🚧 "fin" 5.3
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@ -4274,21 +4274,230 @@ Q.E.D.
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$a_k = 7a_{k - 1}$ for each integer $k \geq 2$. Show that
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$a_n = 3 \cdot 7^{n - 1}$ for every integer $n \geq 1$.
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**Proof (by mathematical induction):**
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Let $P(n)$ be the statement:
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$$ a_n = 3 \cdot 7^{n - 1} $$
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_Basis Step:_
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Prove $P(1)$. That is:
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$$ a_1 = 3 \cdot 7^{1 - 1} $$
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$$ = 3 \cdot 7^{0} $$
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$$ = 3 \cdot 1 $$
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$$ = 3 $$
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Since $a_1 = 3$ is defined in the problem statement, this equality is true.
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Therefore $P(1)$ is true.
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_Inductive _Step:_
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Let $k$ be any integer such that $k \geq 1$.
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Suppose $P(k)$. That is:
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$$ a_k = 3 \cdot 7^{k - 1} $$
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This is the inductive hypothesis.
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Prove $P(k + 1)$. That is:
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$$ a_{k + 1} = 3 \cdot 7^{(k + 1) - 1} $$
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Alternatively:
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$$ a_{k + 1} = 3 \cdot 7^k $$
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By the definition of the given sequence:
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$$ a_{k + 1} = 7a_k $$
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By the inductive hypothesis:
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$$ = 7(3 \cdot 7^{k - 1}) $$
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By the laws of exponents:
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$$ = 3 \cdot 7^k $$
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And this is the right-hand side of the equality to be shown.
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Q.E.D.
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25. A sequence $b_0, b_1, b_2, \dots$ is defined by letting $b_0 = 5$ and
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$b_k = 4 + b_{k - 1}$ for each integer $k \geq 1$. Show that $b_n > 4n$ for
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every integer $n \geq 0$.
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**Proof (by mathematical induction):**
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Let $P(n)$ be the inequality:
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$$ b_n > 4n $$
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_Basis Step:_
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Prove $P(0)$. That is:
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$$ b_0 > 4(0) $$
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$$ 5 > 4(0) $$
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$$ 5 > 0 $$
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This inequality holds. Therefore $P(0)$ is true.
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_Inductive Step:_
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Let $k$ be any integer where $k \geq 0$.
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Suppose $P(k)$. That is:
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$$ b_k > 4k $$
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This is the inductive hypothesis.
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Prove $P(k + 1)$. That is:
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$$ b_{k + 1} > 4(k + 1) $$
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By the definition of the sequence:
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$$ b_{k + 1} = 4 + b_k $$
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By the inductive hypothesis:
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$$ > 4 + 4k $$
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$$ > 4(1 + k) $$
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$$ > 4(k + 1) $$
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Q.E.D.
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26. A sequence $c_0, c_1, c_2, \dots$ is defined by letting $c_0 = 3$ and
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$c_k = (c_{k - 1})^2$ for every integer $k \geq 1$. Show that $c_n = 3^{2n}$
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for each integer $n \geq 0$.
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**Proof (by mathematical induction):**
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Let $P(n)$ be the equality:
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$$ c_n = 3^{2n} $$
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_Basis Step:_
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Prove $P(0)$. That is:
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$$ c_0 = 3^{2(0)} $$
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$$ c_0 = 3^{0} $$
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$$ c_0 = 1 $$
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Stopping here. It is likely Epp has a typo, she means $c_n = 3^{2^n}$, not
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$c_n = 3^{2n}$.
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27. A sequence $d_1, d_2, d_3, \dots$ is defined by letting $d_1 = 2$ and
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$d_k = \dfrac{d_{k - 1}}{k}$ for each integer $k \geq 2$. Show that for
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every integer $n \geq 1$, $d_n = \dfrac{2}{n!}$.
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**Proof (by mathematical induction):**
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Let $P(n)$ be the equation:
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$$ d_n = \frac{2}{n!} $$
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_Basis Step:_
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Prove $P(1)$. That is:
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$$ d_1 = \frac{2}{1!} $$
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$$ d_1 = \frac{2}{1} $$
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$$ d_1 = 2 $$
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Since the problem statement states that $d_1 = 2$, this matches our equality.
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Therefore $P(1)$ is true.
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_Inductive Step:_
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Let $k$ be any integer such that $k \geq 1$.
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Suppose $P(k)$, that is:
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$$ d_k = \frac{2}{k!} $$
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This is the inductive hypothesis.
