From e17badef30484bf3caf3e43a505010650047f0ef Mon Sep 17 00:00:00 2001 From: tomit4 Date: Tue, 16 Jun 2026 19:03:28 -0700 Subject: [PATCH] :construction: Setup for 5.2 --- chapter_5/exercises.md | 344 +++++++++++++++++++++++++++++++++++++ chapter_5/notes.md | 200 +++++++++++++++++++++ chapter_5/test_yourself.md | 19 ++ 3 files changed, 563 insertions(+) diff --git a/chapter_5/exercises.md b/chapter_5/exercises.md index 68b256f..35b7291 100644 --- a/chapter_5/exercises.md +++ b/chapter_5/exercises.md @@ -837,3 +837,347 @@ $$ 2301_{10} = 8FD_{16} $$ 91. Write a formal version of the algorithm you developed for exercise 87. Already done. + +--- + +**Exercise Set 5.2** + +Page 309 + +1. Use the technique illustrated at the beginning of this section to show that + the statements in (a) and (b) are true. + +a. If +$\left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right) = \dfrac{1}{5}$ +then +$\left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right) = \dfrac{1}{6}$. + +b. If +$\left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right) = \dfrac{1}{6}$ +then +$\left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right)\left(1 - \dfrac{1}{7}\right) = \dfrac{1}{7}$. + +2. For each positive integer $n$, let $P(n)$ be the formula + +$$ 1 + 3 + 5 + \dots + (2n - 1) = n^2 $$ + +a. Write $P(1)$. Is $P(1)$ true? + +b. Write $P(k)$. + +c. Write $P(k + 1)$. + +d. In a proof by mathematical induction that the formula holds for every integer +$n \geq 1$, what must be shown in the inductive step? + +3. For each positive integer $n$, let $P(n)$ be the formula + +$$ 1^2 + 2^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6} $$ + +a. Write $P(1)$. Is $P(1)$ true? + +b. Write $P(k)$. + +c. Write $P(k + 1)$. + +d. In a proof by mathematical induction that the formula holds for every integer +$n \geq 1$, what must be shown in the inductive step? + +4. For each integer $n$ with $n \geq 2$, let $P(n)$ be the formula + +$$ \sum_{i = 1}^{n - 1}{i(i + 1)} = \frac{n(n - 1)(n + 1)}{3} $$ + +a. Write $P(1)$. Is $P(1)$ true? + +b. Write $P(k)$. + +c. Write $P(k + 1)$. + +d. In a proof by mathematical induction that the formula holds for every integer +$n \geq 1$, what must be shown in the inductive step? + +5. Fill in the missing pieces in the following proof that + +$$ 1 + 3 + 5 + \dots + (2n - 1) = n^2 $$ + +for every integer $n \geq 1$. + +**Proof:** Let the property $P(n)$ be the equation + +$$ 1 + 3 + 5 + \dots + (2n - 1) = n^2 $$ + +_Show that_ $P(1)$ is true: + +To establish $P(1)$, we must show that when $1$ is substituted in place of $n$, +the left-hand side equals the right-hand side. But when $n = 1$, the left-hand +side is the sum of all the odd integers from $1$ to $2 \cdot 1 - 1$, which is +the sum of the odd integers from $1$ to $1$ and is just $1$. The right-hand side +is __ (a) __, which also equals $1$. So $P(1)$ is true. + +_Show that for every integer $k \geq 1$, if $P(k)$ is true then $P(k + 1)$ is +true:_ + +Let $k$ be any integer with $k \geq 1$. + +_[Suppose $P(k)$ is true. That is:]_ + +Suppose + +$1 + 3 + 5 \cdot + (2k - 1) =$ __ (b) __. + +_[This is the inductive hypothesis.]