🚧 Setup for 5.4
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@ -556,3 +556,355 @@ By inductive hypothesis, the remaining squares in each of the three quadrants
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can be completely covered by L-shaped trominoes. Thus every square in the
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$2^{k + 1} \times 2^{k + 1}$ checkerboard except the one that was removed can be
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completely covered by L-shaped trominoes _[as was to be shown]_.
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---
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Page 324
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**Principle of Strong Mathematical Induction**
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Let $P(n)$ be a property that is defined for integers $n$, and let $a$ and $b$
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be fixed integers with $a \leq b$. Suppose the following two statements are
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true:
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1. $P(a), P(a + 1), \dots$, and $P(b)$ are all true. **(basis step)**
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2. For every integer $k \geq b$, if $P(i)$ is true for each integer $i$ from $a$
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through $k$, then $P(k + 1)$ is true. **(inductive step)**
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Then the statement
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$$ \text{for every integer } n \geq a, P(n) $$
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is true. (The supposition that $P(i)$ is true for each integer $i$ from $a$
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through $k$ is called the **inductive hypothesis**. Another way to state the
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inductive hypothesis is to say that $P(a), P(a + 1), \dots, P(k)$ are all true.)
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---
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Page 325
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**Proof (by strong mathematical induction):**
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Let the property $P(n)$ be the sentence
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$n$ is divisible by a prime number.
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_Show that $P(2)$ is true:_
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To establish $P(2)$, we must show that
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$2$ is divisible by a prime number.
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But this is true because $2$ is divisible by $2$ and $2$ is a prime number.
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_Show that for every integer $k \geq 2$, if $P(i)$ is true for each integer from
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$2$ through $k$, then $P(k + 1)$ is also true:_
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Let $k$ be any integer with $k \geq 2$ and suppose that
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$i$ is divisible by a prime number for each integer $i$ from $2$ through $k$.
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We must show that
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$k + 1$ is divisible by a prime number.
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_Case 1($k + 1$ is prime):_
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In this case $k + 1$ is divisible by a prime number, namely, itself.
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_Case 2($k + 1$ is not prime):_
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In this case $k + 1 = ab$ where $a$ and $b$ are integers with $1 < a< k + 1$ and
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$1 < b < k + 1$. Thus, in particular, $2 \leq a \leq k$, and so by inductive
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hypothesis, $a$ is divisible by a prime number $p$. In addition because
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$k + 1 = ab$, we have that $k + 1$ is divisible by $a$. Hence, since $k + 1$ is
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divisible by $a$ and $a$ is divisible by $p$, by transitivity of divisibility,
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$k + 1$ is divisible by the prime number $p$.
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Therefore, regardless of whether $k + 1$ is prime or not, it is divisible by a
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prime number. _[as was to be shown.]_
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_[Since we have proved both the basis and the inductive step of the strong
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mathematical induction, we conclude that the given statement is true.]_
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---
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Page 326
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**Proof:**
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Let $s_0, s_1, s_2, \dots$ be the sequence defined by specifying that
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$s_0 = 0, s_1 = 4$, and $s_k = 6a_{k - 1} - 5a_{k - 2}$ for every integer
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$k \geq 2$, and let the property $P(n)$ be the formula
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$$ s_n = 5^n - 1 $$
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We will use strong mathematical induction to prove that for every integer
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$n \geq 0$, $P(n)$ is true.
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_Show that $P(0)$ and $P(1)$ are true:_
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To establish $P(0)$ and $P(1)$, we must show that
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$$ s_0 = 5^0 - 1 \text{ and } s_1= 5^1 - 1 $$
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But, by definition of $s_0, s_1, s_2, \dots$, we have that $s_0 = 0$ and
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$s_1 = 4$. Since $5^0 - 1 = 1 - 1 = 0$ and $5^1 - 1 = 5 - 1 = 4$, the values of
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$s_0$ and $s_1$ agree with the values given by the formula.
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_Show that for every integer $k \geq 1$, if $P(i)$ is true for each integer $i$
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from $0$ through $k$, then $P(k + 1)$ is also true:_
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Let $k$ be any integer with $k \geq 1$ and suppose that
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$$ s_i = 5^i - 1 \text{ for each integer } i \text{ with } 0 \leq i \leq k $$
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We must show that
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$$ s_{k + 1} = 5^{k + 1} - 1 $$
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But since $k \geq 1$, we have that $k + 1 \geq 2$, and so
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$$ s_{k + 1} = 6s_k - 5s_{k - 1} $$
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$$ = 6(5^k - 1) - 5(5^{k - 1} - 1) $$
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$$ = 6 \cdot 5^k - 6 - 5^k + 5 $$
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$$ = (6 - 1)5^k - 1 $$
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$$ = 5 \cdot 5^k - 1 $$
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$$ = 5^{k + 1} - 1 $$
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_[as was to be shown]._
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_[Since we have proved both the basis and the inductive step of the strong
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mathematical induction, we conclude that the given statement is true.]_
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---
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Page 328
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**Convention**
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Let us agree to say that a single number $x_1$ is a product with one factor and
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can be computed with zero multiplications.