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Prove $P(k + 1)$, that is:
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$$ d_{k + 1} = \frac{2}{(k + 1)!} $$
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By the given sequence:
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$$ d_{k + 1} = \frac{d_k}{k + 1} $$
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By the inductive hypothesis:
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$$ = \frac{2}{(k + 1)k!} $$
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$$ = \frac{2}{(k + 1)!} $$
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Q.E.D.
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28. Prove that for every integer $n \geq 1$,
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$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2n - 1)}{(2n + 1)(2n + 3) + \dots + (2n + (2n - 1))} $$
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$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2n - 1)}{(2n + 1) + (2n + 3) + \dots + (2n + (2n - 1))} $$
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**Proof (by mathematical induction):**
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Let $P(n)$ be the equality:
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$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2n - 1)}{(2n + 1) + (2n + 3) + \dots + (2n + (2n - 1))} $$
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_Basis Step:_
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Prove $P(1)$, that is:
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$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2(1) - 1)}{(2(1) + 1) + (2(1) + 3) + \dots + (2(1) + (2(1) - 1))} $$
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$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2 - 1)}{(2 + 1) + (2 + 3) + \dots + (2 + (2 - 1))} $$
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$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + 1}{(2 + 1) + (2 + 3) + \dots + (2 + 1)} $$
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$$ \frac{1}{3} = \frac{1}{(2 + 1) + (2 + 3) + \dots + 3} $$
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$$ \frac{1}{3} = \frac{1}{3} $$
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Therefore $P(1)$ is true.
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_Inductive Step:_
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Let $k$ be any integer where $k \geq 1$.
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Suppose $P(k)$, that is:
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$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2k - 1)}{(2k + 1) + (2k + 3) + \dots + (2k + (2k - 1))} $$
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This is the inductive hypothesis.
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Prove $P(k + 1)$, that is:
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$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2(k + 1) - 1)}{(2(k + 1) + 1) + (2(k + 1) + 3) + \dots + (2(k + 1) + (2(k + 1) - 1))} $$
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Starting from the inductive hypothesis:
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$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2k - 1)}{(2k + 1) + (2k + 3) + \dots + (2k + (2k - 1))} $$
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Omitted.
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Exercises 29 and 30 use the definition of string and string length from page 13
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in Section 1.4. Recursive definitions for these terms are given in section 5.9.
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@ -4298,11 +4507,114 @@ in Section 1.4. Recursive definitions for these terms are given in section 5.9.
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$n \geq 1$, if a string $s$ in $L$ has a length $3n$, then $s$ contains an
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even number of _b_'s.
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30. A set $S$ consists of strings obtgained by juxtaposing one or more copies of
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**Proof (by mathematical induction):**
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Suppose a set $L$ consists of strings by juxtaposing one or more of _abb_,
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_bab_, and _bba_.
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Let $P(n)$ be the sentence:
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If a string $s$ in $L$ has length $3n$, then $s$ contains an even number of
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_b_'s.
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_Basis Step:_
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Prove $P(1)$, that is:
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If a string $s$ in $L$ has length $3(1)$, then $s$ contains an even number of
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_b_'s.
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Since:
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$$ L = \{\text{abb}, \text{bab}, \text{bba}\} $$
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All three string $s$ in $L$ have a length of $3$ and all of them have an even
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number of _b_'s.
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Therefore $P(1)$ is true.
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_Inductive Step:_
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Let $k$ be any integer where $k \geq 1$.
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Suppose $P(k)$, that is:
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If a string $s$ in $L$ has length $3k$, then $s$ contains an even number of
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_b_'s.
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This is the inductive hypothesis.
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Prove $P(k + 1)$, that is:
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If a string $s$ in $L$ has length $3(k + 1)$, then $s$ contains an even number
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of _b_'s.
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Now $3(k + 1) = 3k + 3$ and the strings in $L$ are obtained by juxtaposing
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strings already in $L$ with one of _abb_, _bab_, or _bba_. Thus, either the
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initial or the final three characters in $s$ are _abb_, _bab_, or _bba_.
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Moreoever, the other $3k$ characters in $s$ are also in $L$ by definition of
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$L$, and so, by the inductive hypothesis, the other $3k$ characters in $s$
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contain an even number, say $m$, of _b_'s. Because each of _abb_, _bab_, and
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_bba_ contains 2 _b_'s, the total number of _b_'s in $s$ is $m + 2$, which is a
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sum of even integers and hence is even.
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Q.E.D.
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30. A set $S$ consists of strings obtained by juxtaposing one or more copies of
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1110 and 0111. Use mathematical induction to prove that for every integer
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$n \geq 1$, if a string $s$ in $S$ has a length $4n$, then the number of 1's
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in $s$ is a multiple of 3.
