From 159f080fea5451a1aef3d83962314b49e18709e4 Mon Sep 17 00:00:00 2001 From: tomit4 Date: Mon, 15 Jun 2026 15:22:05 -0700 Subject: [PATCH] :construction: Fin chapter 4 --- chapter_4/exercises.md | 408 ++++++++++++++++++++++++++++++++++++++++- chapter_4/notes.md | 2 +- leftoff.txt | 2 +- 3 files changed, 401 insertions(+), 11 deletions(-) diff --git a/chapter_4/exercises.md b/chapter_4/exercises.md index 42d57b7..9c7c487 100644 --- a/chapter_4/exercises.md +++ b/chapter_4/exercises.md @@ -9396,23 +9396,92 @@ _relatively_ prime. After execution, $\text{gcd} = 1$, and because of this, $34,391$ and $6,728$ are _relatively_ prime. -RESUME HERE - 22. Prove that for all positive integers $a$ and $b$, $a \mid b$ if, and only if, $\text{gcd}(a, b) = a$. (Note that to prove "$A$ if, and only if, $B$," you need to prove "if $A$ then $B$" and "if $B$ then $A$.") +**Proof:** + +Suppose $a$ and $b$ are any positive integers such that $a \mid b$. + +By the definition of divisibility, any number is divisible by at least $1$ and +itself. It follows then that $a \mid a$. This makes $a$ a common divisor of $a$ +and $b$. Thus, by definition of the common divisor, $a$ is less than or equal to +the greatest common divisor of $a$ and $b$, $a \leq \text{gcd}(a, b)$. + +Similarly, since $\text{gcd}(a, b)$ is the greatest common divisor of $a$ and +$b$, it must also divide $a$, $\text{gcd}(a, b) \mid a$. Thus, by Theorem 4.4.1, +$\text{gcd}(a, b) \leq a$. + +We know that when $a \leq \text{gcd}(a, b)$ and $\text{gcd}(a, b) \leq a$ are +both true, this means that $\text{gcd}(a, b) = a$. + +Therefore $\text{gcd}(a, b) = a$. _[as was to be shown.]_ + +Then, suppose $a$ and $b$ are any positive integers such that their greatest +common divisor is $\text{gcd}(a, b) = a$. + +By definition of a common divisor, $a \mid \text{gcd}(a, b)$ and +$\text{gcd}(a, b) \mid b$. + +Therefore $a \mid b$. _[as was to be shown.]_ + +Q.E.D. + 23. a. Prove that if $a$ and $b$ are integers, not both zero, and $d = \text{gcd}(a, b)$, then $\dfrac{a}{d}$ and $\dfrac{b}{d}$ are integers with no common divisor that is greater than $1$. +**Proof by contradiction:** + +Suppose not. That is, suppose that $a$ and $b$ are any nonzero integers and +there is some positive integer $d$, where $d = \text{gcd}(a, b)$ and +$\dfrac{a}{d}$ and $\dfrac{b}{d}$ are integers with a common divisor that is +greater than $1$. + +Since $d = \text{gcd}(a, b)$. By the definition of a common divisor, $d \mid a$ +and $d \mid b$. + +Since $d \mid a$ and $d \mid b$, by the definition of divisibility, $a = dk$ and +$b = dm$ for some integers $k$ and $m$. By algebra: + +$$ a = dk \to \frac{a}{d} = k $$ + +$$ b = dm \to \frac{b}{d} = m $$ + +Since $\dfrac{a}{d}$ and $\dfrac{b}{d}$ are integers with a common divisor that +is greater than $1$. This means that $k$ and $m$ have some common divisor, $c$ +such that $c > 1$. Because $c > 1$, this means that $dc > d$. It follows that +$dc$ would then be a common divisor of $a$ and $b$ where $dc > d$, which means +that $d$ is not the greatest common divisor of $a$ and $b$, +$d \neq \text{gcd}(a, b)$. + +This is a contradiction. + +Q.E.D. + b. Write an algorithm that accepts the numerator and denominator of a fraction as input and produces as output the numerator and denominator of that fraction written in lowest terms. (The algorithm may call upon the Euclidean algorithm as needed.) +**Algorithm** + +_[Given a rational number $\dfrac{N}{D}$, output both $N$ and $D$ in lowest +terms.]