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---
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Page 329
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**Proof (by strong mathematical induction):**
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Let the property $P(n)$ be the sentence
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If $x_1, x_2, \dots, x_n$ are $n$ numbers, then no matter how parentheses are
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inserted into their product, the number of multiplications used to compute the
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product is $n - 1$.
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_Show that $P(1)$ is true:_
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To establish $P(1)$, we must show that
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The number 9f multiplications needed to compute the product of $x_1$ is $1 - 1$.
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This is true because, by convention, $x_1$ is a product that can be computed
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with $0$ multiplications and $0 = 1 - 1$.
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_Show that for every integer $k \geq 1$, if $P(i)$ is true for each integer $i$
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from $1$ through $k$, then $P(k + 1)$ is also true:_
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Let $k$ be any integer with $k \geq 1$ and suppose that
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For each integer $i$ from $1$ through $k$, if $x_1, x_2, \dots, x_i$ are
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numbers, then no matter how parentheses are inserted into their product, the
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number of multiplications used to compute the product is $i - 1$.
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We must show that
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If $x_1, x_2, \dots, x_{k + 1}$ are $k + 1$ numbers, then no matter how
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parentheses are inserted into their product, the number of multiplications used
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to compute the product is $(k + 1) - 1 = k$.
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Consider a product of $k + 1$ factors: $x_1, x_2, \dots, x_{k + 1}$. When
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parentheses are inserted in order to compute the product, some multiplication is
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the final one and each of the two factors making up the final multiplication is
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a product of fewer than $k + 1$ factors. Let $L$ be the product of the left-hand
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factors and $R$ be the product of the right-hand factors, and suppose that $L$
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is composed of $l$ factors and $R$ is composed of $r$ factors. Then
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$l + r = k + 1$, the total number of factors in the product, and
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$$ 1 \leq l \leq k \text{ and } 1 \leq r \leq k $$
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By inductive hypothesis, evaluating $L$ takes $l - 1$ multiplications and
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evaluating $R$ takes $r - 1$ multiplications. Because one final multiplication
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is needed to evaluate $L \cdot R$, the number of multiplications needed to
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evaluate the product of all $k + 1$ factors is
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$$ (l - 1) + (r - 1) + 1 = (l + r) - 1 = (k + 1) - 1 = k $$
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_[as was to be shown]._
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_[Since we have proved both the basis and the inductive step of the strong
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mathematical induction, we conclude that the given statement is true.]_
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---
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Page 330
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**Theorem 5.4.1 Existence and Uniqueness of Binary Integer Representations**
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Given any positive integer $n$, $n$ has a unique representation in the form
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$$ n = c_r \cdot 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_2 \cdot 2^2 + c_1 \cdot 2 + c_0 $$
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where $r$ is a nonnegative integer, $c_r = 1$, and $c_j = 1$ or $0$ for each
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$j = 0, 1, 2, \dots, r - 1$.
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**Proof:**
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We give separate proofs by strong mathematical induction to show first the
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existence and second the uniqueness of the binary representation.
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_Existence (proof by strong mathematical induction):_
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Let the property $P(n)$ be the equation
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$$ n = c_r \cdot 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_2 \cdot 2^2 + c_1 \cdot 2 + c_0 $$
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where $r$ is a nonnegative integer, $c_r = 1$, and $c_j = 1$ or $0$ for each
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$j = 0, 1, 2, \dots, r - 1$.
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_Show that $P(1)$ is true:_
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Let $r = 0$ and $c_0 = 1$. Then $1 = c_r \cdot 2^r$, and so $n = 1$ can be
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written in the required form.
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_Show that for every integer $k \geq 1$, if $P(i)$ is true for each integer $i$
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from $1$ through $k$, then $P(k + 1)$ is also true:_
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Let $k$ be an integer with $k \geq 1$. Suppose that for each integer $i$ from
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$1$ through $k$,
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$$ = c_r \cdot 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_2 \cdot 2^2 + c_1 \cdot 2 + c_0 $$
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where $r$ is a nonnegative integer, $c_r = 1$, and $c_j = 1$ or $0$ for each
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$j = 0, 1, 2, \dots, r - 1$ . We must show that $k + 1$ can be written as a sum
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of powers of $2$ in the required form.
