Notes › A First Course in Analysis (Pedrick) Lecture 2

Measurement and the Rationals

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Table of Contents

Measurement Theory

Any given amount is really a chosen unit amount a certain “number” of times
Two amounts are commensurable if they share a unit amount (e.g. $mu$ and $nu$ for $m, n \in \mathbb{Z}$)

Directed Segments

A single point is a segment of length 0 for all units
For segments with more than one point, they exist in pairs called directed segments
- Expressed as arrows, one going to the left and the other going to the right
- Left-directed segments have negatives of their lengths as measures
All directed segments lie on a single line with initial endpoint at the origin $\mathcal{o}$

Need for Division

If $\alpha$ and $\beta$ are mutually commensurable points with shared unit $u$, then we need to define a number $x$ such that $\beta$ may be expressed in terms of $\alpha$
- Since $\alpha = nu$ and $\beta = mu$, then $\beta = x\alpha$ and $mu = x \cdot nu \implies m = xn$
We need to create a number system in which $m = nx$ always has a unique solution
- Exception: division by zero, so $n \ne 0$

Definition 02.1 (Division Equivalence).

The equations $nx = m$ and $qx = p$ $(n, q \ne 0)$ in $\mathbb{Z}$ are equivalent if $mq = pn$. Alternatively, it means $\frac{m}{n} = \frac{p}{q}$.

Exercise 02.2.

EXERCISE 1 [Pedrick, A First Course in Analysis, p. 33]

Exercise 02.3.

EXERCISE 2 [Pedrick, A First Course in Analysis, p. 34] All numbers can be expressed as products of primes. For a fully simplified fraction, $\frac{a}{b}$, $\gcd(a, b) = 1$, since $a$ and $b$ do not share any factors.

Defining $\mathbb{Q}$

The set of fully simplified fractions forms $\mathbb{Q}$

Definition 02.4 (Sum of Fractions).

$$\frac{m}{n} + \frac{p}{q} = \frac{mq + np}{nq}$$

Exercise 02.5.

(EXERCISE 3 [Pedrick, A First Course in Analysis, p. 34])

Exercise 02.6.

(EXERCISE 4 [Pedrick, A First Course in Analysis, p. 34])

Definition 02.7 (Product of Fractions).

$$\frac{m}{n} \cdot \frac{p}{q} = \frac{m \cdot p}{n \cdot q}$$

Exercise 02.8.

(EXERCISE 5 [Pedrick, A First Course in Analysis, p. 35])

Exercise 02.9.

EXERCISE 6. [Pedrick, A First Course in Analysis, p. 35]

Exercise 02.10.

EXERCISE 7. [Pedrick, A First Course in Analysis, p. 35]

Theorem 02.11 (Division of Fractions).

Any equation in $\mathbb{Q}$ of the form

$$\frac{m}{n} \cdot x = \frac{p}{q}, m \ne 0$$

has the unique solution $\frac{np}{mq}$.

Exercise 02.12.

(EXERCISE 8. [Pedrick, A First Course in Analysis, p. 35])

Definition 02.13 (Order of Fractions).

The order between fractions is given by

$$\frac{m}{n} < \frac{p}{q} \implies mq < np$$

if $n, q > 0$.

Exercise 02.14.

(EXERCISE 9. [Pedrick, A First Course in Analysis, p. 35])

Exercise 02.15.

(EXERCISE 10 [Pedrick, A First Course in Analysis, p. 35])

Exercise 02.16.

(EXERCISE 11 [Pedrick, A First Course in Analysis, p. 35])

Exercise 02.17.

(EXERCISE 12 [Pedrick, A First Course in Analysis, p. 35])

Measurement of Directed Segments

If $\alpha$ and $\beta$ are commensurable, and $\beta$ is not a point ($\alpha = mu, \beta = nu, n \ne 0$), you may express $\alpha$ in terms of $\beta$ using $\mathbb{Q}$ and determine the unit $u$
- Only commensurable lengths can be expressed in terms of each other using $\mathbb{Q}$

Theorem 02.18 (Archimedean Property).

$\forall y > 0 \in \mathbb{Q}, \exists m \in \mathbb{N} | y \le m$. Alternative definitions with directed segments: $\forall y > 0 \in \mathbb{Q} \exists k \in \mathbb{N} | y \le k \cdot u$. “There are arbitrarily large natural numbers.”

Corollary 02.19 (Reverse Archimedean Property).

$\forall x > 0 \in \mathbb{Q}, \exists n \in \mathbb{N} | \frac{1}{n} < x$. “The reciprocals of the natural numbers can be arbitrarily small.”

Example 02.20.

(EXERCISE 14 [Pedrick, A First Course in Analysis, p. 37])

Axioms of Ordered Fields

Any number system that satisfies the below properties is called an ordered field
- Number systems that are ordered fields and satisfy the Archimedean property are called Archimedean ordered fields

Axiom 02.21 (Axioms for Addition).

Associative and Commutative
$\exists 0 \in F$ that is neutral under addition.
$\exists$ additive inverse “negative $x$” $\forall x \in F$.

Axiom 02.22 (Axioms for Multiplication).

Associative and Commutative
$\exists 1 \in F$ that is neutral under multiplication.
$\exists$ multiplicative inverse $1/x$ $\forall x \in F, x \ne 0$.

Axiom 02.23 (Distributive Laws).

Multiplication and Addition are connected by the distributive laws.

Axiom 02.24 (Order).

Order of $F$ is transitive and trichotomy holds.
Multiplication by $x > 0$ preserves order.

Example 02.25.

(EXERCISE 15 [Pedrick, A First Course in Analysis, p. 38])

Example 02.26.

