Page 184 **Assumptions** - In this text we assume familiarity with the laws of basic algebra, which are listed in Appendix A. - We also use the three properties of equality: For all objects $A$, $B$, and $C$, (1) $A = A$, (2) if $A = B$, then $B = 1$, and (3) if $A = B$ and $B = C$, then $A = C$. - And we use the principle of substitution: For all objects $A$ and $B$, if $A = B$, then we may substitute $B$ whenever we have $A$. - In addition, we assume that there is no integer between $0$ and $1$ and that the set of all integers is closed under addition, subtraction, and multiplication. This means that sums, differences, and products of integers are integers. --- Page 185 **Definitions** An integer $n$ is **even** if, and only if, $n$ equals twice some integer. An integer $n$ is **odd** if, and only if, $n$ equals twice some integer plus $1$. Symbolically, for any integer $n$ $$ n \text{ is even} \Leftrightarrow n = 2k \text{ for some integer } k $$ $$ n \text{ is odd} \Leftrightarrow n = 2k + 1 \text{ for some integer } k $$ --- Page 186 **Definition** An integer $n$ is **prime** if, and only if, $n > 1$ and for all positive integers $r$ and $s$, if $n = rs$, then either $r$ or $s$ equals $n$. An integer $n$ is **composite** if, and only if, $n > 1$ and $n = rs$ for some integers $r$ and $s$ with $1 < r < n$ and $1 < s < n$. In symbols: For each integer $n$ with $n > 1$, $$ n \text{ is prime} \Leftrightarrow \forall \text{ positive integers } r \text{ and } s, \text{ if } n = rs \text{ then either } r = 1 \text{ and } s = n \text{ or } r = n \text{ and } s = 1 $$ $$ n \text{ is composite} \Leftrightarrow \exists \text{ positive integers } r \text{ and } s \text{ such that } n = rs \text{ and } 1 < r < n \text{ and } 1 < s < n $$ --- Page 188 **Disproof by Counterexample** To disprove a statement of the form "$\forall x \in D, \text{ if } P(x) \text{ then } Q(x)$," find a value of $x$ in $D$ for which the hypothesis $P(x)$ is true and the conclusion $Q(x)$ is false. Such an $x$ is called a **counterexample**. --- Page 189 **Generalizing from the Generic Particular** To show that _every_ element of a set satisfies a certain property, suppose $x$ is a _particular_ but _arbitrarily chosen_ element of the set, and show that $x$ satisfies the property. --- Page 191 **Existential Instantiation** If the existence of a certain kind of object is assumed or has been deduce, then it can be given a name, as long as that name is not currently being used to refer to something else in the same discussion. --- Page 192 **Theorem 4.1.1** The sum of any two even integers is even. **Proof:** Suppose $m$ and $n$ are any _[particular but arbitrarily chosen]_ even integers. _[We must show that $m + n$ is even.]_ By definition of even, $m = 2r$ and $n = 2s$ for some integers $r$ and $s$. Then $$ m + n = 2r + 2s \quad \text{ by substitution} $$ $$ \quad = 2(r + s) \quad \text{ by factoring out a 2} $$ Let $t = r + s$. Note that $t$ is an integer because it is a sum of integers. Hence $$ m + n = 2r \quad \text{where } t \text{ is an integer} $$ It follows by definition of even that $m + n$ is even. _[This is what we needed to show.]_ --- Page 196 Personal Note: The entirety of 4.2 is extremely helpful in breaking down in exactly how to write proofs (for beginners). I'd advise revisiting this entire section frequently. --- Page 206 **Definition** A real number $r$ is **rational** if, and only if, it can be expressed as a quotient of two integers with a nonzero denominator. A real number that is not rational is **irrational**. More formally, if $r$ is a real number, then $$ r \text{ is rational } \Leftrightarrow \exists \text{ integers } a \text{ and } b \text{ such that } r = \frac{a}{b} \text{ and } b \neq 0 $$ --- Page 207 **Zero Product Property** If neither of two real numbers is zero, then their product is also not zero. --- Page 208 **Theorem 4.3.1** Every integer is a rational number. --- Page 209 **Theorem 4.3.2** The sum of any two rational numbers is rational. **Proof:** Suppose $r$ and $s$ are any rational numbers. _[We must show that $r + s$ is rational.]