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_Case 1 ($k + 1$ is even):_
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In this case $\dfrac{(k + 1)}{2}$ is an integer, and by inductive hypothesis,
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since $1 \leq \dfrac{(k + 1)}{2} \leq k$, then
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$$ \frac{k + 1}{2} = c_r \cdot 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_2 \cdot 2^2 + c_1 \cdot 2 + c_0 $$
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where $r$ is a nonnegative integer, $c_r = 1$, and $c_j = 1$ or $0$ for each
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$j = 0, 1, 2, \dots, r - 1$. Multiplying both sides of the equation by $2$ gives
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$$ k + 1= c_r \cdot 2^{r + 1} + c_{r - 1} \cdot 2^r + \dots + c_2 \cdot 2^3 + c_1 \cdot 2^2 + c_0 \cdot 2 $$
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which is the sum of powers of $2$ of the required form.
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_Case 2 ($k + 1$ is odd):_
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In this case $\dfrac{k}{2}$ is an integer, and by inductive hypothesis, since
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$1 \leq \dfrac{k}{2} = k$, then
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$$ \frac{k}{2} = c_r \cdot 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_2 \cdot 2^2 + c_1 \cdot 2 + c_0 $$
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where $r$ is a nonnegative integer, $c_r = 1$, and $c_j = 1$ or $0$ for each
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$j = 0, 1, 2, \dots r - 1$. Multiplying both sides of the equation by $2$ and
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adding $1$ gives
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$$ k + 1 = c_r \cdot 2^{r + 1} + c_{r - 1} \cdot 2^r + \dots + c_2 \cdot 2^3 + c_1 \cdot 2^2 + c_0 \cdot 2 + 1 $$
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which is also a sum of powers of $2$ of the required form.
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The preceding arguments show that regardless 9f whether $k + 1$ is even or odd,
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$k + 1$ has a representation of the required form. _[Or, in other words,
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$P(k + 1)$ is true as was to be shown.]_
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_[Since we have proved the basis step and the inductive step of the strong
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mathematical induction, the existence half of the theorem is true.]_
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_Uniqueness:_
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To prove uniqueness, suppose that there is an integer $n$ with two different
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representations as a sum of nonnegative integer powers of $2$. Equating the two
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representations and canceling all identical terms gives
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$$ 2^r + c{r - 1} \cdot 2^{4 - 1} + \dots + c_1 \cdot 2 + c_0 = 2^s + d_{s - 1} \cdot 2^{s - 1} + \dots + d_1 \cdot 2 + d_0 $$
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where $r$ and $s$ are nonnegative integers and each $c_i$ and each $d_i$ equal
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$0$ or $1$. Without loss of generality, we may assume that $r < s$. Now by the
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formula for the sum of a geomatric sequence (Theorem 5.2.2) and because $r < s$
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(which implies that $r + 1 \leq s$),
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$$ 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_1 \cdot 2 + \c_0 \leq 2^r + 2^{r - 1} + \dots + 2 + 1 = 2^{r + 1} - 1 < 2^s $$
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Thus
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$$ 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_1 \cdot 2 + c_0 < 2^s + d_{s - 1} \cdot 2^{s - 1} + \dots + d_1 \cdot 2 + d_0 $$
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which contradicts equation (5.4.1). Hence the supposition is false, so any
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integer $n$ has only one representation as a sum of nonnegative integer powers
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9f $2$.
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---
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Page 331
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**Well-Ordering Principle for the Integers**
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Let $S$ be a set of integers containing one or more integers all of which are
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greater than some fixed integer. Then $S$ has a least element.
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---
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Page 332
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**Quotient-Remainder Theorem (Existence Part)**
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**Proof:**
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Let $s$ be the set of all nonnegative integers of the form
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$$ n - dk $$
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where $k$ is an integer. This set has at least one element. _[For if $n$ is
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nonnegative, then_
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$$ n - 0 \cdot d = n \geq 0 $$
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_and so $n - 0 \cdot d$ is in $S$. And if $n$ is negative then_
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$$ n - nd = \underbrace{n}_{< 0}\underbrace{(1 - d)}_{\leq 0 \text{ since } d \text{ is a positive integer}} \geq 0 $$
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_and so $n - nd$ is in $S$.]_
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It follows by the well-ordering principle for the integers that $S$ contains a
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least element $r$. Then, for some specific integer value of $k$, say $q$,
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$$ n - dq = r $$
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_[because every integer in $S$ can be written in this form]._ Adding $dq$ to
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both sides gives
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$$ n = dq + r $$
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Furthermore, $r < d$. _[For suppose $r \geq d$. Then_
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$$ n - d(q + 1) = n - dq - d = r - d \geq 0 $$
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_and so $n - d(q + 1)$ would be a nonnegative integer in $S$ that would be
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smaller than $r$. But $r$ is the smallest integer in $S$. This contradiction
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shows that the supposition $r \geq d$ must be false.]_
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The preceding arguments prove that there exists integers $r$ and $q$ for which
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$$ n = dq + r \text{ and } 0 \leq r < d $$
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_[as was to be shown.]_
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