diff --git a/chapter_5/exercises.md b/chapter_5/exercises.md index 5ce373f..ebecd1a 100644 --- a/chapter_5/exercises.md +++ b/chapter_5/exercises.md @@ -4830,3 +4830,344 @@ follows that $3^{k + 1} - 2$ is even, which is what we needed to show." Omitted. --- + +**Exercise Set 5.4** + +Page 333 + +1. Suppose $a_1, a_2, a_3, \dots$ is a sequence defined as follows: + +$$ a_1 = 1, a_2 = 3, a_k = a_{k - 2} + 2a_{k - 1} $$ + +for each integer $k \geq 3$. + +Prove that $a_n$ is odd for every integer $n \geq 1$. + +2. Suppose $b_1, b_2, b_3, \dots$ is a sequence defined as follows: + +$$ b_1 = 4, b_2 = 12, b_k = b_{k - 2} + b_{k - 1} $$ + +for each integer $k \geq 3$. + +Prove that $b_n$ is divisible by $4$ for every integer $n \geq 1$. + +3. Suppose that $c_0, c_1, c_2, \dots$ is a sequence defined as follows: + +$$ c_0 = 2, c_1 = 2, c_2 = 6, c_k = 3c_{k - 3} $$ + +for every integer $k \geq 3$. + +Prove that $c_n$ is even for each integer $n \geq 0$. + +4. Suppose that $d_1, d_2, d_3, \dots$ is sequence defined as follows: + +$$ d_1 = \frac{9}{10}, d_2 = \frac{10}{11}, d_k = d_{k - 1} \cdot d_{k - 2} $$ + +for every integer $k \geq 3$. + +Prove that $0 < d_n \leq 1$ for each integer $n \geq 1$. + +5. Suppose that $e_0, e_1, e_2, \dots$ is a sequence defined as follows: + +$$ e_0 = 12, e_1 = 29, e_k, = 5e_{k - 1} - 6e_{k - 2} $$ + +for each integer $k \geq 2$. + +Prove that $e_n = 5 \cdot 3^n + 7 \cdot 2^n$ for every integer $n \geq 0$. + +6. Suppose that $f_0, f_1, f_2, \dots$ is a sequence defined as follows: + +$$ f_0 = 5, f_1 = 16, f_k = 7f_{k - 1} - 10f_{k - 2} $$ + +for every integer $k \geq 2$. + +Prove that $f_n = 3 \cdot 2^n + 2 \cdot 5^n$ for each integer $n \geq 0$. + +7. Suppose that $g_1, g_2, g_3, \dots$ is a sequence defined as follows: + +$$ g_1 = 3, g_2 = 5, g_k = 3g_{k - 1} - 2g_{k - 2} $$ + +for each integer $k \geq 3$. + +Prove that $g_n = 2^n + 1$ for every integer $n \geq 1$. + +8. Suppose that $h_0, h_1, h_2, \dots$ is a sequence defined as follows: + +$$ h_0 = 1, h_1 = 2, h_2 = 3, h_k = h_{k - 1} + h_{k - 2} + h_{k - 3} $$ + +for each integer $k \geq 3$. + +a. Prove that $h_n \leq 3^n$ for every integer $n \geq 0$. + +b. Suppose that $s$ is any real number such that $s^3 \geq s^2 + s + 1$. (This +implies that $2 > s > 1.83$.) Prove that $h_n \leq s^n$ for every integer +$n \geq 2$. + +9. Define a sequence $a_1, a_2, a_3, \dots$ as follows: $a_1 = 1, a_2 = 3$, and + $a_k = a_{k - 1} + a_{k - 2}$ for every integer $k \geq 3$. (This sequence is + known as the Lucas sequence.) Use strong mathematical induction to prove that + $a_n \leq \left(\dfrac{7}{4}\right)^n$ for every integer $n \geq 1$. + +10. The introductory example solved with ordinary mathematical induction in + Section 5.3 can also be solved using strong mathematical induction. Let + $P(n)$ be "any $n$¢ can be obtained using a combination of $3$¢ and $5$¢ + coins." Use strong mathematical induction to prove that $P(n)$ is true for + every integer $n \geq 8$. + +11. You begin solving a jigsaw puzzle by finding two pieces that match and + fitting them together. Every subsequent step of the solution consists of + fitting together two blocks, each of which is made up of one or more pieces + that have previously been assembled. Use strong mathematical induction to + prove that for every integer $n \geq 1$, the number of steps required to put + together all $n$ pieces of a jigsaw puzzle is $n - 1$. + +12. The sides of a circular track contain a sequence of $n$ cans of gasoline. + For each integer $n \geq 1$, the total amount in the cans is sufficient to + enable a