real | Arbias Hashani

Lecture 1: Sets, Real Numbers, Fields, Ordered Fields

October 30, 2012

The real number system is a set $\{a,b,c, ...\}$ on which the operations of addition and multiplication are defined so that every pair of real numbers (say $a$ and $b$ ) has a unique sum and product, both real numbers, with the following properties:

Given any (and all) $a,b,c \in \mathbb{R}$ , we have

(A) $a+b=b+a$ and $a*b=b*a$ (Commutative laws)

(B) $(a+b)+c=a+(b+c)$ and $(a*b)*c=a*(b*c)$ (Associative laws)

(D) There exist distinct real numbers $0$ and $1$ such that $a+0=a$ and $a*1=a$ , for all $a$ .

(E) For each $a$ there is a real number $-a$ such that $a+(-a)=0$ and if $a \neq 0$ , there is a real number $\frac{1}{a}$ such that $a*\frac{1}{a}=1$

A set on which two operations are defined so as to have properties (A) – (E) is a field.

Motivation: Think about these properties for a second before moving on. (A) establishes what the operations do, (B) then asks what happens when we include more elements, a “double operation” if you like. (C) then mixes the two operands into one. (D) and (E) are closely related. Suppose we want to add an element to our given real number that… gives us our real number? So we are adding nothing, or $0$ . Suppose we have the same idea for our other operand, multiplication, we have the real number $1$ to do this for us.

Then we ask, can we find real numbers that give us what we just discovered, namely $0$ and $1$ from addition and multiplication? For that we have $-a$ and $\frac{1}{a}$ with $a \neq 0$ for the second condition.

Exercise: Can you think of another example of a field?
Exercise: Do the set of integers $\mathbb{Z}$ form a field? Verify through (A) – (E).

The real number system is ordered by the relation $<;$ , which has the following properties:

(F) For each pair of real numbers $a$ and $b$ , exactly one of the following is true: $a <; b$ , $a=b$ , $b<;a$ .

(G) If $a<;b$ and $b<;c$ then $a<;c$ . (Transitive)

(H) If $a<;b$ , then $a+c<;b+c$ for any $c$ and if $b<;c$ then $a*c<;b*c$ .

A field with an order relation satisfying (F) – (H) is an ordered field. The real numbers (and rationals) form an ordered field.

Suppose we get bored of the real numbers and define a new field satisfying everything $\mathbb{R}$ does but is defined as such. For any $a,b \in \mathbb{R}$ we define a new number $z$ as $z = a+b*\imath$ , where $\imath^2 = -1$ . We call this set $\mathbb{C}$ , the set of complex numbers.

Exercise: Verify that $\mathbb{C}$ is a field.
Exercise: Is $\mathbb{C}$ an ordered field? (Use (F) – (H) to prove order or give a counterexample)
(Hard) Exercise: Why have we defined $\imath^2 = -1$ and not $\imath=\sqrt{-1}$ ?

Questions to think about:

Are all fields ordered fields?
Are finite fields ordered?
What do these two above questions (if and when answered) say about ordered fields? (Hint: Infinite number of…)

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