_ + +_[We must show that $P(k + 1)$ is true. That is:]_ + +We must show that __ \(c\) __ = __ (d) __. + +Now the left-hand side of $P(k + 1)$ is + +$$ 1 + 3 + 5 + \dots + (2(k + 1) - 1) $$ + +$$ = 1 + 3 + 5 + \dots + (2k + 1) $$ + +$$ = [1 + 3 + 5 + \dots + (2k - 1)] + (2k + 1) $$ + +the next-to-last term is $2k - 1$ because __ (e) __ + +$$ = k^2 + (2k + 1) $$ + +by __ (f) __ + +$$ = (k + 1)^2 $$ + +which is the right-hand side of $P(k + 1)$ _[as was to be shown]._ + +_[Since we have proved the basis step and the inductive step, we conclude that +the given statement is true.]_ + +_Note:_ This proof was annotated to help make its logical flow more obvious. In +standard mathematical writing, such annotation is omitted. + +Prove each statement in 6-9 using mathematical induction. Do not derive them +from Theorem 5.2.1 or Theorem 5.2.2. + +6. For every integer $n \geq 1$, + +$$ 2 + 4 + 6 + \dots + 2n = n^2 + n $$ + +7. For every integer $n \geq 1$, + +$$ 1 + 6 + 11 + 16 + \dots + (5n - 4) = \frac{n(5n - 3)}{2} $$ + +8. For every integer $n \geq 0$, + +$$ 1 + 2 + 2^2 + \dots + 2^n = 2^{n + 1} + 1 $$ + +9. For every integer $n \geq 3$, + +$$ 4^3 + 4^4 + 4^5 + \dots + 4^n = \frac{4(4^n - 16)}{3} $$ + +Prove each of the statements in 10-18 by mathematical induction. + +10. $1^2 + 2^2 + \dots + n^2 = \dfrac{n(n + 1)(2n + 1)}{6}$, for every integer + $n \geq 1$. + +11. $1^3 + 2^3 + \dots + n^3 = \left[\dfrac{n(n + 1)}{2}\right]^2$, for every + integer $n \geq 1$. + +12. $\dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{n(n + 1)} = \dfrac{n}{n + 1}$, + for every integer $n \geq 1$. + +13. $\sum_{i = 1}^{n - 1}{i(i + 1)} = \dfrac{n(n - 1)(n + 1)}{3}$, for every + integer $n \geq 2$. + +14. $\sum_{i = 1}^{n + 1}{i \cdot 2^i} = n \cdot 2^{n + 2} + 2$, for every + integer $n \geq 0$. + +15. $\sum_{i = 1}^{n}{i(i!)} = (n + 1)! - 1$, for every integer $n \geq 1$. + +16. $\left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{n^2}\right) = \dfrac{n + 1}{2n}$, + for every integer $n \geq 2$. + +17. $\prod_{i = 0}^{n}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{(2n + 2)!}$, + for every integer $n \geq 0$. + +18. $\prod_{i = 2}^{n}{\left(1 - \dfrac{1}{i}\right)} = \dfrac{1}{n}$ for every + integer $n \geq 2$. + +_Hint:_ See the discussion at the beginning of this section. + +19. (For students who have studied calculus) Use mathematical induction, the + product rule from calculus, and the facts that $\dfrac{d(x)}{dx} = 1$ and + that $x^{k + 1} = x \cdot x^k$ to prove that for every integer $n \geq 1$, + $\dfrac{d(x^n)}{dx} = nx^{n - 1}$. + +Use the formula for the sum of the first $n$ integers and/or the formula for the +sum of a geometric sequence to evaluate the sums in 20-29 or to write them in +closed form. + +20. $4 + 8 + 12 + 16 + \dots + 200$ + +21. $5 + 10 + 15 + 20 + \dots + 300$ + +22. + +a. $3 + 4 + 5+ 6 + \dots + 1000$ + +b. $3 + 4 + 5 + 6 + \dots + m$ + +23. + +a. $7 + 8 + 9 + 10 + \dots + 600$ + +b. $7 + 8 + 9 + 10 + \dots + k$ + +24. $1 + 2 + 3 + \dots + (k - 1)$, where $k$ is any integer with $k \geq 2$. + +25. + +a. $1 + 2 + 2^2 + \dots + 2^{25}$ + +b. $2 + 2^2 + 2^3 + \dots + 2^{26}$ + +c. $2 + 2^2 + 2^3 + \dots + 2^n$ + +26. $3 + 3^2 + 3^3 + \dots + 3^n$, where $n$ is any integer with $n \geq 1$. + +27. $5^3 + 5^4 + 5^5 + \dots + 5^k$, where $k$ is any integer with $k \geq 3$. + +28. $1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dots + \dfrac{1}{2^n}$, where $n$ is + any positive integer. + +29. $1 - 2 + 2^2 - 2^3 + \dots + (-1)^n2^n$, where $n$ is any positive integer. + +30. Observe that + +$$ \frac{1}{1 \cdot 3} = \frac{1}{3} $$ + +$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} = \frac{2}{5} $$ + +$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} = \frac{3}{7} $$ + +$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} = \frac{4}{9} $$ + +Guess a general formula and prove it by mathematical induction. + +31. Compute values of the product + +$$ \left(1 + \frac{1}{1}\right)\left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right) \dots \left(1 + \frac{1}{n}\right) $$ + +for small values of $n$ in order to conjecture a general formula for the +product. Prove your conjecture by mathematical induction. + +32. Observe that + +$$ 1 = 1 $$ + +$$ 1 - 4 = -(1 + 2) $$ + +$$ 1 - 4 + 9 = 1 + 2 + 3 $$ + +$$ 1 - 4 + 9 - 16 = -(1 + 2 + 3 + 4) $$ + +$$ 1 - 4 + 9 - 16 + 25 = 1 + 2 + 3 + 4 + 5 $$ + +Guess a general formula and prove it by mathematical induction. + +33. Find a formula in $n$, $a$, $m$, and $d$ for the sum + $(a + md) + (a + (m + 1)d) + (a + (m + 2)d) + \dots + (a + (m + n)d)$, where + $m$ and $n$ are integers, $n \geq 0$, and $a$ and $d$ are real numbers. + Justify your answer. + +34. Find a formula in $a$, $r$, $m$, and $n$ for the sum + +$$ ar^m + ar^{m + 1} + ar^{m + 2} + \dots + ar^{m + n} $$ + +where $m$ and $n$ are integers, $n \geq 0$, and $a$ and $r$ are real numbers. +Justify your answer. + +35. You have two parents, four grandparents, eight great-grandparents, and so + forth. + +a. If all your ancestors were distinct, what would be the total number of your +ancestors for the past 40 generations (counting your parents' generation as +number one)? (_Hint:_ Use the formula for the sum of a geometric sequence.) + +b. Assuming that each generation represents 25 years, how long is 40 +generations? + +c. The total number of people who have ever lived is approximately 10 billion, +which equals $10^{10}$ people. Compare this fact with the answer to part (a). +What can you deduce? + +Find the mistakes in the proof fragments in 36-38. + +36. + +**Theorem:** + +For any integer $n \geq 1$, + +$$ 1^2 + 2^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6} $$ + +**"Proof (by mathematical induction):** + +Certainly the theorem is true for $n = 1$ because $1^2 = 1$ and +$\dfrac{1(1 + 1)(2 \cdot 1 + 1)}{6} = 1$ . So the basis step is true. For the +inductive step, suppose that $k$ is any integer with $k \geq 1$, +$k^2 = \dfrac{k(k + 1)(2k + 1)}{6}$. We must show that +$(k + 1)^2 = \dfrac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6}$." + +37. + +**Theorem:** + +For any integer $n \geq 0$, + +$$ 1 + 2 + 2^2 + \dots + 2^n = 2^{n + 1} - 1 $$ + +**"Proof (by mathematical induction):** + +Let the property $P(n)$ be + +$$ 1 + 2 + 2^2 + \dots + 2^n = 2^{n + 1} - 1 $$ + +_Show that $P(0)$ is true:_ + +The left-hand side of $P(0)$ is $1 + 2 + 2^2 + \dots + 2^0 = 1$ and the +right-hand side is $2^{0 + 1} - 1 = 2 - 1 = 1$ also. So $P(0)$ is true." + +38. + +**Theorem:** + +For any integer $n \geq 1$, + +$$ \sum_{i = 1}^{n}{i(i!)} = (n + 1)! - 1 $$ + +**"Proof (by mathematical induction):** + +Let the property $P(n)$ be + +$$ \sum_{i = 1}^{n}{i(i!)} = (n + 1)! - 1 $$ + +_Show that $P(1)$ is true:_ + +When $n = 1$, + +$$ \sum_{i = 1}^{i}{i(i!)