_ Then, by definition of rational, $r = \dfrac{a}{b}$ and $s = \dfrac{c}{d}$ for some integers $a$, $b$, $c$, and $d$ with $b \neq 0$ and $d \neq 0$. Thus $$ r + s = \frac{a}{b} + \frac{c}{d} \quad \text{ by substitution} $$ $$ \quad = \frac{ad + bc}{bd} \quad \text{ by basic algebra} $$ Let $p = ad + bc$ and $q = bd$. Then $p$ and $q$ are integers because products and sums of integers are integers and because $a$, $b$, $c$, and $d$ are integers. Also $q \neq 0$ by the zero product property. Thus $$ r + s = \frac{p}{q} \text{ where } p \text{ and } q \text{ are integers and } q \neq 0 $$ Therefore, $r + s$ is rational by the definition of a rational number _[as was to be shown]_. --- Page 210 **Corollary 4.2.3** The double of a rational number is rational. --- Page 213 **Definition** If $n$ and $d$ are integers then $n$ is **divisible by** $d$ if, and only if, $n$ equals $d$ times some integer and $d \neq 0$. Instead of "$n$ is divisible by $d$," we can say that $n$ **is a multiple of** $d$, or $d$ **is a factor of** $n$, or $d$ **is a divisor of** $n$, or $d$ **divides** $n$. The notation $d \mid n$ is read "$d$ divides $n$." Symbolically, if $n$ and $d$ are integers: $$ d \mid n \Leftrightarrow \exists \text{ an integer, say } k, \text{ such that } n = dk \text{ and } d \neq 0 $$ The notation $d \nmid n$ is read "$d$ does not divide $n$." --- Page 214 **Theorem 4.4.1 A Positive Divisor of a Positive Integer** For all integers $a$ and $b$, if $a$ and $b$ are positive and $a$ divides $b$ then $a \leq b$. **Proof:** Suppose $a$ and $b$ are positive integers such that $a$ divides $b$. _[We must show that $a \leq b$.]_ By definition of divisibility, there exists an integer $k$ so that $b = ak$. By property T25 of Appendix A, $k$ must be positive because both $a$ and $b$ are positive. It follows that $$ 1 \leq k $$ because every positive integer is greater than or equal to $1$. Multiplying both sides by $a$ gives $$ a \leq ka = b $$ because multiplying both sides of an inequality by a positive number preserves the inequality by property T20 of Appendix A. Thus $a \leq b$ _[as was to be shown]._ --- Page 214 **Theorem 4.4.2 Divisors of 1** The only divisors of $1$ are $1$ and $-1$. **Proof:** Since $1 \cdot 1 = 1$ and $(-1)(-1) = 1$, both $1$ and $-1$ are divisors of $1$. Now suppose $m$ is any integer that divides $1$. Then there exists an integer $n$ such that $1 = mn$. By Theorem T25 in Appendix A, either both $m$ and $n$ are positive or both $m$ and $n$ are negative. If both $m$ and $n$ are positive, then $m$ is a positive integer divisor of $1$. By Theorem 4.4.1, $m \leq 1$, and, since the only positive integer that is less than or equal to $1$ is $1$ itself, it follows that $m = 1$. On the other hand, if both $m$ and $n$ are negative, then by Theorem T12 in Appendix A, $(-m)(-n) = mn = 1$. In this case $-m$ is a positive integer divisor of $1$, and so, by the same reasoning, $-m = 1$ and thus $m = -1$. Therefore there are only two possibilities: either $m = 1$ or $m = -1$. So the only divisors of $1$ are $1$ and $-1$. --- Page 215 For all integers $n$ and $d$, $d \nmid n \Leftrightarrow \frac{n}{d}$ is not an integer. --- Page 216 **Theorem 4.4.3 Transitivity of Divisibility** For all integers $a$, $b$, and $c$, if $a$ divides $b$ and $b$ divides $c$, then $a$ divides $c$. **Proof:** Suppose $a$, $b$, and $c$ are any _[particular but arbitrarily chosen]_ integers such that $a$ divides $b$ and $b$ divides $c$. _[We must show that $a$ divides $c$.]_ By definition of divisibility, $$ b = ar \text{ and } c = bs \quad \text{ for some integers } r \text{ and } s $$ By substitution $$ c = bs $$ $$ \quad = (ar)s $$ $$ \quad = a(rs) \quad \text{ by basic algebra} $$ Let $k = rs$. Then $k$ is an integer since it is a product of integers, and therefore $$ c = ak \quad \text{ where } k \text{ is an integer} $$ Thus $a$ divides $c$ by definition of divisibility. _[This is what was to be shown.]_ --- Page 217 **Theorem 4.4.4 Divisibility by a Prime** Any integer $n > 1$ is divisible by a prime number. **Proof:** Suppose $n$ is a _[particular but _arbitrarily chosen]_ integer that is greater than $1$. _[We must show that there is a prime number that divides $n$.]