certain car to make one complete circuit of the track. In addition, + all the gasoline could fit into the car's gas tank at one time. Use + mathematical induction to prove that it is possible to find an initial + location for placing the car so that it will be able to traverse the entire + track by using the various amounts of gasoline in the cans that it + encounters along the way. + +13. Use strong mathematical induction to prove the existence part of the unique + factorization of integers theorem (Theorem 4.4.5). In other words, prove + that every integer greater than $1$ is either a prime number of a product of + prime numbers. + +14. Any product of two or more integers is a result of successive + multiplications of two integers at a time. For instance, here are a few of + the ways in which $a_1a_2a_3a_4$ might be computed: $(a_1a_2)(a_3a_4)$ or + $(((a_1a_2)a_3)a_4)$ or $a_1((a_2a_3)a_4)$. Use strong mathematical + induction to prove that any product of two or more odd integers is odd. + +15. Define the "sum" of one integer to be that integer, and use strong + mathematical induction to prove that for every integer $n \geq 1$, any sum + of $n$ even integers is even. + +16. Use strong mathematical induction to prove that for every integer + $n \geq 2$, if $n$ is even, then any sum of $n$ odd integers is even, and if + $n$ is odd, then any sum of $n$ odd integers is odd. + +17. Compute $4^1, 4^2, 4^3, 4^4, 4^5, 4^6, 4^7,$ and $4^8$. Make a conjecture + about the units digit of $4^n$ where $n$ is a positive integer. Use strong + mathematical induction to prove your conjecture. + +18. Compute $9^0, 9^1, 9^2, 9^3, 9^4,$ and $9^5$. Make a conjecture about the + units digit of $9^n$ where $n$ is a positive integer. Use strong + mathematical induction to prove your conjecture. + +19. Suppose that $a_1, a_2, a_3, \dots$ is a sequence defined as follows: + +$a_1 = 1$ $a_k = 2 \cdot a_{\frac{k}{2}}$ + +for every integer $k \geq 2$. + +Prove that $a_n \leq n$ for each integer $n \geq 1$. + +20. Suppose that $b_1, b_2, b_3, \dots$ is a sequence defined as follows: + +$b_1 = 0, b_2 = 3, b_k = 5 \cdot b_{\frac{k}{2}} + 6$ + +for every integer $k \geq 3$. + +Prove that $b_n$ is divisible by $3$ for each integer $n \geq 1$. + +21. Suppose that $c_1, c_2, c_3, \dots$ is a sequence defined as follows: + +$$ c_0 = 1, c_1 = 1, c_k = c_{\frac{k}{2}} + c_{\frac{k}{2}} $$ + +for every integer $k \geq 2$. + +Prove that $c_n = n$ for each integer $n \geq 1$. + +22. One version of the game NIM starts with two piles of objects such as coins, + stones, or matchsticks. In each turn a player is required to remove from one + to three objects from one of the piles. The two players take turns doing + this until both piles are empty. The loser is the first player who can't + make a move. Use strong mathematical induction to show that if both piles + contain the same number of objects at the start of the game, the player who + goes second can always win. + +23. Define a game $G$ as follows: Begin with a pile of $n$ stones and $0$ + points. In the first move split the pile into two possibly unequal + sub-piles, multiply the number of stones in one sub-pile times the number of + stones in the other sub-pile, and add the product to your score. In the + second move, split each of the newly created piles into a pair of possibly + unequal sub-piles, multiply the number of stones in each sub-pile times the + number of stones in the paired sub-pile, and add the new products to your + score. Continue by successively splitting each newly created pile of stones + that has