} = (1 + 1)! - 1 $$ + +So + +$$ 1(1!) = 2! - 1$$ + +and + +$$ 1 = 1 $$ + +Thus $P(1)$ is true." + +39. Use Theorem 5.2.1 to prove that if $m$ and $n$ are any positive integers and + $m$ is odd, then $\sum_{k = 0}^{m - 1}{(n + k)}$ is divisible by $m$. Does + the conclusion hold if $m$ is even? Justify your answer. + +40. Use Theorem 5.2.1 and the result of exercise 10 to prove that if $p$ is any + prime number with $p \geq 5$, then the sum of the squares of any $p$ + consecutive integers is divisible by $p$. diff --git a/chapter_5/notes.md b/chapter_5/notes.md index 3882628..c5e07a1 100644 --- a/chapter_5/notes.md +++ b/chapter_5/notes.md @@ -118,3 +118,203 @@ all $0$'s and $1$'s, and $a = \left(r[i - 1]r[i - 2] \dots r[2]r[1]r[0]\right)_2$.]_ **Output:** $r[0], r[1], r[2], \dots, r[i - 1]$ _[a sequence of integers]_ + +--- + +Page 300 + +**Principle of Mathematical Induction** + +Let $P(n)$ be a property that is defined for integers $n$, and let $a$ be a +fixed integer. Suppose the following two statements are true: + +1. $P(a)$ is true. + +2. For every integer $k \geq a$, if $P(k)$ is true then $P(k + 1)$ is true. + +Then the statement + +$$ \text{for every integer } n \geq a, P(a) $$ + +is true. + +--- + +Page 301 + +**Method of Proof by Mathematical Induction** + +Consider the statement of the form, "For every integer $n \geq a$, a property +$P(n)$ is true." To prove such a statement, perform the following two steps: + +**Step 1 (basis step):** + + Show that $P(a)$ is true. + +**Step 2 (inductive step):** + +Show that for every integer $k \geq a$, if $P(k)$ is true then $P(k + 1)$ is +true. To perform this step, + +**suppose** that $P(k)$ is true, where $k$ is any particular but arbitrarily +chosen integer with $k \geq a$. _[This supposition is called the **inductive +hypothesis**.]_ + +Then **show** that $P(k + 1)$ is true. + +--- + +Page 303 + +**Theorem 5.2.1 Sum of the First $n$ Integers** + +For every integer $n \geq 1$, + +$$ 1 + 2 + \dots + n = \frac{n(n + 1)}{2} $$ + +**Proof (by mathematical induction):** + +Let the property $P(n)$ be the equation + +$$ 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2} $$ + +_Show that $P(1)$ is true:_ + +To establish $P(1)$, we must show that + +$$ 1 = \frac{1(1 + 1)}{2} $$ + +But the left-hand side of this equation is $1$ and the right-hand side is + +$$ \frac{1(1 + 1)}{2} = \frac{2}{2} = 1 $$ + +also. Hence $P(1)$ is true. + +_Show that for every integer $k \geq 1$, if $P(k)$ is true then $P(k + 1)$ is +also true:_ + +_[Suppose that $P(k)$ is true for a particular but arbitrarily chosen integer +$k \geq 1$. That is:]_ + +Suppose that $k$ is any integer with $k \geq 1$ such that + +$$ 1 + 2 + 3 + \dots + k = \frac{k(k + 1)}{2} $$ + +_[We must show that $P(k + 1)$ is true. That is:]_ We must show that + +$$ 1 + 2 + 3 + \dots + (k + 1) = \frac{(k + 1)[(k + 1) + 1]}{2} $$ + +or, equivalently, that + +$$ 1 + 2 + 3 + \dots + (k + 1) = \frac{(k + 1)(k + 2)}{2} $$ + +_[We will show that the left-hand side and the right-hand side of $P(k + 1)$ are +equal to the same quantity and thus are equal to each other.]_ + +The left-hand side of $P(k + 1)$ is + +$$ 1 + 2 + 3 + \dots + (k + 1) $$ + +$$ = 1 + 2 + 3 + \dots + k + (k + 1) $$ + +$$ = \frac{k(k + 1)}{2} + (k + 1) $$ + +$$ = \frac{k(k + 1)}{2} + \frac{2(k + 1)}{2} $$ + +$$ = \frac{k^2 + k}{2} + \frac{2k + 2}{2} $$ + +$$ = \frac{k^2 + 3k + 2}{2} $$ + +And the right-hand side of $P(k + 1)$ is + +$$ \frac{(k + 1)(k + 2)}{2} = \frac{k^2 + 3k + 2}{2} $$ + +Thus the two sides of $P(k + 1)$ are equal to the same quantity and so they are +equal to each other. Therefore, the equation $P(k + 1)$ is true _[as was to be +shown]_. + +_[Since we have proved both the basis step and the inductive step, we conclude +that the theorem is true.]