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.