at least two stones into a pair of sub-piles, multiplying the + number of stones in each sub-pile times the number of stones in the paired + sub-pile, and adding the new products to your score. The game $G$ ends when + no pile contains more than one stone. + +a. Play $G$ starting with $10$ stones and using the following initial moves. In +move $1$ split the pile of $10$ stones into two sub-piles with $3$ and $7$ +stones respectively, compute $3 \cdot 7 = 21$, and find that your score is $21$. +In move $2$ split the pile of $3$ stones into two sub-piles, with $1$ and $2$ +stones respectively, and split the pile of $7$ stones into two sub-piles, with +$4$ and $3$ stones respectively, compute $1 \cdot 2 = 2$ and $4 \cdot 3 = 12$, +and find that your score is $21 + 2 + 12 = 35$. In move $3$ split the pile of +$4$ stones into two sub-piles, each with $2$ stones, and split the pile of $3$ +tones into two sub-piles, with $1$ and $2$ stones respectively, and find +your new score. Continue splitting piles and computing your score until no pile +has more than one stone. Show your final score along with a record of the +numbers of stones in the piles you created with your moves. + +b. Play $G$ again starting with $10$ stones, but use a different initial move +from the one in part (a). Show your final score along with a record of the +numbers of stones in the piles you created with your moves. + +c. Show that you can use strong mathematical induction to prove that for every +integer $n \geq 1$, given the set-up of game $G$, no matter how you split the +piles in the various moves, your final score is $\dfrac{n(n - 1)}{2}$. The basis +step may look a little strange because a pile consisting of one stone cannot be +split into any sub-piles. Another way to say this is that it can only be split +into zero piles, and that gives an answer that agrees with the general formula +for the final score. + +24. Imagine a situation in which eight people, numbered consecutively 1-8, are + arranged in a circle. Starting from person #1, every second person in the + circle is eliminated. The elimination process continues until only one + person remains. In the first round the people numbered $2$, $4$, $6$, and + $8$ are eliminated, in the second round the people numbered $3$ and $7$ are + eliminated, and in the third round person #5 is eliminated, so after the + third round only person #1 remains, as shown on the next page. + +See page 336 for image. + +a. Given a set of sixteen people arranged in a circle and numbered, +consecutively 1-16, list the numbers of the people who are eliminated in each +round if every second person is eliminated and the elimination process continues +until only one person remains. Assume that the starting point is person #1. + +b. Use ordinary mathematical induction to prove that for every integer +$n \geq 1$, given any set of $2^n$ people arranged in a circle and numbered +consecutively $1$ through $2^n$, if one starts from person #1 and goes +repeatedly around the circle successively eliminating every second person, +eventually only person #1 will remain. + +c. Use the result of part (b) to prove that for any nonnegative integers $n$ and +$m$ with $2^n \leq 2^n + m < 2^{n + 1}$, if $r = 2^n + m$, then given any set of +$r$ people arranged in a circle and numbered consecutively $1$ through $r$, if +one starts from person #1 and goes repeatedly around the circle successively +eliminating every second person, eventually only person #$(2m + 1)$ will remain. + +25. Find the mistake in the following "proof" that purports to show that every + nonnegative integer power of every nonzero real number is $1$. + +"**Proof:** + +Let $r$ be any nonzero real number and let the property $P(n)$ be the equation +$r^n = 1$. + +_Show that $P(0)$ is true:_ + +$P(0)$ is true because $r^0 = 1$ by definition of zeroth power. + +_Show that for every integer $k \geq 0$, if $P(i)$ is true for each integer $i$ +from $0$ through $k$, then $P(k + 1)$ is also true:_ + +Let $k$ be any integer $k \geq 0$ and suppose that $r^i = 1$ for each integer +$i$ from $0$ through $k$. This is the inductive hypothesis. + +We must show that $r^{k + 1} = 1$. Now + +$$ r^{k + 1} = r^{k + k - (k - 1)} $$ + +because $k + k - (k - 1) = k + k - k + 1 = k + 1$ + +$$ = \frac{r^k \cdot r^k}{r^{k - 1}} $$ + +by the laws of exponents + +$$ = \frac{1 \cdot 1}{1} $$ + +by inductive hypothesis + +$$ = 1 $$ + +Thus $r^{k + 1} = 1$ _[as was to be shown]._ + +_[Since we have proved both the basis and the inductive step of the strong +mathematical induction, we conclude that the given statement is true.]"_ + +26. Use the well-ordering principle for the integers to prove Theorem 4.4.4: + Every integer greater than $1$ is divisible by a prime number. + +27. Use the well-ordering principle for the integers to prove the existence part + of the unique factorization of integers theorem. In other words, prove that + every integer greater than $1$ is either prime or a product of prime + numbers. + +28. + +a. The Archimedean property for the rational numbers states that for every +rational number $r$, there is an integer $n$ such that $n > r$. Prove this +property. + +b. Prove that given any rational number $r$, the number $-r$ is also rational. + +c. Use the results of parts (a) and (b) to prove that given any rational number +$r$, there is an integer $m$ such that $m < r$. + +29. Use the results of exercise 28 and the well-ordering principle for the + integers to show that given any rational number $r$, there is an integer $m$ + such that $m \leq r < m + 1$. + +30. Use the well-ordering principle to prove that given any integer $n \geq 1$, + there exists an odd integer $m$ and a nonnegative integer $k$ such that + $n = 2^k \cdot m$. + +31. Give examples to illustrate the proof of Theorem 5.4.1. + +32. Suppose $P(n)$ is a property such that + + 1. $P(0)$, $P(1)$, $P(2)$ are all true, + + 2. for each integer $k \geq 0$, if $P(k)$ is true, then $P(3k)$ is true. + Must it follow that $P(n)$ is true for every integer $n \geq 0$? If yes, + explain why; if no, give a counterexample. + +33. Prove that if a statement can be proved by strong mathematical induction, + then it can be proved by ordinary mathematical induction. To do this, let + $P(n)$ be a property that is defined for each integer $n$, and suppose the + following two statements are true: + + 1. $P(a), P(a + 1), P(a + 2) \dots, P(b)$. + + 2. For any integer $k \geq b$, if $P(i)$ is true for each integer $i$ from + $a$ through $k$, then $P(k + 1)$ is true. + +The principle of strong mathematical induction would allow us to conclude +immediately that $P(n)$ is true for every integer $n \geq a$. Can we reach the +same conclusion using the principle of ordinary mathematical induction? Yes! To +see this, let $Q(n)$ be the property + +$P(j)$ is true for each integer $j$ with $a \leq j \leq n$. + +Then use ordinary mathematical induction to show that $Q(n)$ is true for every +integer $n \geq b$. That is, prove: + + 1. $Q(b)$ is true. + + 2. For each integer $k \geq b$, if $Q(k)$ is true then $Q(k + 1)$ is true. + +34. It is a fact that every integer $n \geq 1$ can be written in the