_ If $n$ is prime, then $n$ is divisible by a prime number (namely itself), and we are done. If $n$ is not prime, the as discussed in Example 4.1.2b, $n = r_0s_0$ where $r_0$ and $s_0$ are integers and $1 < r_0 < n$ and $1 < s_0 < n$. It follows by definition of divisibility that $r_0 \mid n$. If $r_0$ is prime, then $r_0$ is a prime number that divides $n$, and we are done. If $r_0$ is not prime, then $r_0 = r_1s_1$ where $r_1$ and $s_1$ are integers and $1 < r_1 < r_0$ and $1 < s_1 < r_0$. It follows by the definition of divisibility that $r_q \mid r_0$. But we already know that $r_0 \mid n$. Consequently, by transitivity of divisibility, $r_1 \mid n$. If $r_1$ is prime, then $r_1$ is a prime number that divides $n$, and we are done. If $r_1$ is not prime, then $r_1 = r_2s_2$ where $r_2$ and $s_2$ are integers and $1 < r_2 < r_1$ and $1 < s_2 < r_1$. It follows by the definition of divisibility that $r_2 \mid r_1$. But we already know that $r_1 \mid n$. Consequently, by transitivity of divisibility, $r_2 \mid n$. If $r_2$ is prime, then $r_2$ is a prime number that divides $n$, and we are done. If $r_2$ is not prime, then we may repeat the previous process by factoring $r_2$ as $r_3s_3$. We may continue in this way, factoring successive factors of $n$ until we find a prime factor. We must succeed in a finite number of steps because each new factor is both less than the previous one (which is less than $n$) and greater than $1$, and there are fewer than $n$ integers strictly between $1$ and $n$. Thus we obtain a sequence $$ r_0, r_1, r_2, \dots, r_k $$ where $k \geq 0$, $1 < r_k < r_{k - 1} < \dots < r_2 < r_1 < r_0 < n$, and $r_i \mid n$ for each $i = 0, 1, 2, \dots, k$. The condition for termination is that $r_k$ should be prime. Hence $r_k$ is a prime number that divides $n$. _[This is what we were to show.]_ --- Page 218 **Proposed Divisibility Property:** For all integers $a$ and $b$, if $a \mid b$ and $b \mid a$ then $a = b$. **Counterexample:** Let $a = 2$ and $b = -2$. Then $-2 = (-1) \cdot 2$ and $2 = (-1) \cdot (-2)$, and thus $$ a \mid b \text{ and } b \mid a, \text{ but } a \neq b \text{ because } 2 \neq -2 $$ Therefore, the statement is false. --- Page 219 **Theorem 4.4.5 Unique Factorization of Integers Theorem (Fundamental Theorem of Arithmetic)** Given any integer $n > 1$, there exist a positive integer $k$, distinct prime numbers $p_1, p_2, \dots, p_k$, and positive integers $e_1, e_2, \dots, e_k$ such that $$ n = p_1^{e_1}p_2^{e^2}p_3^{e^3}\dots p_k^{e_k} $$ and any other expression for $n$ as a product of prime numbers is identical to this except, perhaps, for the order in which the factors are written. --- Page 219 **Definition** Given any integer $n > 1$, the **standard factored form** of $n$ is an expression of the form $$ n = p_1^{e_1}p_2^{e^2}p_3^{e^3}\dots p_k^{e_k} $$ where $n$ is a positive integer, $p_1,p_2,\dots , p_k$ are prime numbers, $e_1,e_2,\dots ,e_k$ are positive integers, and $p_1 < p_2 < \dots < p_k$. --- Page 223 **Theorem 4.5.1 The Quotient Remainder Theorem** Given any integer $n$ and positive integer $d$, there exists unique integers $q$ and $r$ such that $$ n = dq + r \quad \text{ and } \quad 0 \leq r < d $$ --- Page 224 **Definition** Given an integer $n$ and a positive integer $d$, $$ n\ div\ d = \text{ the integer quotient obtained when } n \text{ is divided by } d \text{ and } $$ $$ n \mod d = \text{ the nonnegative integer remainder obtained when } n \text{ is divided by } d $$ Symbolically, if $n$ and $d$ are integers and $d > 0$ then $$ n\ div\ d = \quad \text{ and } \quad n \mod d = r \Leftrightarrow n = dq + r $$ where $q$ and $r$ are integers and $0 \leq r < d$. --- **Theorem 4.5.2 The Parity Property** Any two consecutive integers have opposite parity. **Proof:** Suppose that two _[particular but arbitrarily chosen]_ consecutive integers are given; call them $m$ and $m + 1$. _[We must show that one of $m$ and $m + 1$ is even and that the other is odd.]_ By the parity property, either $m$ is even or $m$ is odd. _[We break the proof into two cases depending on whether $m$ is even or odd.]_ _Case 1 ($m$ is even):_ In this case, $m = 2k$ for some integer $k$, and so $m + 1 = 2k + 1$, which is odd _[by the definition of odd.]_ Hence in this case, one of $m$ and $m + 1$ is even and the other is odd. _Case 2 ($m$ is odd):_ In this case, $m = 2k + 1$ for some integer $k$, and so $m + 1 = (2k + 1) + 1 = 2k + 2 = 2(k + 1)$. But $k + 1$ is an integer because it is a sum of two integers. Therefore, $m + 1$ equals twice some integer, and thus $m + 1$ is even. Hence in this case also, one of $m$ and $m + 1$ is even and the other is odd. It follows that regardless of which case actually occurs for the particular $m$ and $m + 1$ is even and the other is odd. _[This is what was to be shown.]