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**Proof (by mathematical induction):**
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Suppose a set $S$ contains strings obtained by juxtaposing one or more copies of
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1110 and 0111.
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Let $P(n)$ be the sentence:
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If a string $s$ in $S$ has length $4n$, then the number of $1$'s in $s$ is a
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multiple of $3$.
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_Basis Step:_
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Prove $P(1)$, that is:
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If a string $s$ in $S$ has length $4(1)$, then the number of $1$'s in $s$ is a
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multiple of $3$.
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Since $S$ consists only of strings obtained by juxtaposing 1110 and 0111, then
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at a minimum, the strings in $S$ must have a length of $4$. This means that the
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only two strings in $S$ that have a length of $4$ are 1110 and 0111. The number
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of $1$'s in $s$ is a multiple of $3$ in both cases.
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Therefore $P(1)$ is true.
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_Inductive Step:_
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Let $k$ be any integer where $k \geq 1$.
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Suppose $P(k)$, that is:
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If a string $s$ in $S$ has length $4k$, then the number of $1$'s in $s$ is a
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multiple of $3$.
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This is the inductive hypothesis.
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Prove $P(k + 1)$, that is:
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If a string $s$ in $S$ has length $4(k + 1)$, then the number of $1$'s in $s$ is
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a multiple of $3$.
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Now $4(k + 1) = 4k + 4$ and the strings in $S$ are obtained by juxtaposing
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strings already in $S$ with one of 1110 or 0111. Thus, the number of $1$'s is a
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multiple of $3$ in both cases. Moreover, the other $4k$ digits in $s$ are also
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in $S$ by the definition of $S$, and so, by inductive hypothesis, the other $4k$
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characters in $s$ contain an odd number, say $m$ of $1$'s. Because each of 1110
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and 0111 contain 3 $1$'s, the total number of $1$'s in $s$ is $m + 1$, which is
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the sum of odd integers and hence is odd.
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Q.E.D.
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31. Use mathematical induction to give an alternative proof for the statement
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proved in Example 4.9.9:
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@ -4310,32 +4622,46 @@ For any positive integer $n$, a complete graph on $n$ vertices has
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$\dfrac{n(n - 1)}{2}$ edges. _Hint:_ Let $P(n)$ be the sentence, "the number of
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edges in a complete graph on $n$ vertices is $\dfrac{n(n - 1)}{2}$."
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Omitted.
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32. Some $5 \times 5$ checkerboards with one square removed can be completely
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covered by L-shaped trominoes, whereas other $5 \times 5$ checkerboards
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cannot. Find examples of both kinds of checkerboards. Justify your answers.
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Omitted.
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33. Consider a $4 \times 6$ checkerboard. Draw a covering of the board by
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L-shaped trominoes.
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Omitted.
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34.
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a. Use mathematical induction to prove that for each integer $n \geq 1$, any
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checkerboard with dimensions $2 \times 3n$ can be completely covered with
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L-shaped trominoes.
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Omitted.
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b. Let $n$ be any integer greater than or equal to $1$. Use the result of part
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(a) to prove by mathematical induction that for every integer $m$, any
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checkerboard with dimensions $2m \times 3n$ can be completely covered with
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L-shaped trominoes.
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Omitted.
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35. Let $m$ and $n$ be any integers that are greater than or equal to $1$.
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a. Prove that a necessary condition for an $m \times n$ checkerboard to be
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completely coverable by L-shaped trominoes is that $mn$ be divisible by $3$.
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Omitted.
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b. Prove that having $$ be divisible by $3$ is not a sufficient condition for an
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$m \times n$ checkerboard to be completely covered by L-shaped trominoes.
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Omitted.
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36. In a round-robin tournament each team plays every other team exactly once
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with ties not allowed. If the teams are labeled $T_1, T_2, \dots, T_n$, then
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the outcome of such a tournament can be represented by a directed graph, in
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@ -4357,10 +4683,14 @@ to obtain an ordering $T_1, T_2, \dots, T_k$. Consider three cases: $T'$ beats
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$T_1$, $T'$ loses to the first $m$ teams (where $1 \leq m \leq k - 1$) and beats
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the $(m + 1)$st team, and $T'$ loses to all the other teams.)
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Omitted.
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37. On the outside rim of a circular disk the integers from $1$ through $30$ are
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painted in random order. Show that no matter what this order is, there must
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be three successive integers whose sum is at least 45.
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Omitted.
