[
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\frac{n^2 - n}{2}.
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\]
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For \(n=2\), there is exactly 1 line segment joining the two points. The formula gives
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\[
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\frac{2^2 - 2}{2} = \frac{4-2}{2} = 1.
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\]
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So the statement holds for \(n=2\).
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Assume that for some \(n \ge 2\), the number of line segments joining \(n\) points is
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\[
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\frac{n^2 - n}{2}.
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\]
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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
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\[
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\frac{n^2 - n}{2} + n.
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\]
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Simplify the expression:
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\[
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\frac{n^2 - n}{2} + n = \frac{n^2 - n + 2n}{2} = \frac{n^2 + n}{2} = \frac{n(n+1)}{2}.
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\]
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Therefore, by the PMI, the number of line segments joining all pairs of \(n\) points is
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\[
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\frac{n^2 - n}{2}
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\]
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for all \(n \ge 2\).
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`)
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== 10
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#mitext(`
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The Tower of Hanoi problem with \( n \) disks can be solved in exactly \( 2^n - 1 \) moves.
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For \( n = 1 \):
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There is only one disk, and it can be moved directly from the source peg to the destination peg in one move.
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Since \( 2^1 - 1 = 1 \), the statement holds for \( n = 1 \).
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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.
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1. Move the top \( k \) disks from the source peg to the auxiliary peg.
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By the inductive hypothesis, this takes \( 2^k - 1 \) moves.
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2. Move the largest disk (the \((k+1)^\text{th}\) disk) from the source peg to the destination peg.
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This requires 1 move.
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3. Move the \( k \) disks from the auxiliary peg to the destination peg.
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Again by the inductive hypothesis, this requires \( 2^k - 1 \) moves.
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\[
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\text{Total Moves} = (2^k - 1) + 1 + (2^k - 1) = 2 \cdot 2^k - 1 = 2^{k+1} - 1.
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\]
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This completes the inductive step.
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By the PMI, the Tower of Hanoi with \( n \) disks can be solved in \( 2^n - 1 \) moves.
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`)
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== 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.
|
||
`)
|