980 lines
27 KiB
Markdown
980 lines
27 KiB
Markdown
Page 184
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**Assumptions**
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- In this text we assume familiarity with the laws of basic algebra, which are
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listed in Appendix A.
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- We also use the three properties of equality: For all objects $A$, $B$, and
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$C$, (1) $A = A$, (2) if $A = B$, then $B = 1$, and (3) if $A = B$ and
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$B = C$, then $A = C$.
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- And we use the principle of substitution: For all objects $A$ and $B$, if
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$A = B$, then we may substitute $B$ whenever we have $A$.
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- In addition, we assume that there is no integer between $0$ and $1$ and that
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the set of all integers is closed under addition, subtraction, and
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multiplication. This means that sums, differences, and products of integers
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are integers.
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---
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Page 185
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**Definitions**
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An integer $n$ is **even** if, and only if, $n$ equals twice some integer. An
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integer $n$ is **odd** if, and only if, $n$ equals twice some integer plus $1$.
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Symbolically, for any integer $n$
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$$ n \text{ is even} \Leftrightarrow n = 2k \text{ for some integer } k $$
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$$ n \text{ is odd} \Leftrightarrow n = 2k + 1 \text{ for some integer } k $$
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---
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Page 186
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**Definition**
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An integer $n$ is **prime** if, and only if, $n > 1$ and for all positive
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integers $r$ and $s$, if $n = rs$, then either $r$ or $s$ equals $n$. An integer
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$n$ is **composite** if, and only if, $n > 1$ and $n = rs$ for some integers $r$
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and $s$ with $1 < r < n$ and $1 < s < n$.
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In symbols: For each integer $n$ with $n > 1$,
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$$ n \text{ is prime} \Leftrightarrow \forall \text{ positive integers } r \text{ and } s, \text{ if } n = rs \text{ then either } r = 1 \text{ and } s = n \text{ or } r = n \text{ and } s = 1 $$
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$$ n \text{ is composite} \Leftrightarrow \exists \text{ positive integers } r \text{ and } s \text{ such that } n = rs \text{ and } 1 < r < n \text{ and } 1 < s < n $$
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---
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Page 188
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**Disproof by Counterexample**
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To disprove a statement of the form
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"$\forall x \in D, \text{ if } P(x) \text{ then } Q(x)$," find a value of $x$ in
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$D$ for which the hypothesis $P(x)$ is true and the conclusion $Q(x)$ is false.
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Such an $x$ is called a **counterexample**.
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---
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Page 189
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**Generalizing from the Generic Particular**
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To show that _every_ element of a set satisfies a certain property, suppose $x$
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is a _particular_ but _arbitrarily chosen_ element of the set, and show that $x$
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satisfies the property.
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---
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Page 191
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**Existential Instantiation**
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If the existence of a certain kind of object is assumed or has been deduce, then
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it can be given a name, as long as that name is not currently being used to
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refer to something else in the same discussion.
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---
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Page 192
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**Theorem 4.1.1**
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The sum of any two even integers is even.
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**Proof:** Suppose $m$ and $n$ are any _[particular but arbitrarily chosen]_
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even integers. _[We must show that $m + n$ is even.]_ By definition of even,
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$m = 2r$ and $n = 2s$ for some integers $r$ and $s$. Then
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$$ m + n = 2r + 2s \quad \text{ by substitution} $$
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$$ \quad = 2(r + s) \quad \text{ by factoring out a 2} $$
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Let $t = r + s$. Note that $t$ is an integer because it is a sum of integers.
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Hence
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$$ m + n = 2r \quad \text{where } t \text{ is an integer} $$
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It follows by definition of even that $m + n$ is even. _[This is what we needed
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to show.]_
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---
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Page 196
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Personal Note: The entirety of 4.2 is extremely helpful in breaking down in
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exactly how to write proofs (for beginners). I'd advise revisiting this entire
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section frequently.
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---
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Page 206
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**Definition**
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A real number $r$ is **rational** if, and only if, it can be expressed as a
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quotient of two integers with a nonzero denominator. A real number that is not
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rational is **irrational**. More formally, if $r$ is a real number, then
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$$ r \text{ is rational } \Leftrightarrow \exists \text{ integers } a \text{ and } b \text{ such that } r = \frac{a}{b} \text{ and } b \neq 0 $$
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---
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Page 207
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**Zero Product Property**
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If neither of two real numbers is zero, then their product is also not zero.
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---
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Page 208
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**Theorem 4.3.1**
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Every integer is a rational number.