form + +$$ c_r \cdot 3^r + c_{r - 1} \cdot 3^{r - 1} + \dots + c_2 \cdot 3^2 + c_1 \cdot 3 + c_0 $$ + +where $c_r = 1$ or $2$ and $c_i = 0, 1,$ or $2$ for each integer +$i = 0, 1, 2, \dots, r - 1$. Sketch a proof of this fact. + +35. Use mathematical induction to prove the existence part of the + quotient-remainder theorem. In other words, use mathematical induction to + prove that given any integer $n$ and any positive integer $d$, there exists + integers $q$ and $r$ such that $n = dq + r$ and $0 \leq r < d$. + +36. Prove that if a statement can be proved using ordinary mathematical + induction, then it can be proved by the well-ordering principle. + +37. Use the principle of ordinary mathematical induction to prove the + well-ordering principle for the integers. diff --git a/chapter_5/notes.md b/chapter_5/notes.md index a6c69d4..c7b93ce 100644 --- a/chapter_5/notes.md +++ b/chapter_5/notes.md @@ -556,3 +556,355 @@ By inductive hypothesis, the remaining squares in each of the three quadrants can be completely covered by L-shaped trominoes. Thus every square in the $2^{k + 1} \times 2^{k + 1}$ checkerboard except the one that was removed can be completely covered by L-shaped trominoes _[as was to be shown]_. + +--- + +Page 324 + +**Principle of Strong Mathematical Induction** + +Let $P(n)$ be a property that is defined for integers $n$, and let $a$ and $b$ +be fixed integers with $a \leq b$. Suppose the following two statements are +true: + +1. $P(a), P(a + 1), \dots$, and $P(b)$ are all true. **(basis step)** + +2. For every integer $k \geq b$, if $P(i)$ is true for each integer $i$ from $a$ + through $k$, then $P(k + 1)$ is true. **(inductive step)** + +Then the statement + +$$ \text{for every integer } n \geq a, P(n) $$ + +is true. (The supposition that $P(i)$ is true for each integer $i$ from $a$ +through $k$ is called the **inductive hypothesis**. Another way to state the +inductive hypothesis is to say that $P(a), P(a + 1), \dots, P(k)$ are all true.) + +--- + +Page 325 + +**Proof (by strong mathematical induction):** + +Let the property $P(n)$ be the sentence + +$n$ is divisible by a prime number. + +_Show that $P(2)$ is true:_ + +To establish $P(2)$, we must show that + +$2$ is divisible by a prime number. + +But this is true because $2$ is divisible by $2$ and $2$ is a prime number. + +_Show that for every integer $k \geq 2$, if $P(i)$ is true for each integer from +$2$ through $k$, then $P(k + 1)$ is also true:_ + +Let $k$ be any integer with $k \geq 2$ and suppose that + +$i$ is divisible by a prime number for each integer $i$ from $2$ through $k$. + +We must show that + +$k + 1$ is divisible by a prime number. + +_Case 1($k + 1$ is prime):_ + +In this case $k + 1$ is divisible by a prime number, namely, itself. + +_Case 2($k + 1$ is not prime):_ + +In this case $k + 1 = ab$ where $a$ and $b$ are integers with $1 < a< k + 1$ and +$1 < b < k + 1$. Thus, in particular, $2 \leq a \leq k$, and so by inductive +hypothesis, $a$ is divisible by a prime number $p$. In addition because +$k + 1 = ab$, we have that $k + 1$ is divisible by $a$. Hence, since $k + 1$ is +divisible by $a$ and $a$ is divisible by $p$, by transitivity of divisibility, +$k + 1$ is divisible by the prime number $p$. + +Therefore, regardless of whether $k + 1$ is prime or not, it is divisible by a +prime number. _[as was to be shown.]_ + +_[Since we have proved both the basis and the inductive step of the strong +mathematical induction, we conclude that the given statement is true.]