🚧 Setup for 8.4
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@ -459,3 +459,469 @@ $A$,
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$$ (a, b) R (c, d) \Leftrightarrow ad = bc $$
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The fact is that $R$ is an equivalence relation.
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---
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Page 549
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**Theorem 8.4.1 Modular Equivalences**
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Let $a$, $b$, and $n$ be any integers and suppose $n > 1$. The following
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statements are all equivalent:
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1. $n | (a - b)$
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2. $a \equiv b (\mod n)$
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3. $a = b + kn$ for some integer $k$
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4. $a$ and $b$ have the same (nonnegative) remainder when divided by $n$
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5. $a \mod n = b \mod n$
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**Proof:**
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We will show that
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$(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (5) \Rightarrow (1)$.
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It will follow by the transitivity of if-then that all five statements are
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equivalent.
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So let $a$, $b$, and $n$ be any integers with $n > 1$.
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_Proof that $(1) \Rightarrow (2)$:_
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Suppose that $n | (a - b)$. By definition of congruence module $n$, we can
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immediately conclude that $a \equiv b (\mod n)$.
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_Proof that $(2) \Rightarrow (3)$:_
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Suppose $a \equiv b (\mod n)$. By definition of congruence modulo $n$,
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$n | (a - b)$. Thus, by definition of divisibility, $a - b = kn$, for some
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integer $k$. Adding $b$ to both sides gives that $a = b + kn$.
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_Proof that $(3) \Rightarrow (4)$:_
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Suppose that $a = b + kn$, for some integer $k$. Use the quotient-remainder
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theorem to divide $a$ by $n$ to obtain
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$$ a = qn + r \text{ where } q \text{ and } r \text{ are integers and } 0 \leq r < n $$
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So $r$ is the remainder obtained when $a$ is divided by $n$. Substituting
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$b + kn$ for $a$ in the equation $a = qn + r$ gives that
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$$ b + kn = qn + r $$
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and subtracting $kn$ from both sides and factoring out $n$ yields
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$$ b = (q - k)n + r $$
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Now since $0 \leq r < n$, the uniqueness property of the quotient-remainder
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theorem guarantees that $r$ is also the remainder obtained when $b$ is divided
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by $n$. Thus $a$ and $b$ have the same remainder when divided by $n$.
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_Proof that $(4) \Rightarrow (5)$:_
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Suppose that $a$ and $b$ have the same remainder when divided by $n$. It follows
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immediately from the definition of the $\mod$ function that
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$a \mod n = b \mod n$.
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_Proof that $(5) \Rightarrow (1)$:_
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Suppose that $a \mod n = b \mod n$. By definition of the $\mod$ function, $a$
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and $b$ have the same remainder when divided by $n$. Thus, by the
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quotient-remainder theorem, we can write
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$$ a = q_1n + r \text{ and } b = q_2n + r \text{ where } q_1, q_2 \text{ and } r \text{ are integers and } 0 \leq r < n $$
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It follows that
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$$ a - b = (q_1n + r) - (q_2n + r) = (q_1 - q_2)n $$
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Therefore, since $q_1 - q_2$ is an integer, $n | (a - b)$.
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---
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Page 550
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**Definition**
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Given integers $a$ and $n$ with $n > 1$, **the residue of $a$ modulo $n$** is
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$a \mod n$, the nonnegative remainder obtained when $a$ is divided by $n$. The
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numbers $0, 1, 2, \dots, n - 1$ are called a **complete set of residues modulo
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$n$**. To **reduce a number modulo $n$** means to set it equal to its residue
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modulo $n$. If a modulus $n > 1$ is fixed throughout a discussion and an integer
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$a$ is given, the words "modulo $n$" are often dropped and we simply speak of
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**the residue of $a$**.
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---
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Page 550
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**Theorem 8.4.2 Congruence Modulo $n$ Is an Equivalence Relation**
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If $n$ is any integer with $n > 1$, congruence modulo $n$ is an equivalence
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relation on the set of all integers. The distinct equivalence classes of the
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relation are the sets $[0], [1], [2], \dots, [n - 1]$, where for each
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$a = 0, 1, 2, \dots, n - 1$,
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$$ [a] = \{m \in \mathbb{Z} | m \equiv a (\mod n)\} $$
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or, equivalently,
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$$ [a] = \{m \in \mathbb{Z} | m = a + kn \text{ for some integer } k\} $$
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**Proof:**
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Suppose $n$ is any integer with $n > 1$. We must show that congruence modulo $n$
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is reflexive, symmetric, and transitive.
