1244 lines
35 KiB
Markdown
1244 lines
35 KiB
Markdown
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 $W$.
|