_ + +--- + +Page 326 + +**Proof:** + +Let $s_0, s_1, s_2, \dots$ be the sequence defined by specifying that +$s_0 = 0, s_1 = 4$, and $s_k = 6a_{k - 1} - 5a_{k - 2}$ for every integer +$k \geq 2$, and let the property $P(n)$ be the formula + +$$ s_n = 5^n - 1 $$ + +We will use strong mathematical induction to prove that for every integer +$n \geq 0$, $P(n)$ is true. + +_Show that $P(0)$ and $P(1)$ are true:_ + +To establish $P(0)$ and $P(1)$, we must show that + +$$ s_0 = 5^0 - 1 \text{ and } s_1= 5^1 - 1 $$ + +But, by definition of $s_0, s_1, s_2, \dots$, we have that $s_0 = 0$ and +$s_1 = 4$. Since $5^0 - 1 = 1 - 1 = 0$ and $5^1 - 1 = 5 - 1 = 4$, the values of +$s_0$ and $s_1$ agree with the values given by the formula. + +_Show that for every integer $k \geq 1$, if $P(i)$ is true for each integer $i$ +from $0$ through $k$, then $P(k + 1)$ is also true:_ + +Let $k$ be any integer with $k \geq 1$ and suppose that + +$$ s_i = 5^i - 1 \text{ for each integer } i \text{ with } 0 \leq i \leq k $$ + +We must show that + +$$ s_{k + 1} = 5^{k + 1} - 1 $$ + +But since $k \geq 1$, we have that $k + 1 \geq 2$, and so + +$$ s_{k + 1} = 6s_k - 5s_{k - 1} $$ + +$$ = 6(5^k - 1) - 5(5^{k - 1} - 1) $$ + +$$ = 6 \cdot 5^k - 6 - 5^k + 5 $$ + +$$ = (6 - 1)5^k - 1 $$ + +$$ = 5 \cdot 5^k - 1 $$ + +$$ = 5^{k + 1} - 1 $$ + +_[as was to be shown]._ + +_[Since we have proved both the basis and the inductive step of the strong +mathematical induction, we conclude that the given statement is true.]_ + +--- + +Page 328 + +**Convention** + +Let us agree to say that a single number $x_1$ is a product with one factor and +can be computed with zero multiplications. + +--- + +Page 329 + +**Proof (by strong mathematical induction):** + +Let the property $P(n)$ be the sentence + +If $x_1, x_2, \dots, x_n$ are $n$ numbers, then no matter how parentheses are +inserted into their product, the number of multiplications used to compute the +product is $n - 1$. + +_Show that $P(1)$ is true:_ + +To establish $P(1)$, we must show that + +The number 9f multiplications needed to compute the product of $x_1$ is $1 - 1$. + +This is true because, by convention, $x_1$ is a product that can be computed +with $0$ multiplications and $0 = 1 - 1$. + +_Show that for every integer $k \geq 1$, if $P(i)$ is true for each integer $i$ +from $1$ through $k$, then $P(k + 1)$ is also true:_ + +Let $k$ be any integer with $k \geq 1$ and suppose that + +For each integer $i$ from $1$ through $k$, if $x_1, x_2, \dots, x_i$ are +numbers, then no matter how parentheses are inserted into their product, the +number of multiplications used to compute the product is $i - 1$. + +We must show that + +If $x_1, x_2, \dots, x_{k + 1}$ are $k + 1$ numbers, then no matter how +parentheses are inserted into their product, the number of multiplications used +to compute the product is $(k + 1) - 1 = k$. + +Consider a product of $k + 1$ factors: $x_1, x_2, \dots, x_{k + 1}$. When +parentheses are inserted in order to compute the product, some multiplication is +the final one and each of the two factors making up the final multiplication is +a product of fewer than $k + 1$ factors. Let $L$ be the product of the left-hand +factors and $R$ be the product of the right-hand factors, and suppose that $L$ +is composed of $l$ factors and $R$ is composed of $r$ factors. Then +$l + r = k + 1$, the total number of factors in the product, and + +$$ 1 \leq l \leq k \text{ and } 1 \leq r \leq k $$ + +By inductive hypothesis, evaluating $L$ takes $l - 1$ multiplications and +evaluating $R$ takes $r - 1$ multiplications. Because one final multiplication +is needed to evaluate $L \cdot R$, the number of multiplications needed to +evaluate the product of all $k + 1$ factors is + +$$ (l - 1) + (r - 1) + 1 = (l + r) - 1 = (k + 1) - 1 = k $$ + +_[as was to be shown]._ + +_[Since we have proved both the basis and the inductive step of the strong +mathematical induction, we conclude that the given statement is true.]