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---
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Page 209
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**Theorem 4.3.2**
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The sum of any two rational numbers is rational.
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**Proof:**
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Suppose $r$ and $s$ are any rational numbers. _[We must show that $r + s$ is
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rational.]_ Then, by definition of rational, $r = \dfrac{a}{b}$ and
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$s = \dfrac{c}{d}$ for some integers $a$, $b$, $c$, and $d$ with $b \neq 0$ and
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$d \neq 0$. Thus
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$$ r + s = \frac{a}{b} + \frac{c}{d} \quad \text{ by substitution} $$
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$$ \quad = \frac{ad + bc}{bd} \quad \text{ by basic algebra} $$
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Let $p = ad + bc$ and $q = bd$. Then $p$ and $q$ are integers because products
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and sums of integers are integers and because $a$, $b$, $c$, and $d$ are
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integers. Also $q \neq 0$ by the zero product property. Thus
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$$ r + s = \frac{p}{q} \text{ where } p \text{ and } q \text{ are integers and } q \neq 0 $$
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Therefore, $r + s$ is rational by the definition of a rational number _[as was
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to be shown]_.
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---
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Page 210
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**Corollary 4.2.3**
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The double of a rational number is rational.
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---
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Page 213
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**Definition**
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If $n$ and $d$ are integers then
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$n$ is **divisible by** $d$ if, and only if, $n$ equals $d$ times some integer
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and $d \neq 0$.
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Instead of "$n$ is divisible by $d$," we can say that
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$n$ **is a multiple of** $d$, or
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$d$ **is a factor of** $n$, or
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$d$ **is a divisor of** $n$, or
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$d$ **divides** $n$.
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The notation $d \mid n$ is read "$d$ divides $n$." Symbolically, if $n$ and $d$
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are integers:
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$$ d \mid n \Leftrightarrow \exists \text{ an integer, say } k, \text{ such that } n = dk \text{ and } d \neq 0 $$
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The notation $d \nmid n$ is read "$d$ does not divide $n$."
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---
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Page 214
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**Theorem 4.4.1 A Positive Divisor of a Positive Integer**
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For all integers $a$ and $b$, if $a$ and $b$ are positive and $a$ divides $b$
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then $a \leq b$.
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**Proof:**
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Suppose $a$ and $b$ are positive integers such that $a$ divides $b$. _[We must
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show that $a \leq b$.]_ By definition of divisibility, there exists an integer
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$k$ so that $b = ak$. By property T25 of Appendix A, $k$ must be positive
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because both $a$ and $b$ are positive. It follows that
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$$ 1 \leq k $$
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because every positive integer is greater than or equal to $1$. Multiplying both
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sides by $a$ gives
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$$ a \leq ka = b $$
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because multiplying both sides of an inequality by a positive number preserves
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the inequality by property T20 of Appendix A. Thus $a \leq b$ _[as was to be
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shown]._
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---
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Page 214
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**Theorem 4.4.2 Divisors of 1**
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The only divisors of $1$ are $1$ and $-1$.
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**Proof:**
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Since $1 \cdot 1 = 1$ and $(-1)(-1) = 1$, both $1$ and $-1$ are divisors of $1$.
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Now suppose $m$ is any integer that divides $1$. Then there exists an integer
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$n$ such that $1 = mn$. By Theorem T25 in Appendix A, either both $m$ and $n$
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are positive or both $m$ and $n$ are negative. If both $m$ and $n$ are positive,
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then $m$ is a positive integer divisor of $1$. By Theorem 4.4.1, $m \leq 1$,
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and, since the only positive integer that is less than or equal to $1$ is $1$
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itself, it follows that $m = 1$. On the other hand, if both $m$ and $n$ are
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negative, then by Theorem T12 in Appendix A, $(-m)(-n) = mn = 1$. In this case
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$-m$ is a positive integer divisor of $1$, and so, by the same reasoning,
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$-m = 1$ and thus $m = -1$. Therefore there are only two possibilities: either
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$m = 1$ or $m = -1$. So the only divisors of $1$ are $1$ and $-1$.
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---
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Page 215
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For all integers $n$ and $d$, $d \nmid n \Leftrightarrow \frac{n}{d}$ is not an
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integer.
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---
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Page 216
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**Theorem 4.4.3 Transitivity of Divisibility**
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For all integers $a$, $b$, and $c$, if $a$ divides $b$ and $b$ divides $c$, then
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$a$ divides $c$.