_ --- Page 227 **Method of Proof by Division into Cases** To prove a statement of the form "If $A_1$ or $A_2$ or $\dots$ or $A_n$, then $C$," prove all of the following: $$ \text{If } A_1, \text{ then } C \\ \text{If } A_2, \text{ then } C \\ \vdots \\ \text{If } A_n, \text{ then } C \\ $$ This process shows that $C$ is true regardless of which of $A_1$, $A_2$, $\dots$, $A_n$ happens to be the case. --- Page 229 **Theorem 4.5.3** The square of any odd integer has the form $8m + 1$ for some integer $m$. **Proof:** Suppose $n$ is a _[particular but arbitrarily chosen]_ odd integer. By the quotient-remainder theorem with the divisor equal to $4$, $n$ can be written in one of the forms $$ 4q \quad \text{ or } \quad 4q + 1 \quad \text{ or } \quad 4q + 2 \quad \text{ or } \quad 4q + 3 $$ for some integer $q$. In fact, since $n$ is odd and $4q$ and $4q + 2$ are even, $n$ must have one of the forms $$ 4q + 1 \quad \text{ or } \quad 4q + 3 $$ _Case 1($n = 4q + 1$ for some integer $q$)_ _[We must find an integer $m$ such that $n^2 = 8m + 1$.]_ Since $n = 4q + 1$, $$ n^2 = (4q + 1)^2 \quad \text{ by substitution} $$ $$ \quad = (4q + 1)(4q + 1) \quad \text{ by definition of square} $$ $$ \quad = 16q^2 + 8q + 1 $$ $$ \quad = 8(2q^2 + 1) + 1 \quad \text{ by the laws of algebra} $$ Let $m = 2q^2 + q$. Then $m$ is an integer since $2$ and $q$ are integers and sums and products of integers are integers. Thus, substituting, $$ n^2 = 8m + 1 \quad \text{ where } m \text{ is an integer} $$ _Case 2 ($n = 4q + 3$ for some integer $q$):_ _[We must find an integer $m$ such that $n^2 = 8m + 1$.]_ Since $n = 4q + 3$, $$ n^2 = (4q + 3)^2 \quad \text{ by substitution} $$ $$ \quad = (4q + 3)(4q + 3) \quad \text{ by definition of square} $$ $$ \quad = 16q^2 + 24q + 9 $$ $$ \quad = 16q^2 + 24q + (8 + 1) $$ $$ \quad = 8(2q^2 + 3q + 1) + 1 \quad \text{ by the laws of algebra} $$ _[The motivation for the choice of algebra steps was the desire to write the expression in the form $8 \cdot \text{ some integer } + 1$.]_ Let $m = 2q^2 + 3q + 1$. Then $m$ is an integer since $1$, $2$, $3$, and $q$ are integers and sums and products of integers are integers. Thus, substituting, $$ n^2 = 8m + 1 \quad \text{ where } m \text{ is an integer} $$ Cases 1 and 2 show that given any odd integer, whether of the form $4q + 1$ or $4q + 3$, $n^2 = 8m + 1$ for some integer $m$. _[This is what we needed to show.]_ --- Page 231 **Definition** For any real number $x$, the **absolute value of** $x$, denoted $|x|$, is defined as follows: $$ |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} $$ --- Page 231 **Lemma 4.5.4** For every real number, $r$, $-|r| \leq r \leq |r|$ **Proof:** Suppose $r$ is any real number. We divide into cases according to whether $r = 0$, $r > 0$, or $r < 0$. _Case 1($r = 0$):_ In this case, by definition of absolute value, $|r| = r = 0$ since $0 = -0$, we have that $-0 = -|r| = 0 = r = |r|$, and so it is true that $$ -|r| \leq r \leq |r| $$ _Case 2 ($r > 0$):_ In this case, by definition of absolute value, [$\&|\text{pipe}|r|\text{pipe}||=|r\&$]. Also, since $r$ is positive and $-|r|$ is negative, $-|r| < r$. Thus it is true that $$ -|r| \leq r \leq |r| $$ _Case 3 ($r < 0$):_ In this case, by definition of absolute value, $|r| = -r$. Multiplying both sides by $-1$ gives that $-|r| = r$. Also, since $r$ is negative and $|r|$ is positive, $r < |r|$. Thus it is also true in this case that $$ -|r| \leq r \leq |r| $$ Hence, in every case, $$ -|r| \leq r \leq |r| $$ _[as was to be shown]._ --- Page 231 **Lemma 4.5.5** For ever real number $r$, $|-r| = |r|$. **Proof:** Suppose $r$ is any real number. By Theorem T23 in Appendix A, if $r > 0$, then $-r < 0$, and if $r < 0$, then $-r > 0$. Thus $$ |-r| = \begin{cases} -r & \text{if } -r > 0 \\ 0 & \text{if } -r = 0 \\ -(-r) & \text{if } -r < 0 \end{cases} $$ $$ \quad = \begin{cases} -r & \text{if } -r > 0 \\ 0 & \text{if } r = 0 \\ r & \text{if } -r < 0 \end{cases} $$ $$ \quad = \begin{cases} -r & \text{if } r < 0 \\ 0 & \text{if } r = 0 \\ r & \text{if } r > 0 \end{cases} $$ $$ \quad = \begin{cases} r & \text{if } r \geq 0 \\ -r & \text{if } r < 0 \\ \end{cases} $$ $$ \quad = |r| $$ --- Page 231 **Theorem 4.5.6 The Triangle Inequality** For all real numbers $x$ and $y$, $|x + y| \leq |x| + |y|$. **Proof:** Suppose $x$ and $y$ are real numbers. _Case 1 ($x + y \geq 0$):_ In this case, $|x + y| = x + y$, and so, by Lemma 4.5.4, $$ x \leq |x| \quad \text{ and } y \leq |y| $$ Hence, by Theorem T26 of Appendix A, $$ |x + y| = x + y \leq |x| + |y| $$ _Case 2 ($x + y < 0$):_ In this case, $|x + y| = -(x + y) = (-x) + (-y)$, and so, by Lemmas 4.5.4 and 4.5.5, $$ -x \leq |-x| = |x| \quad \text{ and } \quad -y \leq |-y| = |y| $$ It follows, by Theorem T26 of Appendix A, that $$ |x + y| = (-x) + (-y) \leq |x| + |y| $$ Hence in both cases $|x + y| = |x| + |y|$ _[as was to be shown]._ --- Page 234 **Definition** Given any real number $x$, the **floor of** $x$, denoted $\lfloor x \rfloor$, is defined as follows: $\lfloor x \rfloor =$ that unique integer $n$ such that $n \leq x < n + 1$. Symbolically, if $x$ is a real number and $n$ is an integer, then $$ \lfloor x \rfloor = n \Leftrightarrow n \leq x < n + 1 $$ --- Page 235 **Definition** Given any real number $x$, the **ceiling of** $x$, denoted $\lceil x \rceil$, is defined as follows: $\lceil x \rceil =$ that unique integer $n$ such that $n - 1 < x \leq n$. Symbolically, if $x$ is a real number and $n$ is an integer, then $$ \lceil x \rceil = n \Leftrightarrow n - 1 < x \leq n $$ --- Page 237 **Theorem 4.6.1** For every real number $x$ and every integer $m$, $\lfloor x + m \rfloor = \lfloor x \rfloor + m$. **Proof:** Suppose any real number $x$ and any integer $m$ are given. _[We must show that $\lfloor x + m \rfloor = \lfloor x \rfloor + m$.]_ Let $n = \lfloor x \rfloor$. By definition of floor, $n$ is an integer and $$ n \leq x < n + 1 $$ Add $m$ to all three parts to obtain $$ n + m \leq x + m < n + m + 1 $$ _[since adding a number to both sides of an inequality does not change the direction of the inequality]._ Now $n + m$ is an integer _[since $n$ and $m$ are integers and a sum of integers is an integer]_, and so, by definition of floor, the left-hand side of the equation to be shown is $$ \lfloor x + m \rfloor = n + m $$ But $n = \lfloor x \rfloor$. Hence, by substitution, $$ n + m = \lfloor x \rfloor + m $$ which is the right-hand side of the equation to be shown. Thus $\lfloor x + m \rfloor = \lfloor x \rfloor + m$ _[as was to be shown]._ --- Page 238 **Theorem 4.6.2 The Floor of $\dfrac{n}{2}$** For any integer $n$, $$ \left\lfloor \frac{n}{2} \right\rfloor = \begin{cases} \dfrac{n}{2} & \text{if } n \text{ is even} \\ \dfrac{n - 1}{2} & \text{if } n \text{ is odd} \\ \end{cases} $$ **Proof:** Suppose $n$ is a _[particular but arbitrarily chosen]_ integer. By the quotient remainder theorem, either $n$ is odd or $n$ is even. _Case 1 ($n$ is odd):_ In this case, $n = 2k + 1$ for some integer $k$. _[We must show that $\left\lfloor \dfrac{n}{2} \right\rfloor = \dfrac{(n - 1)}{2}$.]_ But the left-hand side of the equation to be shown is $$ \left\lfloor \frac{n}{2} \right\rfloor = \left\lfloor \frac{2k + 1}{2} \right\rfloor = \left\lfloor \frac{2k}{2} + \frac{1}{2} \right\rfloor = \left\lfloor k + \frac{1}{2} \right\rfloor = k $$ because $k$ is an integer and $k \leq k + \dfrac{1}{2} < k + 1$. And the right-hand side of the equation to be shown is $$ \frac{n - 1}{2} = \frac{(2k + 1) - 1}{2} = \frac{2k}{2} = k $$ also. So since both the left-hand and right-hand sides equal $k$, they are equal to each other. That is, $\left\lfloor \dfrac{n}{2} \right\rfloor = \dfrac{n - 1}{2}$ _[as was to be shown]._ _Case 2 ($n$ is even):_ In this case, $n = 2k$ for some integer $k$. _[We must show that $\left\lfloor \dfrac{n}{2} \right\rfloor$]_ The rest of the proof of this case is left as an exercise. --- Page 239 **Theorem 4.6.3** If $n$ is any integer and $d$ is a positive integer, and if $q = \left\lfloor \frac{n}{d} \right\rfloor$ and $r = n - d \cdot \left\lfloor \frac{n}{d} \right\rfloor$, then $$ n = dq + r \quad \text{ and } \quad 0 \leq r < d $$ **Proof:** Suppose $n$ is any integer, $d$ is a positive integer, $q = \left\lfloor \dfrac{n}{d} \right\rfloor$, and $r = n - d \cdot \left\lfloor \frac{n}{d} \right\rfloor$. _[We must show that $n = dq + r$ and $0 \leq r < d$.]_ By substitution, $$ dq + r = d \cdot \left\lfloor \frac{n}{d} \right\rfloor + \left(n - d \cdot \left\lfloor \frac{n}{d} \right\rfloor\right) = n $$ So it remains only to show that $0 \leq r < d$. But $q = \left\lfloor \frac{n}{d} \right\rfloor$. Thus, by definition of floor, $$ q \leq \frac{n}{d} < q + 1 $$ Then $$ dq \leq n < dq + d \quad \text{ by multiplying all parts by } d $$ and so $$ 0 \leq n - dq < d \quad \text{ by subtracting } dq \text{ from all parts.} $$ But $$ r = n - d\left\lfloor \frac{n}{d} \right\rfloor = n - dq $$ Hence $$ 0 \leq r < d \quad \text{ by substitution} $$ _[This is what was to be shown.]