_ + +_[Note that $EA()$ here represents calling the Euclidean Algorithm]_ + +**Input:** $\dfrac{N}{D}$ _[a rational number.]_ + +**Algorithm Body:** + +$d := EA(N, D)\\ n := \dfrac{N}{d}\\ m := \dfrac{D}{d}$ + +**Output:** $n$ and $m$. + 24. Complete the proof of Lemma 4.10.2 by proving the following: If $a$ and $b$ are any integers with $b \neq 0$ and $q$ and $r$ are any integers such that @@ -9422,20 +9491,114 @@ then $$ \text{gcd}(b, r) \leq \text{gcd}(a, b) $$ +**Continuation of Lemma 4.10.2:** + +2. $\text{\textbf{gcd}}(b, r) \leq \text{\textbf{gcd}}(a, b)$: + +a. _[We will first show that any common divisor of $b$ and $r$ is also a common +divisor of $a$ and $b$.]_ + +Let $b$ and $r$ be integers where $b \neq 0$, and let $c$ be a common divisor of +$b$ and $r$. Then $c \mid b$ and $c \mid r$ and so, by definition of +divisibility, $b = nc$ and $r = mc$, for some integers $n$ and $m$. Substitute +into the equation + +$$ a = bq + r $$ + +to obtain + +$$ a = (nc)q + mc $$ + +$$ a = (nq + m)c $$ + +Now, $nq + m$ is an integer, and so, by definition of divisibility, $c \mid a$. +Because we already know that $c \mid b$, we can conclude that $c$ is a common +divisor of $a$ and $b$ _[as was to be shown]._ + +b. _[Next, we show that $\text{gcd}(b, r) \leq \text{gcd}(a, b)$.]_ + +Now the greatest common divisor of $b$ and $r$ is defined because $b$ and $r$ +are not both zero. Also, by part (a), every common divisor of $b$ and $r$ is a +common divisor of $a$ and $b$, and so the greatest common divisor of $b$ and $r$ +is a common divisor of $a$ and $b$. But then $\text{gcd}(b, r)$ (being one of +the common divisors of $a$ and $b$) is less than or equal to the greatest common +divisor of $a$ and $b$: + +$$ \text{gcd}(b, r) \leq \text{gcd}(a, b) $$ + 25. a. Prove: If $a$ and $d$ are positive integers and $q$ and $r$ are integers such -that $a = dq + r$ and $0 < r < d$, then +that $a = dq + r$ and $0 \leq r < d$, then $$ -a = d(-(q + 1)) + (d - r) $$ and -$$ 0 < d - r < d $$ +$$ 0 < d - r \leq d $$ + +**Proof:** + +Suppose $a$ and $d$ are positive integers, and that $q$ and $r$ are integers +such that $a = dq + r$ and $0 \leq r < d$. + +By algebra: + +$$ a = dq + r $$ + +$$ -a = -(dq + r) $$ + +$$ -a = -(dq + d - d + r) $$ + +$$ -a = -(d(q + 1) - d + r) $$ + +$$ -a = -d(q + 1) + d - r $$ + +$$ -a = d(-(q + 1)) + (d - r) $$ + +Then, also by algebra: + +$$ 0 \leq r < d $$ + +$$ 0 \geq -r > -d $$ + +Then add $d$ to all sides: + +$$ 0 + d \geq -r + d > -d + d $$ + +$$ d \geq -r + d > 0 $$ + +Rearranged: + +$$ 0 < d - r \leq d $$ + +Therefore $-a = d(-(q + 1)) + (d - r)$ and $0 < d - r \leq d$. + +Q.E.D. b. Indicate how to modify Algorithm 4.10.1 to allow for the input $a$ to be negative. +**Algorithm 4.10.1 Division Algorithm** + +_[Given an integer $a$ and a positive integer $d$, the aim of the algorithm is +to find integers $q$ and $r$ that satisfy the conditions $a = dq + r$ and +$0 \leq r < d$. This is done by subtracting $d$ repeatedly from $a$ until the +result is less than $d$ but is nonnegative._ + +$$ 0 \leq a - d - d - d - \dots - d = a - dq < d$$ + +_The total number of $d$'s that are subtracted is the quotient $q$. The quantity +$a - dq$ equals the remainder $r$.]