_ + +--- + +Page 330 + +**Theorem 5.4.1 Existence and Uniqueness of Binary Integer Representations** + +Given any positive integer $n$, $n$ has a unique representation in the form + +$$ n = c_r \cdot 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_2 \cdot 2^2 + c_1 \cdot 2 + c_0 $$ + +where $r$ is a nonnegative integer, $c_r = 1$, and $c_j = 1$ or $0$ for each +$j = 0, 1, 2, \dots, r - 1$. + +**Proof:** + +We give separate proofs by strong mathematical induction to show first the +existence and second the uniqueness of the binary representation. + +_Existence (proof by strong mathematical induction):_ + +Let the property $P(n)$ be the equation + +$$ n = c_r \cdot 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_2 \cdot 2^2 + c_1 \cdot 2 + c_0 $$ + +where $r$ is a nonnegative integer, $c_r = 1$, and $c_j = 1$ or $0$ for each +$j = 0, 1, 2, \dots, r - 1$. + +_Show that $P(1)$ is true:_ + +Let $r = 0$ and $c_0 = 1$. Then $1 = c_r \cdot 2^r$, and so $n = 1$ can be +written in the required form. + +_Show that for every integer $k \geq 1$, if $P(i)$ is true for each integer $i$ +from $1$ through $k$, then $P(k + 1)$ is also true:_ + +Let $k$ be an integer with $k \geq 1$. Suppose that for each integer $i$ from +$1$ through $k$, + +$$ = c_r \cdot 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_2 \cdot 2^2 + c_1 \cdot 2 + c_0 $$ + +where $r$ is a nonnegative integer, $c_r = 1$, and $c_j = 1$ or $0$ for each +$j = 0, 1, 2, \dots, r - 1$ . We must show that $k + 1$ can be written as a sum +of powers of $2$ in the required form. + +_Case 1 ($k + 1$ is even):_ + +In this case $\dfrac{(k + 1)}{2}$ is an integer, and by inductive hypothesis, +since $1 \leq \dfrac{(k + 1)}{2} \leq k$, then + +$$ \frac{k + 1}{2} = c_r \cdot 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_2 \cdot 2^2 + c_1 \cdot 2 + c_0 $$ + +where $r$ is a nonnegative integer, $c_r = 1$, and $c_j = 1$ or $0$ for each +$j = 0, 1, 2, \dots, r - 1$. Multiplying both sides of the equation by $2$ gives + +$$ k + 1= c_r \cdot 2^{r + 1} + c_{r - 1} \cdot 2^r + \dots + c_2 \cdot 2^3 + c_1 \cdot 2^2 + c_0 \cdot 2 $$ + +which is the sum of powers of $2$ of the required form. + +_Case 2 ($k + 1$ is odd):_ + +In this case $\dfrac{k}{2}$ is an integer, and by inductive hypothesis, since +$1 \leq \dfrac{k}{2} = k$, then + +$$ \frac{k}{2} = c_r \cdot 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_2 \cdot 2^2 + c_1 \cdot 2 + c_0 $$ + +where $r$ is a nonnegative integer, $c_r = 1$, and $c_j = 1$ or $0$ for each +$j = 0, 1, 2, \dots r - 1$. Multiplying both sides of the equation by $2$ and +adding $1$ gives + +$$ k + 1 = c_r \cdot 2^{r + 1} + c_{r - 1} \cdot 2^r + \dots + c_2 \cdot 2^3 + c_1 \cdot 2^2 + c_0 \cdot 2 + 1 $$ + +which is also a sum of powers of $2$ of the required form. + +The preceding arguments show that regardless 9f whether $k + 1$ is even or odd, +$k + 1$ has a representation of the required form. _[Or, in other words, +$P(k + 1)$ is true as was to be shown.]_ + +_[Since we have proved the basis step and the inductive step of the strong +mathematical induction, the existence half of the theorem is true.]