_ --- Page 241 **Method of Proof by Contradiction** 1. Suppose the statement to be proved is false. That is, suppose the negation of the statement is true. 2. Show that this supposition leads logically to a contradiction. 3. Conclude that the statement to be proved is true. --- Page 242 **Theorem 4.7.1** There is no greatest integer. **Proof:** _[We take the negation of the theorem and suppose it to be true.]_ Suppose not. That is, suppose there is a greatest integer $N$. _[We must deduce a contradiction.]_ Then $N \geq n$ for every integer $n$. Let $M = N + 1$. Now $M$ is an integer since it is a sum of integers. Also $M > N$ since $M = N + 1$. Thus $M$ is an integer that is greater than $N$. So $N$ is the greatest integer and $N$ is not the greatest integer, which is a contradiction. _[This contradiction shows that the supposition is false and hence, that the theorem is true.]_ --- **Theorem 4.7.2** There is no integer that is both even and odd. **Proof:** _[We take the negation of the theorem and suppose it to be true.]_ Suppose not. That is, suppose that there is at least one integer $n$ that is both even and odd. _[We must deduce a contradiction.]_ By definition of even, $n = 2a$ for some integer $a$, and by definition of odd, $n = 2b + 1$ for some integer $b$. Consequently, $$ 2a = 2b + 1 \quad \text{ by equating the two expressions for } n $$ and so $$ 2a - 2b = 1 \\ 2(a - b) = 1 \\ a j b = \frac{1}{2} \quad \text{ by algebra} $$ Now since $a$ and $b$ are integers, the difference $a - b$ must also be an integer. But $a - b = \dfrac{1}{2}$, and $\dfrac{1}{2}$ is not an integer. Thus $a - b$ is an integer and $a - b$ is not an integer, which is a contradiction. _[This contradiction shows that the supposition is false and, hence, that the theorem is true.]_ --- Page 244 **Theorem 4.7.3** The sum of any rational number and any irrational number is irrational. **Proof:** _[We take the negation of the theorem and suppose it to be true.]_ Suppose not. That is, suppose there is a rational number $r$ and an irrational number $s$ such that $r + s$ is rational. _[We must deduce a contradiction.]_ By definition of rational, $r = \dfrac{a}{b}$ and $r + s = \dfrac{c}{d}$ for some integers $a$, $b$, $c$, and $d$ with $b \neq 0$ and $d \neq 0$. By substitution, $$ \frac{a}{b} + s = \frac{c}{d} $$ and so, $$ s = \frac{c}{d} - \frac{a}{b} \quad \text{ by subtracting } \frac{a}{b} \text{ from both sides} $$ $$ = \frac{bc - ad}{bd} $$ Now $bc - ad$ and $bd$ are both integers _[since $a$, $b$, $c$, and $d$ are integers and since products and differences of integers are integers]_, and $bd \neq 0$ _[by the zero product property.]_ Hence $s$ is a quotient of the two integers $bc - ad$ and $bd$ with $bd \neq 0$. Thus, by definition of rational, $s$ is rational, which contradicts the supposition that $s$ is irrational. _[Hence the supposition is false and the theorem is true.]_ --- Page 245 **Method of Proof by Contraposition** 1. Express the statement to be proved in the form $$ \forall x \text{ in } D, \text{ if } P(x) \text{ then } Q(x) $$ (This step may be done mentally.) 2. Rewrite this statement in the contrapositive form $$ \forall x \text{ in } D, \text{ if } Q(x) \text{ is false then } P(x) \text{ is false} $$ (This step may be done mentally.) 3. Prove the contrapositive by a direct proof. a. Suppose $x$ is a (particular but arbitrarily chosen) element of $D$ such that $Q(x)$ is false. b. Show that $P(x)$ is false. --- Page 245 **Proposition 4.7.4** For every integer $n$, if $n^2$ is even then $n$ is even. **Proof (by contraposition):** Suppose $n$ is any odd integer. _[We must show that $n^2$ is odd.]_ By definition of odd, $n = 2k + 1$ for some integer $k$. By substitution and algebra, $$ n^2 =(2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1 $$ Now $2k^2 + 2k$ is an integer because products and sums of integers are integers. So $n^2 = 2 \cdot (\text{an integer}) + 1$, and thus, by definition of odd, $n^2$ is odd _[as was to be shown]._ --- Page 246 **Proposition 4.7.4** For every integer $n$, if $n^2$ is even then $n$ is even. **Proof (by contradiction):** _[We take the negation of the theorem and suppose it to be true.]_ Suppose not. That is, suppose there is an integer $n$ such that $n^2$ is even and $n$ is not even. _[We must deduce a contradiction.]