_ + +--- + +Page 304 + +**Definition** + +If a sum with a variable number of terms is shown to equal an expression that +does not contain an ellipsis or a summation symbol, we say that the sum is +written **in closed form.** + +--- + +Page 306 + +**Theorem 5.2.2 Sum of a Geometric Sequence** + +For any real number $r$ except $1$, and any integer $n \geq 0$, + +$$ \sum_{i = 0}^{n}{r^i} = \frac{r^{n + 1} - 1}{r - 1} $$ + +**Proof (by mathematical induction):** + +Suppose $r$ is a particular but arbitrarily chosen real number that is not equal +to $1$, and let the property $P(n)$ be the equation + +$$ \sum_{i = 0}^{n}{r^i} = \frac{r^{n + 1} - 1}{r - 1} $$ + +We must show that $P(n)$ is true for every integer $n \geq 0$. We do this by +mathematical induction on $n$. + +_Show that $P(0)$ is true:_ + +To establish $P(0)$, we must show that + +$$ \sum_{i = 0}^{0}{r^i} = \frac{r^{0 + 1} - 1}{r - 1} $$ + +The left-hand side of this equation is $r^0 = 1$ and the right-hand side is + +$$ \frac{r^{0 + 1} - 1}{r - 1} = \frac{r - 1}{r - 1} = 1 $$ + +also because $r^1 = r$ and, since $r \neq 1$, $r - 1 \neq 0$. Hence $P(0)$ is +true. + +_Show that for every integer $k \geq 0$, if $P(k)$ is true then $P(k + 1)$ is +also true:_ + +_[Suppose that $P(k)$ is true for a particular but arbitrarily chosen integer +$k \geq 0$. That is:]_ + +Let $k$ be any integer with $k \geq 0$, and suppose that + +$$ \sum_{i = 0}^{k}{r^j} = \frac{r^{k + 1} - 1}{r - 1} $$ + +_[We must show that $P(k + 1)$ is true. That is:]_ We must show that + +$$ \sum_{i = 0}^{k + 1}{r^j} = \frac{r^{(k + 1) + 1} - 1}{r - 1} $$ + +or, equivalently, that + +$$ \sum_{i = 0}^{k + 1}{r^j} = \frac{r^{k + 2} - 1}{r - 1} $$ + +_[We will show that the left-hand side of $P(k + 1)$ equals the right-hand +side.]_ + +The left-hand side of $P(k + 1)$ is + +$$ \sum_{i = 0}^{k + 1}{r^j} = \sum_{i = 0}^{k}{r^i + r^{k + 1}} $$ + +$$ = \frac{r^{k + 1} - 1}{r - 1} + r^{k + 1} $$ + +$$ = \frac{r^{k + 1} - 1}{r - 1} + \frac{r^{k + 1}(r - 1)}{r - 1} $$ + +$$ = \frac{(r^{k + 1} - 1) + r^{k + 1}(r - 1)}{r - 1} $$ + +$$ = \frac{r^{k + 1} - 1 + r^{k + 2} - r^{k + 1}}{r - 1} $$ + +$$ = \frac{r^{k + 2} - 1}{r - 1} $$ + +which is the right-hand side of $P(k + 1)$ _[as was to be shown]._ + +_[Since we have proved the basis step and the inductive step, we conclude that +the theorem is true.]_ diff --git a/chapter_5/test_yourself.md b/chapter_5/test_yourself.md index dc554ab..490a7c7 100644 --- a/chapter_5/test_yourself.md +++ b/chapter_5/test_yourself.md @@ -30,3 +30,22 @@ $$ \sum_{k = m}^{n}{a_k + cb_k} $$ _____. $$ \prod_{k = m}^{n}{a_kb_k} $$ + +--- + +**Test Yourself** + +Page 309 + +1. Mathematical induction is a method for proving that a property defined for + integers $n$ is true for all values of $n$ that are _____. + +2. Let $P(n)$ be a property defined for integers $n$ and consider constructing a + proof by mathematical induction for the statement "P(n) is true for all + $n \geq a$." + +a. In the basis step one must show _____. + +b. In the inductive step one supposes that _____ for a particular but +arbitrarily chosen value of an integer $k \geq a$. This supposition is called +the _____. One then has to show that _____.