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_Proof of reflexivity:_
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Suppose $a$ is any integer. To show that $a \equiv a (\mod n)$, we must show
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that $n | (a - a)$. Now $a - a = 0$, and $n | 0$ because $0 = n \cdot 0$.
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Therefore $a \equiv a (\mod n)$.
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_Proof of symmetry:_
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Suppose $a$ and $b$ are any integers such that $a \equiv b(\mod n)$. We must
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show that $b \equiv a (\mod n)$. Now since $a \equiv b(\mod n)$, then
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$n | (a - b)$. Thus, by definition of divisibility, $a - b = nk$, for some
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integer $k$. Multiplying both sides of this equation by $-1$ to obtain
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$$ -(a - b) = -nk $$
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or, equivalently,
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$$ b - a = n(-k) $$
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Thus, by definition of divisibility $n | (b - a)$, and so, by definition of
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congruence modulo $n$, $b \equiv a (\mod n)$.
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_Proof of transitivity:_
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This is left as exercise 5 at the end of the section.
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_Proof that the distinct equivalence classes are
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$[0], [1], [2], \dots, [n - 1]$:_
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This is left as exercise 6 at the end of the section.
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---
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Page 551
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**Theorem 8.4.3 Modular Arithmetic**
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Let $a$, $b$, $c$, $d$, and $n$ be integers with $n > 1$, and suppose
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$$ a \equiv c (\mod n) \text{ and } b \equiv d(\mod n) $$
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Then
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1. $(a + b) \equiv (c + d)(\mod n)$
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2. $(a - b) \equiv (c - d)(\mod n)$
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3. $ab \equiv cd(\mod n)$
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4. $a^m \equiv c^m(\mod n)$ for every positive integer $m$
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**Proof:**
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Because we will make greatest use of part 3 of this theorem, we prove it here
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and leave the proofs of the remaining parts of the theorem to exercises 9-11 at
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the end of the section.
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_Proof of Part 3:_Proof Suppose $a$, $b$, $c$, $d$, and $n$ are integers with
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$n > 1$, and suppose $a \equiv b(\mod n)$ and $c \equiv d(\mod n)$. By Theorem
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8.4.1, there exists integers $s$ and $t$ such that
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$$ a = c + sn \text{ and } b = d + tn $$
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Then
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$$ ab = (c + sn)(d + tn) $$
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$$ = cd + ctn + snd + sntn $$
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$$ = cd + n(ct + sd + stn) $$
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Let $k = ct + sd + stn$. Then $k$ is an integer because it is a sum of products
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of integers, and $ab = cd + nk$. Thus by Theorem 8.4.1, $ab \equiv cd(\mod n)$.
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---
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Page 552
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**Corollary 8.4.4**
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Let $a$, $b$, and $n$ be integers with $n > 1$. Then
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$$ ab \equiv [(a \mod n)(b \mod n)](\mod n) $$
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or, equivalently,
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$$ ab \mod n = [(a \mod n)(b \mod n)]\mod n $$
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In particular, if $m$ is a positive integer, then
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$$ a^m \equiv [(a \mod n)^m](\mod n) $$
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---
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Page 555
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**Definition**
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An integer $d$ is said to be a **linear combination of integers** $a$ and $b$
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if, and only if, there exist integers $s$ and $t$ such that $as + bt = d$.
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---
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Page 555
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**Theorem 8.4.5 Writing a Greatest Common Divisor as a Linear Combination**
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For all integers $a$ and $b$, not both zero, if $d = \text{gcd}(a, b)$, then
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there exist integers $s$ and $t$ such that $as + bt = d$.
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**Proof:**
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Given integers $a$ and $b$, not both zero, and given $d = \text{gcd}(a, b)$, let
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$$ S = \{x | x \text{ is a positive integer and } x = as + bt \text{ for some integers } s \text{ and } t\} $$
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Note that $S$ is a nonempty set because (1) if $a > 0$ then
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$1 \cdot a + 0 \cdot b \in S$, (2) if $a < 0$ then
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$(-1) \cdot a + 0 \cdot b \in S$, and (3) if $a = 0$ then, by assumption,
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$b \neq 0$, and hence $0 \cdot a + 1 \cdot b \in S$ or
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$0 \cdot a + (-1) \cdot b \in S$. Thus, because $S$ is a nonempty subset of
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positive integers, by the well-ordering principle for the integers there is a
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least element $c$ in $S$. By definition of $S$,
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$$ c = as + bt \text{ for some integers } s \text{ and } t $$
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We will show that (1) $c \geq d$, and (2) $c \leq d$, and we will therefore be
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able to conclude that $c = d = \text{gcd}(a, b)$.