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**Proof:**
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Suppose $a$, $b$, and $c$ are any _[particular but arbitrarily chosen]_ integers
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such that $a$ divides $b$ and $b$ divides $c$. _[We must show that $a$ divides
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$c$.]_ By definition of divisibility,
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$$ b = ar \text{ and } c = bs \quad \text{ for some integers } r \text{ and } s $$
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By substitution
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$$ c = bs $$
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$$ \quad = (ar)s $$
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$$ \quad = a(rs) \quad \text{ by basic algebra} $$
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Let $k = rs$. Then $k$ is an integer since it is a product of integers, and
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therefore
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$$ c = ak \quad \text{ where } k \text{ is an integer} $$
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Thus $a$ divides $c$ by definition of divisibility. _[This is what was to be
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shown.]_
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---
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Page 217
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**Theorem 4.4.4 Divisibility by a Prime**
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Any integer $n > 1$ is divisible by a prime number.
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**Proof:**
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Suppose $n$ is a _[particular but _arbitrarily chosen]_ integer that is greater
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than $1$. _[We must show that there is a prime number that divides $n$.]_ If $n$
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is prime, then $n$ is divisible by a prime number (namely itself), and we are
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done. If $n$ is not prime, the as discussed in Example 4.1.2b,
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$n = r_0s_0$ where $r_0$ and $s_0$ are integers and $1 < r_0 < n$ and
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$1 < s_0 < n$.
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It follows by definition of divisibility that $r_0 \mid n$.
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If $r_0$ is prime, then $r_0$ is a prime number that divides $n$, and we are
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done. If $r_0$ is not prime, then
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$r_0 = r_1s_1$ where $r_1$ and $s_1$ are integers and $1 < r_1 < r_0$ and
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$1 < s_1 < r_0$.
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It follows by the definition of divisibility that $r_q \mid r_0$. But we already
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know that $r_0 \mid n$. Consequently, by transitivity of divisibility,
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$r_1 \mid n$.
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If $r_1$ is prime, then $r_1$ is a prime number that divides $n$, and we are
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done. If $r_1$ is not prime, then
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$r_1 = r_2s_2$ where $r_2$ and $s_2$ are integers and $1 < r_2 < r_1$ and
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$1 < s_2 < r_1$.
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It follows by the definition of divisibility that $r_2 \mid r_1$. But we already
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know that $r_1 \mid n$. Consequently, by transitivity of divisibility,
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$r_2 \mid n$.
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If $r_2$ is prime, then $r_2$ is a prime number that divides $n$, and we are
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done. If $r_2$ is not prime, then we may repeat the previous process by
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factoring $r_2$ as $r_3s_3$.
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We may continue in this way, factoring successive factors of $n$ until we find a
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prime factor. We must succeed in a finite number of steps because each new
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factor is both less than the previous one (which is less than $n$) and greater
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than $1$, and there are fewer than $n$ integers strictly between $1$ and $n$.
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Thus we obtain a sequence
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$$ r_0, r_1, r_2, \dots, r_k $$
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where $k \geq 0$, $1 < r_k < r_{k - 1} < \dots < r_2 < r_1 < r_0 < n$, and
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$r_i \mid n$ for each $i = 0, 1, 2, \dots, k$. The condition for termination is
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that $r_k$ should be prime. Hence $r_k$ is a prime number that divides $n$.
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_[This is what we were to show.]_
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---
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Page 218
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**Proposed Divisibility Property:** For all integers $a$ and $b$, if $a \mid b$
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and $b \mid a$ then $a = b$.
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**Counterexample:** Let $a = 2$ and $b = -2$. Then $-2 = (-1) \cdot 2$ and
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$2 = (-1) \cdot (-2)$, and thus
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$$ a \mid b \text{ and } b \mid a, \text{ but } a \neq b \text{ because } 2 \neq -2 $$
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Therefore, the statement is false.
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---
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Page 219
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**Theorem 4.4.5 Unique Factorization of Integers Theorem (Fundamental Theorem of
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Arithmetic)**
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Given any integer $n > 1$, there exist a positive integer $k$, distinct prime
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numbers $p_1, p_2, \dots, p_k$, and positive integers $e_1, e_2, \dots, e_k$
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such that
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$$ n = p_1^{e_1}p_2^{e^2}p_3^{e^3}\dots p_k^{e_k} $$
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and any other expression for $n$ as a product of prime numbers is identical to
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this except, perhaps, for the order in which the factors are written.