(EXERCISE 16 [Pedrick, A First Course in Analysis, p. 38])

Example 02.27.

(EXERCISE 17 [Pedrick, A First Course in Analysis, p. 38])

Example 02.28.

(EXERCISE 18 [Pedrick, A First Course in Analysis, p. 38])

Measurement Density

Definition 02.29 (Density).

A number system, $V$, is dense in a field, $F$, if: $\forall x < y \in F, \exists m \in V | x < m < y$

Example 02.30.

(EXERCISE 19 [Pedrick, A First Course in Analysis, p. 38])

Theorem 02.31 (Density of $\mathbb{Q}$ in Archimedean Ordered Fields).

$\mathbb{Q}$ is dense in any Archimedean ordered field $F$. Proof [Pedrick, A First Course in Analysis, p. 39]

Exercise 02.32.

(EXERCISE 20 [Pedrick, A First Course in Analysis, p. 40])

Exercise 02.33.

(EXERCISE 21 [Pedrick, A First Course in Analysis, p. 40])

Exercise 02.34.

(EXERCISE 22 [Pedrick, A First Course in Analysis, p. 40])

Exercise 02.35.

(EXERCISE 23 [Pedrick, A First Course in Analysis, p. 40])

Expanding the Definitions of Exponentiation

Definition 02.36 (Extension of Powers to $\mathbb{Z}$).

$$x^{0} = 1 \quad \text{and} \quad x^{-n}=\frac{1}{x^n}$$

Definition 02.37 (Extension of Powers to $\mathbb{Q}$).

$$x^{p/q} = \sqrt[q]{x^p}, q \ne 0$$

$x^{1/q} = \sqrt[q]{x}$ is the member(s) of $F$ whose $q$th power is $x$.
- If $q$ is odd, there is one positive member in $F$.
- If $q$ is even, there are two members in $F$, one positive and one negative, in pairs. We use the current notation to refer to the positive member unless explicitly denoted otherwise.
For this extension, we assume the member is defined to begin with. If $\sqrt[q]{x^p}$ is not a member of $F$ (i.e., irrational), we do not consider it.

Interval Notation

Definition 02.38 (Interval).

An interval is a subset defined using the order of $F$ for $a \le b$ in $F$. The interval may either be open, closed, left-closed right-open, or left-open right-closed. Open means you don’t consider the endpoint, and vice-versa. The length of the interval is $b-a$.

We use brackets to denote closed interval and parenthesis for open interval
- For example, the notation for the open interval on $(a, b) \in F$ is:

$$(a, b) =\{x \in F : a < x < b\}$$

If $a=b$, the closed interval reduces to a point, and all other cases are $\emptyset$
You can also use $\infty$ in interval notation to denote “half-lines” or a “full line” (which is really just the entire set of points on $F$):
- For example, the left half-line of the point $a$ (excluding the point $a$ itself) is:

$$(\infty, a) = \{x \in F:x

Definition 02.39 (Absolute Value).

We denote the absolute value of $x \in F$ by:

$$|x| = \begin{cases} x, & x>0 \\ -x, & x<0 \end{cases}$$

Exercise 02.40.

(EXERCISE 24 [Pedrick, A First Course in Analysis, p. 41])

Since $|x|$ is the distance of $x$ from the origin, we can conclude that $|b-a|$ is the distance between $a, b \in F$
$\forall x \in F, -|x| \le x \le |x|$, and one of the equality signs must hold

Exercise 02.41.

(EXERCISE 25 [Pedrick, A First Course in Analysis, p. 41])

Theorem 02.42 (Triangle Inequality).

$$|x + y| \le |x| + |y|$$

Proof [Pedrick, A First Course in Analysis, p. 41] (using the result from the previous Exercise)

Exercise 02.43.

(EXERCISE 26 [Pedrick, A First Course in Analysis, p. 41])

Exercise 02.44.

(EXERCISE 27 [Pedrick, A First Course in Analysis, p. 41])

Exercise 02.45.

(EXERCISE 28 [Pedrick, A First Course in Analysis, p. 41])

Boundedness

Definition 02.46 (Boundedness).

$S \subset F$ is bounded above if:

$$\exists b \in F | x \in S \implies x\le b.$$

Alternatively:

$$\exists b \in F | x \le b \forall x \in S.$$

Any such $b$ is called an upper bound of $F$.
Any member of $F$ larger than an upper bound of $S$ is also an upper bound of $S$.

Similar logic applies for bounded below:

$$\exists b \in F | x \ge b \forall x \in S.$$

If $S$ is bounded above and below in $F$, it is bounded.

The Archimedean property really just says that $\mathbb{N}$ is not bounded above in $F$ for any choice of unit

Example 02.47.

(EXERCISE 31 [Pedrick, A First Course in Analysis, p. 42])

Example 02.48.

(EXERCISE 32 [Pedrick, A First Course in Analysis, p. 42])

Example 02.49.

(EXERCISE 33 [Pedrick, A First Course in Analysis, p. 42])

Example 02.50.

(EXERCISE 34 [Pedrick, A First Course in Analysis, p. 42])

Example 02.51.

(EXERCISE 35 [Pedrick, A First Course in Analysis, p. 42])

Example 02.52.

(EXERCISE 36 [Pedrick, A First Course in Analysis, p. 42])

Division Algorithm

Definition 02.53 (Division Algorithm).

For a dividend $n \in F$ and a divisor $d > 0 \in F$, there exists a quotient $q \in \mathbb{Z}$ and a remainder $0 \le r < d$ such that:

$$n = qd + r.$$

Exercise 02.54.

(EXERCISE 37 [Pedrick, A First Course in Analysis, p. 43])

Sources

Pedrick, A First Course in Analysis

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