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_(1) Proof that $c \geq d$:_
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_[In this part of the proof, we show that $d$ is a divisor of $c$ and thus that
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$d \leq c$.]_ Because $d = \text{gcd}(a, b)$, by definition of greatest common
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divisor, $d | a$ and $d | b$. Hence $a = dx$ and $b = dy$ for some integers $x$
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and $y$. Then
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$$ c = as + bt $$
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$$ = (dx)s + (dy)t $$
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$$ = d(xs + y) $$
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Now $xs + yt$ is an integer because it is a sum of products of integers. Thus,
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by definition of divisibility, $d | c$. Both $c$ and $d$ are positive, and
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hence, by Theorem 4.4.1, $c \geq d$.
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_(2) Proof that $c \leq d$:_
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_[In this part of the proof, we show that $c$ is a divisor of both $a$ and $b$
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and therefore that $c$ is less than or equal to the greatest common divisor of
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$a$ and $b$, which is $d$.]_ Apply the quotient-remainder theorem to the
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division of $a$ by $c$ to obtain
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$$ a = cq + r \text{ for some integers } q \text{ and } r \text{ with } 0 \leq r < c $$
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Thus for some integers $q$ and $r$ with $0 \leq r < c$,
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$$ r = q - cq $$
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Now $c = as + bt$. Therefore, for some integers $q$ and $r$ with $0 \leq r < c$,
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$$ r = a - (as + bt)q $$
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$$ = a(1 - sq) - btq $$
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Thus $r$ is a linear combination of $a$ and $b$. If $r > 0$, then $r$ would be
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in $S$, and so $r$ would be a smaller element of $S$ than $c$, which would
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contradict the fact that $c$ is the least element of $S$. Hence $r = 0$. By
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substitution into (8.4.4),
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$$ a = cq $$
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and therefore $c | a$.
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An almost identical argument establishes that $c | b$ and is left as exercise 30
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at the end of the section.
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Because $c | a$ and $c | b$, $c$ is a common divisor of $a$ and $b$. Hence $c$
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is less than or equal to the greatest common divisor of $a$ and $b$. In other
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words, $c \leq d$.
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From (1) and (2), we conclude that $c = d$. It follows that $d$, the greatest
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common divisor of $a$ and $b$, is equal to $as + bt$.
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---
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Page 557
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**Definition**
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Given any integer $a$ and any positive integer $n$, if there exists an integer
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$s$ such that $as \equiv 1(\mod n)$, then $s$ is called **an inverse for $a$
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modulo $n$.**
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---
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Page 557
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**Definition**
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Integers $a$ and $b$ are **relatively prime** if, and only if,
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$\text{gcd}(a, b) = 1$. Integers $a_1, a_2, a_3, \dots, a_n$ are **pairwise
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relatively prime** if, and only if, $\text{gcd}(a_i, a_j) = 1$ for all integers
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$i$ and $j$ with $1 \leq i$, $j \leq n$, and $i \neq j$.
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---
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Page 557
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**Corollary 8.4.6**
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If $a$ and $b$ are relatively prime integers, then there exist integers $s$ and
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$t$ such that $as + bt = 1$.
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---
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Page 558
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**Corollary 8.4.7 Existence of Inverses Modulo $n$**
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For all integers $a$ and $n$, if $\text{gcd}(a, n) = 1$, then there exists an
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integer $s$ such that $as \equiv 1(\mod n)$, and so $s$ is an inverse for $a$
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modulo $n$.
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**Proof:**
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Suppose $a$ and $n$ are integers and $\text{gcd}(a, n) = 1$. By Corollary 8.4.6,
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there exist integers $s$ and $t$ such that
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$$ as + nt = 1 $$
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Subtracting $nt$ from both sides gives that
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$$ as = 1 - nt = 1 + (-t)n $$
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Thus, by definition of congruence modulo $n$,
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$$ as \equiv 1(\mod n) $$
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---
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Page 562
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**Theorem 8.4.8 Euclid's Lemma**
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For all integers $a$, $b$, and $c$, if $\text{gcd}(a, c) = 1$ and $a | bc$, then
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$a | b$.