_ By the quotient-remainder theorem with divisor equal to $2$, any integer is even or odd. Hence, since $n$ is not even it is odd, and thus, by definition of odd, $n = 2k + 1$ for some integer $k$. By substitution and algebra, $$ n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1 $$ Now $2k^2 + 2k$ is an integer because products and sums of integers are integers. So $n^2 = 2 \cdot (\text{an integer}) + 1$, and thus, by definition of odd, $n^2$ is odd. Therefore, $n^2$ is both even and odd. This contradicts Theorem 4.7.2, which states that no integer can be both even and odd. _[This contradiction shows that the supposition is false and, hence, that the proposition is true.]_ --- Page 252 **Theorem 4.8.1 Irrationality of $\sqrt{2}$** $\sqrt{2}$ is irrational. **Proof (by contradiction):** _[We take the negation and suppose it to be true.]_ Suppose not. That is, suppose $\sqrt{2}$ is rational. Then there are integers $m$ and $n$ with no common factors such that $$ \sqrt{2} = \frac{m}{n} $$ _[by dividing $m$ and $n$ by any common factors if necessary]._ _[We must derive a contradiction.]_ Squaring both sides of equation (4.8.1) gives $$ 2 = \frac{m^2}{n^2} $$ Or, equivalently, $$ m^2 = 2n^2 $$ Note that equation (4.8.2) implies that $m^2$ is even (by definition of even). It follows that $m$ is even (by Proposition 4.7.4). We file this fact away for future reference and also deduce (by definition of even) that $$ m = 2k \quad \text{ for some integer } k $$ Substituting equation (4.8.3) into equation (4.8.2), we see that $$ m^2 = (2k)^2 = 4k^2 = 2n^2 $$ Dividing both sides of the right-most equation by $2$ gives $$ n^2 = 2k^2 $$ Consequently, $n^2$ is even, and so $n$ is even (by Proposition 4.7.4). But we also know that $m$ is even. _[This is the fact we filed away.]_ Hence both $m$ and $n$ have a common factor of $2$. But this contradicts the supposition that $m$ and $n$ have no common factors. _[Hence the supposition is false and so the theorem is true.]_ --- Page 253 **Proposition 4.8.2** $1 + 3\sqrt{2}$ is irrational. **Proof (by contradiction):** Suppose not. Suppose $1 + 3\sqrt{2}$ is rational. _[We must derive a contradiction.]_ Then by definition of rational, $$ 1 + 3\sqrt{2} = \frac{a}{b} \quad \text{ for some integers } a \text{ and } b \text{ with } b \neq 0 $$ It follows that $$ 3\sqrt{2} = \frac{a}{b} - 1 \quad \text{ by subtracting } 1 \text{ from both sides} $$ $$ = \frac{a}{b} - \frac{b}{b} \quad \text{ by substitution} $$ $$ = \frac{a - b}{b} $$ Hence $$ \sqrt{2} = \frac{a - b}{3b} $$ But $a - b$ and $3b$ are integers (since $a$ and $b$ are integers and differences and products of integers are integers), and $3b \neq 0$ by the zero product property. Hence $\sqrt{2}$ is a quotient of the two integers $a - b$ and $3b$ with $3b \neq 0$, and so $\sqrt{2}$ is rational (by definition of rational). This contradicts the fact that $\sqrt{2}$ is irrational. _[The contradiction shows that the supposition is false.]_ Hence $1 + 3\sqrt{2}$ is irrational. --- Page 254 **Proposition 4.8.3** For any integer $a$ and any prime number $p$, if $p \mid a$ then $p \cancel{\mid} (a + 1)$. **Proof (by contradiction):** Suppose not. That is, suppose there exists an integer $a$ and a prime number $p$ such that $p \mid a$ and $p \mid (a + 1)$. Then, by definition of divisibility, there exists integers $r$ and $s$ such that $a = pr$ and $a + 1 = ps$. It follows that $$ 1 = (a + 1) - a = ps - pr = p(s - r) $$ and so (since $s - r$ is an integer) $p \mid 1$. But, by Theorem 4.4.2, the only integer divisors of $1$ are $1$ and $-1$, and $p > 1$ because $p$ is prime. Thus $p \leq 1$ and $p > 1$, which is a contradiction. _[Hence the supposition is false, and the proposition is true.]_ --- Page 254 **Theorem 4.8.4 Infinitude of the Primes** The set of prime numbers is infinite. **Proof (by contradiction):** Suppose not. That is, suppose the set of prime numbers is finite. _[WE must deduce a contradiction.]_ Then some prime number $p$ is the largest of all the prime numbers, and hence we can list the prime numbers in ascending order. $$ 2, 3< 5, 7, 11, \dots, p $$ Let $N$ be the product of all the prime numbers plus $1$: $$ N = (2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \dots p) + 1 $$ Then $N > 1$, and so, by Theorem 4.4.4, $N$ is divisible by some prime number $q$. Because $q$ is prime, $q$ must equal one of the prime numbers $2, 3, 5< 7, 11, \dots, p$. Thus, by definition of divisibility, $q$ divides $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \dots p$, and so, by Proposition 4.8.3, $q$ does not divide $(2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \dots p) + 1$, which equals $N$. Hence $N$ is divisible by $q$ and $N$ is not divisible by $q$, and we have reached a contradiction. _[Therefore, the supposition is false and the theorem is true.]