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