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**Proof:**
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Suppose $a$, $b$, and $c$ are integers, $\text{gcd}(a, c) = 1$, and $a | bc$.
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_[We must show that $a | b$.]_ By Theorem 8.4.5, there exist integers $s$ and
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$t$ so that
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$$ as + ct = 1 $$
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Multiply both sides of this equation by $b$ to obtain
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$$ bas + bct = b $$
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Since $a | bc$, by definition of divisibility there exists an integer $k$ such
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that
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$$ bc = ak $$
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Substituting (8.4.8) into (8.4.7), rewriting, and factoring out an $a$ gives
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that
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$$ b = bas + (ak)t = a(bs + kt) $$
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Let $r = bs + kt$. Then $r$ is an integer (because $b$, $s$, $k$, and $t$ are
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all integers), and $b = ar$. Thus $a | b$ by definition of divisibility.
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Page 562
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**Theorem 8.4.9 Cancellation Theorem for Modular Congruence**
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For all integers $a$, $b$, and $c$, and $n$ with $n > 1$, if
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$\text{gcd}(c, n) = 1$ and $ac = bc(\mod n)$, then $a \equiv b(\mod n)$.
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**Proof:**
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Suppose $a$, $b$, $c$, and $n$ are integers, $\text{gcd}(c, n) = 1$, and
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$ac \equiv bc(\mod n)$. _[We must show that $a \equiv b(\mod n)$.]_ By
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definition of congruence modulo $n$,
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$$ n | (ac - bc) $$
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and so, since
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$$ ac - bc = (a - b)c $$
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then
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$$ n | (a - b)c $$
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Because $\text{gcd}(c, n) = 1$, we may apply Euclid's lemma to obtain
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$$ n | (a - b) $$
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and so, by definition of congruence modulo $n$,
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$$ a \equiv b(\mod n) $$
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---
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Page 563
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**Theorem 8.4.10 Fermat's Little Theorem**
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If $p$ is any prime number and $a$ is any integer such that $p \cancel{|} a$,
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then $a^{p - 1} \equiv 1(\mod p)$.
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**Proof:**
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Suppose $p$ is any prime number and $a$ is any integer such that
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$p \cancel{|} a$. Note that $a \neq 0$ because otherwise $p$ would divide $a$.
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Consider the set of integers
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$$ S = \{a, 2a, 3a, \dots, (p - 1)a\} $$
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We claim that no two elements of $S$ are congruent modulo $p$. For suppose
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$sa \equiv ra(\mod p)$ for some integers $s$ and $r$ with
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$1 \leq r < s \leq p - 1$. Then, by definition of congruence modulo $p$,
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$$ p | (sa - ra) \text{ or, equivalently, } p | (s - r)a $$
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Now $p \cancel{|} a$ by hypothesis, and because $p$ is prime,
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$\text{gcd}(a, p) = 1$. Thus, by Euclid's lemma, $p | (s - r)$, But this is
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impossible because $0 < s - r < p$.
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Consider the function $F$ from $S$ to the set $T = \{1, 2, 3, \dots, (p - 1)\}$
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that sends each element of $S$ to its residue modulo $p$. Then $F$ is one-to-one
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because no two elements of $S$ are congruent modulo $p$. In Section 9.4 we prove
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that if a function from one finite set to another is one-to-one, then it is also
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onto. Hence $F$ is onto, and so $p - 1$ residues of the $p - 1$ elements of $S$
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are exactly the numbers $1, 2, 3 \dots, (p - 1)$.
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It follows by Theorem 8.4.3(3) that
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$$ a \cdot 2a \cdot 3a \cdots (p - 1)a \equiv [1 \cdot 2 \cdot 3 \cdots (p - 1)](\mod p) $$
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or, equivalently,
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$$ a^{p - 1}(p - 1)! \equiv (p - 1)!(\mod p) $$
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Now because $p$ is prime, $p$ and $(p - 1)!$ are relatively prime. Thus, by the
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cancellation theorem for modular congruence (Theorem 8.4.9),
|
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|
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$$ a^{p - 1} \equiv 1(\mod p) $$
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|
|
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Loading…
Add table
Add a link
Reference in a new issue