_ --- Page 258 **Definition** The total degree of a graph is the sum of the degrees of all the vertices of the graph. --- Page 259 **Theorem 4.9.1 The Handshake Theorem** If $G$ is any graph, then the sum of the degrees of all the vertices of $G$ equals twice the number of edges of $G$. Specifically, if the vertices of $G$ are $v_1, v^2, \dots v_n$, where $n$ is a nonnegative integer, then $$ \text{the total degree of } G = \text{deg}(v_1) + \text{deg}(v_2) + \dots + \text{deg}(v_n) $$ $$ = 2 \cdot (\text{the number of edges of } G) $$ **Proof:** Let $G$ be a particular but arbitrarily chosen graph, and suppose that $G$ has $n$ vertices $v_1, v_2, \dots v_n$ and $m$ edges, where $n$ is a positive integer and $m$ is a nonnegative integer. We claim that each edge of $G$ contributes $2$ to the total degree of $G$. For suppose $e$ is an arbitrarily chosen edge with endpoints $v_i$ and $v_j$. This edge contributes $1$ to the degree of $v_i$ and $1$ to the degree of $v_j$. As shown below, this is true even if $i = j$, because an edge that is a loop is counted twice in computing the degree of the vertex on which it is incident. (see Page 259) Therefore, $e$ contributes $2$ to the total degree of $G$. Since $e$ was arbitrarily chosen, this shows that _each_ edge of $G$ contributes $2$ to the total degree of $G$. Thus $$ \text{the total degree of } G = 2 \cdot (\text{the number of edges of } G) $$ --- Page 259 **Corollary 4.9.2** The total degree of a graph is even. **Proof:** By Theorem 4.9.1 the total degree of $G$ equals $2$ times the number of edges of $G$, which is an integer, and so the total degree of $G$ is even. --- Page 260 **Proposition 4.9.3** In any graph there is an even number of vertices of odd degree. **Proof:** Suppose $G$ is any graph, and suppose $G$ has $n$ vertices of odd degree and $m$ vertices of even degree, where $n$ is a positive integer and $m$ is a nonnegative integer. _[We must show that $n$ is even.]_ Let $E$ be the sum of the degrees of all the vertices of even degree, $O$ the sum of the degrees of all the vertices of odd degree, and $T$ the total degree of $G$. If $u_1, u_2, \dots, u_m$ are the vertices of even degree and $v_1, v_2, \dots, v_n$ are the vertices of odd degree, then $$ E = \text{deg}(u_1) + \text{deg}(u_2) + \dots + \text{deg}(u_m), $$ $$ O = \text{deg}(v_1) + \text{deg}(v_2) + \dots + \text{deg}(v_m), \text{ and} $$ $$ T = \text{deg}(u_1) + \dots + \text{deg}(u_m) + \text{deg}(v_1) + \dots + \text{deg}(v_n) = E + 0 $$ Now $T$, the total degree of $G$, is an even integer by Corollary 4.9.2. Also $E$ is even since either $E$ is zero, which is even, or $E$ is a sum of even numbers. Now since $$ T = E + O $$ then $$ O = T - E $$ Hence $O$ is a difference of two even integers, and so $O$ is even. By assumption, $\text{deg}(v_i)$ is odd for every integer $i = 1, 2, \dots, n$. Thus $O$, an even integer, is a sum of the $n$ odd integers $\text{deg}(v_1), \text{deg}(v_2), \dots, \text{deg}(v_n)$. But if a sum of $n$ odd integers is even, then $n$ is even. Therefore, $n$ is even _[as was to be shown]._ --- Page 262 **Definition and Notation** A **simple graph** is a graph that does not have any loops or parallel edges. In a simple graph, an edge with endpoints $v$ and $w$ is denoted $\{v, w\}$. --- Page 263 **Definition** Let $n$ be a positive integer. A **complete graph on $n$ vertices**, denoted $K_n$, is a simple graph with $n$ vertices and exactly one edge connecting each pair of distinct vertices. --- Page 264 **Definition** Let $m$ and $n$ be positive integers. A **complete bipartite graph on $(m, n)$ vertices**, denoted $K_{m, n}$, is a simple graph whose vertices are divided into two distinct subsets, $V$ with $m$ vertices and $W$ with $n$ vertices, in such a way that 1. every vertex of $K_{m, n}$ belongs to one of $V$ or $W$, but no vertex belongs to both $V$ and $W$; 2. there is exactly one edge from each vertex of $V$ to each vertex of $W$; 3. there is no edge from any one vertex of $V$ to any other vertex of $V$; 4. there is no edge from any one vertex of $W$ to any other vertex of $W$.