_ + +_Uniqueness:_ + +To prove uniqueness, suppose that there is an integer $n$ with two different +representations as a sum of nonnegative integer powers of $2$. Equating the two +representations and canceling all identical terms gives + +$$ 2^r + c{r - 1} \cdot 2^{4 - 1} + \dots + c_1 \cdot 2 + c_0 = 2^s + d_{s - 1} \cdot 2^{s - 1} + \dots + d_1 \cdot 2 + d_0 $$ + +where $r$ and $s$ are nonnegative integers and each $c_i$ and each $d_i$ equal +$0$ or $1$. Without loss of generality, we may assume that $r < s$. Now by the +formula for the sum of a geomatric sequence (Theorem 5.2.2) and because $r < s$ +(which implies that $r + 1 \leq s$), + +$$ 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_1 \cdot 2 + \c_0 \leq 2^r + 2^{r - 1} + \dots + 2 + 1 = 2^{r + 1} - 1 < 2^s $$ + +Thus + +$$ 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_1 \cdot 2 + c_0 < 2^s + d_{s - 1} \cdot 2^{s - 1} + \dots + d_1 \cdot 2 + d_0 $$ + +which contradicts equation (5.4.1). Hence the supposition is false, so any +integer $n$ has only one representation as a sum of nonnegative integer powers +9f $2$. + +--- + +Page 331 + +**Well-Ordering Principle for the Integers** + +Let $S$ be a set of integers containing one or more integers all of which are +greater than some fixed integer. Then $S$ has a least element. + +--- + +Page 332 + +**Quotient-Remainder Theorem (Existence Part)** + +**Proof:** + +Let $s$ be the set of all nonnegative integers of the form + +$$ n - dk $$ + +where $k$ is an integer. This set has at least one element. _[For if $n$ is +nonnegative, then_ + +$$ n - 0 \cdot d = n \geq 0 $$ + +_and so $n - 0 \cdot d$ is in $S$. And if $n$ is negative then_ + +$$ n - nd = \underbrace{n}_{< 0}\underbrace{(1 - d)}_{\leq 0 \text{ since } d \text{ is a positive integer}} \geq 0 $$ + +_and so $n - nd$ is in $S$.]_ + +It follows by the well-ordering principle for the integers that $S$ contains a +least element $r$. Then, for some specific integer value of $k$, say $q$, + +$$ n - dq = r $$ + +_[because every integer in $S$ can be written in this form]._ Adding $dq$ to +both sides gives + +$$ n = dq + r $$ + +Furthermore, $r < d$. _[For suppose $r \geq d$. Then_ + +$$ n - d(q + 1) = n - dq - d = r - d \geq 0 $$ + +_and so $n - d(q + 1)$ would be a nonnegative integer in $S$ that would be +smaller than $r$. But $r$ is the smallest integer in $S$. This contradiction +shows that the supposition $r \geq d$ must be false.]_ + +The preceding arguments prove that there exists integers $r$ and $q$ for which + +$$ n = dq + r \text{ and } 0 \leq r < d $$ + +_[as was to be shown.]_ diff --git a/chapter_5/test_yourself.md b/chapter_5/test_yourself.md index 414881e..72afe70 100644 --- a/chapter_5/test_yourself.md +++ b/chapter_5/test_yourself.md @@ -58,6 +58,8 @@ $P(k)$ is true; inductive hypothesis; $P(k + 1)$ is true. --- +**Test Yourself** + Page 320 1. Mathematical induction differs from the kind of induction used in the natural @@ -69,3 +71,22 @@ deductive using inductive reasoning. prove + +--- + +**Test Yourself** + +Page 333 + +1. In a proof by strong mathematical induction the basis step may require + checking a property $P(n)$ for more _____ value of $n$. + +2. Suppose that in the basis step for a proof by strong mathematical induction + the property $P(n)$ was checked for every integer $n$ from $a$ through $b$. + Then in the inductive step one assumes that for any integer $k \geq b$, the + property $P(n)$ is true for all values of $i$ from _____ through _____ and + one shows that _____ is true. + +3. According to the well-ordering principle for the integers, if a set $S$ of + integers contains at least _____ and if there is some integer that is less + than or equal to every _____, then _____.