_ + +**Input:** _$a$ [an integer], $d$ [a positive integer]_ + +**Algorithm Body:** + +$\text{\textbf{if }}(a \geq 0) \text{\textbf{then}}\\ \ \ r := a, q := 0\\ \ \ \text{\textbf{while }}(r \geq d)\\ \ \ \ \ r := r - d\\ \ \ \ \ q := q + 1\\ \ \ \text{\textbf{end while}}\\ \text{\textbf{else}}\\ \ \ r := -a, q := 0\\ \ \ \text{\textbf{while }}(r \geq d)\\ \ \ \ \ r := r - d\\ \ \ \ \ q := q + 1\\ \ \ \text{\textbf{end while}}\\ \ \ \text{\textbf{if }}(r == 0) \text{\textbf{ then}}\\ \ \ \ \ q := -q\\ \ \ \text{\textbf{else}}\\ \ \ \ \ q := -(q + 1)\\ \ \ \ \ r := d - r\\ \ \ \text{\textbf{end if}}\\ \text{\textbf{end if}}$ + +**Output:** $q$, $r$ + 26. a. Prove that if $a$, $d$, $q$, and $r$ are integers such that $a = dq + r$ and @@ -9443,13 +9606,71 @@ $0 \leq r < d$, then $$ q = \left\lfloor \frac{a}{d} \right\rfloor \quad \text{ and } r = a - \left\lfloor \frac{a}{d} \right\rfloor \cdot d$$ -b. In a computer language with a built-in floor function, $\text{div}$ and -$\mod$ can be calculated as follows: +**Proof:** -$$ a \text{div } d = \left\lfoor \frac{a}{d} \right\rfloor \quad \text{ and } \quad a \mod d = a - \left\lfloor \frac{a}{d} \right\rfloor \cdot d $$ +Suppose $a$, $d$, $q$, and $r$ are any integers such that $a = dq + r$ and +$0 \leq r < d$. + +By algebra: + +$$ a = dq + r $$ + +$$ r = a - dq $$ + +Then by substitution: + +$$ 0 \leq r < d $$ + +$$ 0 \leq a - dq < d $$ + +$$ dq \leq a < dq + d $$ + +$$ dq \leq a < d(q + 1) $$ + +$$ q \leq \frac{a}{d} < q + 1 $$ + +Thus, by the definition of floor, $q = \left\lfloor \dfrac{a}{d} \right\rfloor$. + +Then, by substitution: + +$$ a = dq + r $$ + +$$ a = d\left\lfloor \frac{a}{d} \right\rfloor + r $$ + +$$ r = a - d\left\lfloor \frac{a}{d} \right\rfloor $$ + +$$ r = a - \left\lfloor \frac{a}{d} \right\rfloor \cdot d $$ + +Therefore $q = \left\lfloor \dfrac{a}{d} \right\rfloor$ and +$r = a - \left\lfloor \dfrac{a}{d} \right\rfloor \cdot d$. + +Q.E.D. + +b. In a computer language with a built-in floor function, $\text{div}$ and +$\text{mod}$ can be calculated as follows: + +$$ a \text{ div } d = \left\lfloor \frac{a}{d} \right\rfloor \quad \text{ and } \quad a \mod d = a - \left\lfloor \frac{a}{d} \right\rfloor \cdot d $$ Rewrite the steps of Algorithm 4.10.2 for a computer language with a built-in -floor function but without $\text{div}$ and $\mod$. +floor function but without $\text{div}$ and $\text{mod}$. + +**Algorithm Body:** + +$a := A, b := B, r:= B$ + +_[If $b \neq 0$, compute $a \mod b$, the remainder of the integer division of +$a$ by $b$, and set $r$ equal to this value. Then repeat the process using $b$ +in place of $a$ and $r$ in place of $b$.]_ + +$\text{\textbf{while }} (b \neq 0)\\ \ \ \ \ r := a - \left\lfloor \dfrac{a}{b} \right\rfloor \cdot b$ + +_[The value of $a \mod b$ can be obtained by calling the division algorithm.]_ + +$\ \ \ \ a := b\\ \ \ \ \ b := r\\ \text{\textbf{end while}}$ + +_[After execution of the $\text{\textbf{while}}$ loop, $\text{gcd}(A, B) = a$.]_ + +$\text{gcd} := a$ 27. An alternative to the Euclidean algorithm uses subtraction rather than division to compute greatest common divisors. (After all, division is @@ -9494,10 +9715,78 @@ $\text{gcd} = \text{gcd}(A, B)$.]