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38. Suppose that $n$ _a_'s and $n$ _b_'s are distributed around the outside of a
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circle. Use mathematical induction to prove that for any integer $n \geq 1$,
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given any such arrangement, it is possible to find a starting point so that
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@ -4371,11 +4701,15 @@ the $(m + 1)$st team, and $T'$ loses to all the other teams.)
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See Page 322 for image.
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Omitted.
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39. For a polygon to be **convex** means that given any two points on or inside
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the polygon, the line joining the points lies entirely inside the polygon.
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Use mathematical induction to prove that for every integer $n \geq 3$, the
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angles of any $n$-sided convex polygon add up to $180(n - 2)$ degrees.
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Omitted.
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40.
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a. Prove that in an $8 \times 8$ checkerboard with alternating black and white
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@ -4383,11 +4717,15 @@ squares, if the squares in the top right and bottom left corners are removed the
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remaining board cannot be covered with dominoes. (_Hint:_ Mathematical induction
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is not needed for this proof.)
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Omitted.
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b. Use mathematical induction to prove that for each positive integer $n$, if a
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$2n \times 2n$ checkerboard with alternating black and white squares has one
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white square and one black square removed anywhere on the board, the remaining
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squares can be covered with dominoes.
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Omitted.
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41. A group of people are positioned so that the distance between any two people
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is different from the distance between any other two people. Suppose that
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the group contains an odd number of people and each person sends a message
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@ -4396,12 +4734,16 @@ squares can be covered with dominoes.
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inspired by the article "Odd Pie Fights" by L. Carmony, _The Mathematics
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Teacher_, **72**(1), 1979, 61-64.]
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Omitted.
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42. Show that for any integer $n$, it is possible to find a group of $n$ people
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who are all positioned so that the distance between any two people is
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different from the distance between any other two people, so that each
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person sends a message to their nearest neighbor, and so that every person
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in the group receives a message from another person in the group.
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Omitted.
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43. Define a game as follows: You begin with an urn that contains a mixture of
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white and black balls, and during the game you have access to as many
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additional white and black balls as you might need. In each move you remove
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@ -4420,13 +4762,19 @@ urn, then three balls, and then four balls. For each case keep track of the
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number of white and black balls you start with and the color of the ball at the
|
||||
end of the game.
|
||||
|
||||
Omitted.
|
||||
|
||||
b. Does the number of white balls seem to be predictive? Does the number of
|
||||
black balls seem to be predictive? Make a conjecture about the color of the ball
|
||||
at the end of the game given the numbers of white and black balls at the
|
||||
beginning.
|
||||
|
||||
Omitted.
|
||||
|
||||
c. Use mathematical induction to prove the conjecture you made in part (b).
|
||||
|
||||
Omitted.
|
||||
|
||||
44. Let $P(n)$ be the following sentence: Given any graph $G$ with $n$ vertices
|
||||
satisfying the condition that every vertex of $G$ has degree at most $M$,
|
||||
then the vertices of $G$ can be colored with at most $M + 1$ colors in such
|
||||
|
|
@ -4438,6 +4786,8 @@ must be true for $n = a$ and the argument of the inductive step must be correct
|
|||
for every integer $k \geq a$. IN 45 and 46 find the mistakes in the "proofs" by
|
||||
mathematical induction.
|
||||
|
||||
Omitted.
|
||||
|
||||
45.
|
||||
|
||||
**"Theorem:"** For any integer $n \geq 1$, all the numbers in a set of $n$
|
||||
|
|
@ -4459,6 +4809,8 @@ in $B$ equal $a_1$ and all the numbers in $C$ equal $a_1$ (since both sets have
|
|||
only $k$ numbers). But every number in $A$ is in $B$ or $C$, so all the numbers
|
||||
in $A$ equal $a_1$; hence all are equal to each other."
|
||||
|
||||
Omitted.
|
||||
|
||||
46.
|
||||
|
||||
**"Theorem:"** For every integer $n \geq 1$, $3^n - 2$ is even.
|
||||
|
|
@ -4474,3 +4826,7 @@ $$ = (3^k - 2) + 3^k \cdot 2 $$
|
|||
Now $3^k - 2$ is even by inductive hypothesis and $3^k \cdot 2$ is even by
|
||||
inspection. Hence the sum of the two quantities is even (by Theorem 4.1.1). It
|
||||
follows that $3^{k + 1} - 2$ is even, which is what we needed to show."
|
||||
|
||||
Omitted.
|
||||
|
||||
---
|
||||
|
|
|
|||
|
|
@ -1 +1 @@
|
|||
321
|
||||
324
|
||||
|
|
|
|||
Loading…
Add table
Add a link
Reference in a new issue