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---
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Page 219
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**Definition**
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Given any integer $n > 1$, the **standard factored form** of $n$ is an
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expression of the form
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$$ n = p_1^{e_1}p_2^{e^2}p_3^{e^3}\dots p_k^{e_k} $$
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where $n$ is a positive integer, $p_1,p_2,\dots , p_k$ are prime numbers,
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$e_1,e_2,\dots ,e_k$ are positive integers, and $p_1 < p_2 < \dots < p_k$.
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---
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Page 223
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**Theorem 4.5.1 The Quotient Remainder Theorem**
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Given any integer $n$ and positive integer $d$, there exists unique integers $q$
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and $r$ such that
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$$ n = dq + r \quad \text{ and } \quad 0 \leq r < d $$
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---
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Page 224
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**Definition**
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Given an integer $n$ and a positive integer $d$,
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$$ n\ div\ d = \text{ the integer quotient obtained when } n \text{ is divided by } d \text{ and } $$
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$$ n \mod d = \text{ the nonnegative integer remainder obtained when } n \text{ is divided by } d $$
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Symbolically, if $n$ and $d$ are integers and $d > 0$ then
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$$ n\ div\ d = \quad \text{ and } \quad n \mod d = r \Leftrightarrow n = dq + r $$
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where $q$ and $r$ are integers and $0 \leq r < d$.
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---
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**Theorem 4.5.2 The Parity Property**
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Any two consecutive integers have opposite parity.
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**Proof:**
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Suppose that two _[particular but arbitrarily chosen]_ consecutive integers are
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given; call them $m$ and $m + 1$. _[We must show that one of $m$ and $m + 1$ is
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even and that the other is odd.]_ By the parity property, either $m$ is even or
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$m$ is odd. _[We break the proof into two cases depending on whether $m$ is even
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or odd.]_
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_Case 1 ($m$ is even):_ In this case, $m = 2k$ for some integer $k$, and so
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$m + 1 = 2k + 1$, which is odd _[by the definition of odd.]_ Hence in this case,
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one of $m$ and $m + 1$ is even and the other is odd.
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_Case 2 ($m$ is odd):_ In this case, $m = 2k + 1$ for some integer $k$, and so
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$m + 1 = (2k + 1) + 1 = 2k + 2 = 2(k + 1)$. But $k + 1$ is an integer because it
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is a sum of two integers. Therefore, $m + 1$ equals twice some integer, and thus
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$m + 1$ is even. Hence in this case also, one of $m$ and $m + 1$ is even and the
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other is odd.
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It follows that regardless of which case actually occurs for the particular $m$
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and $m + 1$ is even and the other is odd. _[This is what was to be shown.]_
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---
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Page 227
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**Method of Proof by Division into Cases**
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To prove a statement of the form "If $A_1$ or $A_2$ or $\dots$ or $A_n$, then
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$C$," prove all of the following:
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$$
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\text{If } A_1, \text{ then } C \\
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\text{If } A_2, \text{ then } C \\
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\vdots \\
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\text{If } A_n, \text{ then } C \\
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$$
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This process shows that $C$ is true regardless of which of $A_1$, $A_2$,
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$\dots$, $A_n$ happens to be the case.
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---
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Page 229
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**Theorem 4.5.3**
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The square of any odd integer has the form $8m + 1$ for some integer $m$.
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**Proof:** Suppose $n$ is a _[particular but arbitrarily chosen]_ odd integer.
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By the quotient-remainder theorem with the divisor equal to $4$, $n$ can be
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written in one of the forms
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$$ 4q \quad \text{ or } \quad 4q + 1 \quad \text{ or } \quad 4q + 2 \quad \text{ or } \quad 4q + 3 $$
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for some integer $q$. In fact, since $n$ is odd and $4q$ and $4q + 2$ are even,
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$n$ must have one of the forms
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$$ 4q + 1 \quad \text{ or } \quad 4q + 3 $$
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_Case 1($n = 4q + 1$ for some integer $q$)_ _[We must find an integer $m$ such
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that $n^2 = 8m + 1$.]_ Since $n = 4q + 1$,
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$$ n^2 = (4q + 1)^2 \quad \text{ by substitution} $$
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$$ \quad = (4q + 1)(4q + 1) \quad \text{ by definition of square} $$
|
|
|
|
$$ \quad = 16q^2 + 8q + 1 $$
|
|
|
|
$$ \quad = 8(2q^2 + 1) + 1 \quad \text{ by the laws of algebra} $$
|
|
|
|
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.]_
|