_ a. Prove Lemma 4.10.3. -b. Trace the execution of Algorithm 4.10.3 for $A = 360$ and $B = 336$. +**Proof:** + +Suppose $a$ and $b$ are any integers such that $a \geq b > 0$. + +Let $c$ be some integer such that $c$ is common divisor of both $a$ and $b$ such +that $c \mid a$ and $c \mid b$. + +Since $c \mid a$ and $c \mid b$, $a = nc$ and $b = mc$ for some integers $n$ and +$m$. + +By substitution: + +$$ a - b = nc - mc $$ + +$$ a - b = c(n - m) $$ + +Now, $n - m$ is an integer by the difference of integers, and so by the +definition of divisibility, $c \mid (a - b)$. + +Thus, it follows that $\text{gcd}(a, b) \leq \text{gcd}(b, a - b)$. + +Let $k$ be some integer such that $k$ is common divisor of both $a - b$ and $b$ +such that $k \mid (a - b)$ and $k \mid b$. + +Since $k \mid (a - b)$ and $k \mid b$, $a - b = nk$ and $b = mk$ for some +integers $n$ and $m$. + +By substitution: + +$$ a - b = nk $$ + +$$ a = nk + b $$ + +$$ a = nk + mk $$ + +$$ a = k(n + m) $$ + +Since $m + n$ is an integer by the sum of integers, the definition of +divisibility tells us that $k \mid a$. + +Since we already know that $k \mid b$, it follows that $k$ is a common divisor +of both $a$ and $b$. + +Thus it follows that $\text{gcd}(a, b) \geq \text{gcd}(b, a - b)$. + +It has now been established that $\text{gcd}(a, b) \leq \text{gcd}(b, a - b)$ +and also that $\text{gcd}(a, b) \geq \text{gcd}(b, a- b)$. + +Therefore $\text{gcd}(a, b) = \text{gcd}(b, a - b)$ + +Q.E.D. + +b. Trace the execution of Algorithm 4.10.3 for $A = 630$ and $B = 336$. + +| | | | | | | | | | | | +| ------------ | --- | --- | --- | --- | --- | --- | --- | -- | -- | -- | +| $A$ | 630 | | | | | | | | | | +| $B$ | 336 | | | | | | | | | | +| $a$ | 630 | 294 | 294 | 252 | 210 | 168 | 126 | 84 | 42 | 0 | +| $b$ | 336 | 336 | 42 | 42 | 42 | 42 | 42 | 42 | 42 | 42 | +| $\text{gcd}$ | | | | | | | | | | 42 | c. Trace the execution of Algorithm 4.10.3 for $A = 768$ and $B = 348$. +| | | | | | | | | | | | | | | +| ------------ | --- | --- | --- | --- | --- | --- | -- | -- | -- | -- | -- | -- | -- | +| $A$ | 768 | | | | | | | | | | | | | +| $B$ | 348 | | | | | | | | | | | | | +| $a$ | 768 | 420 | 72 | 72 | 72 | 72 | 72 | 12 | 12 | 12 | 12 | 12 | 0 | +| $b$ | 348 | 348 | 348 | 276 | 204 | 132 | 60 | 60 | 48 | 36 | 24 | 12 | 12 | +| $\text{gcd}$ | | | | | | | | | | | | | 12 | + Exercises 28-32 refer to the following definition. **Definition:** The **least common multiple** of two nonzero integers $a$ and @@ -9511,18 +9800,119 @@ b. for all positive integers $m$, if $a \mid m$ and $b \mid m$, then $c \leq m$. a. $\text{lcm}(12, 18)$ +$\text{lcm}(12, 18) = 36$ + b. $\text{lcm}(2^2 \cdot 3 \cdot 5, 2^3 \cdot 3^2)$ +$$ \text{lcm}(12, 18) = 2^3 \cdot 3^2 \cdot 5 = 360 $$ + c. $\text{lcm}(2800, 6125)$ +$$ 2800 = 100 \cdot 28 = 10^2 \cdot 7 \cdot 4 = 2^2 \cdot 5^2 \cdot 7 \cdot 2^2 = 2^4 \cdot 5^2 \cdot 7 $$ + +$$ 6125 = 25 \cdot 245 = 5^2 \cdot 5 \cdot 49 = 5^3 \cdot 7^2 $$ + +$$ 2800 = 2^4 \cdot 5^2 \cdot 7 $$ + +$$ 6125 = 5^3 \cdot 7^2 $$ + +$$ \text{lcm}(2800, 6125) = 2^4 \cdot 5^3 \cdot 7^2 = 98000 $$ + 29. Prove that for all positive integers $a$ and $b$, $\text{gcd}(a, b) = \text{lcm}(a, b)$ if, and only if, $a = b$. +**Proof:** + +Suppose $a$ and $b$ are any positive integers such that +$\text{gcd}(a, b) = \text{lcm}(a, b)$. + +Let $k$ be some integer such that $k = \text{gcd}(a, b) = \text{lcm}(a, b)$. + +Since $k = \text{gcd}(a, b)$, $k \mid \text{gcd}(a, b)$, and thus $k \leq a$ and +$k \leq b$. + +And since $k = \text{lcm}(a, b)$, $\text{lcm}(a, b) \mid k$ and thus $a \leq k$ +and $b \leq k$. + +Since $k \leq a$ and $a \leq k$, this means that $k = a$. + +And since $k \leq b$ and $b \leq k$, this means that $k = b$. + +By the law of transitivity, $a = k = b$. + +Thus $a = b$. + +Now, suppose $a$ and $b$ are any positive integers such that $a = b$. + +By the definition of greatest common divisor, $\text{gcd}(a, a) = a$. + +By the definition of least common multiple, $\text{lcm}(a, a) = a$. + +This means that $\text{gcd}(a, a) = \text{lcm}(a, a)$. + +Since $a = b$, $b$ can be substituted for $a$: + +$$ \text{gcd}(a, b) = \text{lcm}(a, b) $$ + +Thus $\text{gcd}(a, b) = \text{lcm}(a, b)$ + +Therefore it has been shown that if $\text{gcd}(a, b) = \text{lcm}(a, b)$, then +$a = b$, and it has also been shown that if $a = b$, then +$\text{gcd}(a, b) = \text{lcm}(a, b)$. + +Q.E.D. + 30. Prove that for all positive integers $a$ and $b$, $a \mid b$ if, and only if, $\text{lcm}(a, b) = b$. +**Proof:** + +Suppose $a$ and $b$ are any positive integers such that $a \mid b$. + +By the definition of a multiple, $b \mid b$ means that $b$ is a multiple of $b$. + +This means that since $a \mid b$ and $b \mid b$, that $b$ is a common multiple +of both $a$ and $b$. + +Let $m$ be some positive integer such that $b \mid m$ and $a \mid m$. Since $b$ +and $m$ are positive integers, and since $b \mid m$, this means that $b \leq m$. + +Thus $\text{lcm}(a, b) = b$. + +Then suppose $a$ and $b$ are any positive integers such that +$\text{lcm}(a, b) = b$. + +By the definition of a common multiple, we know that $a \mid \text{lcm}(a, b)$. +Since we also know that $\text{lcm}(a, b) = b$, by substitution: + +$$ a \mid b $$ + +Thus $a \mid b$. + +Therefore it has been shown that if $a \mid b$, then $\text{lcm}(a, b) = b$ and +it has also been shown that if $\text{lcm}(a, b) = b$, then $a \mid b$. + +Q.E.D. + 31. Prove that for all integers $a$ and $b$, $\text{gcd}(a, b) \mid \text{lcm}(a, b)$. +**Proof:** + +Suppose $a$ and $b$ are any integers. + +Since $a$ is an integer, this means that $\text{gcd}(a, b) \mid a$ and +$a \mid \text{lcm}(a, b)$. + +By the transitive property of divisibility, this means that: + +$$ \text{gcd}(a, b) \mid \text{lcm}(a, b) $$ + +Therefore $\text{gcd}(a, b) \mid \text{lcm}(a, b)$. + +Q.E.D. + 32. Prove that for all positive integers $a$ and $b$, $\text{gcd}(a, b) \cdot \text{lcm}(a, b) = ab$. + +Omitted. diff --git a/chapter_4/notes.md b/chapter_4/notes.md index e1d1d6e..0fc807f 100644 --- a/chapter_4/notes.md +++ b/chapter_4/notes.md @@ -1480,7 +1480,7 @@ _[If $b \neq 0$, compute $a \mod b$, the remainder of the integer division of $a$ by $b$, and set $r$ equal to this value. Then repeat the process using $b$ in place of $a$ and $r$ in place of $b$.]_ -$\text{\textbf{while }} (b \neq 0)\\ \ \ \ \ r:= a \mod b$ +$\text{\textbf{while }} (b \neq 0)\\ \ \ \ \ r := a \mod b$ _[The value of $a \mod b$ can be obtained by calling the division algorithm.]_ diff --git a/leftoff.txt b/leftoff.txt index bbb81cf..1473a88 100644 --- a/leftoff.txt +++ b/leftoff.txt @@ -1 +1 @@ -279 +281