Definition of an ordered field: An ordered field is a field containing a subset of elements closed under addition and multiplication and having the property that every element in the field is either 0, in the subset, or has its additive inverse in the subset.
Before going through ordered fields it is important to note what an axiom is, just in case: An axiom is a statement that is held to be self-evidently true. This, these ordered fields are, by definition, all axioms.
Examples of ordered fields We will begin with the ones for addition:
A1. For all x, y ∈ R, x + y ∈ R and if x = q and y = z, then x + y = w + z
A2. For all x, y ∈ R, x+y=y+x
A3. For all x,y,z ∈ R, x+(y+z) = (x+y)+z
A4. There is a unique real number 0 such that x+0=x for all x ∈ R
A5. For each x ∈ R, there i a unique real number -x such that x+(-x) =0
Now here are the ones for multiplication
M1. For all x,y ∈ R, x⋅y ∈ R and if x=w and y=z, then x⋅y=w⋅z
M2. For all x,y ∈ R, x⋅y=y⋅x
M3. For all x,y,z ∈ R, x⋅(y⋅z)=(x⋅y)⋅z
M4. There is a unique real number 1 such that 1 =/= 0 and x⋅1 = x for all x ∈ R
M5. For each x ∈ R with x =/= 0 there is a unique real number 1/x such that x⋅(1/x)=1.
DL. For all x,y,z ∈ R, x⋅(y+z) x⋅y + x⋅z
These 11 axioms are called "field axioms" because they describe something called a "field" in algebra. Things like A2 and M2 are called commutative laws while A3 and M3 are called associative laws. DL stands for Distributive Law. Thanks to A1 and M1, we can think of addition and multiplication as functions that maps RxR into R.
The other basic operations are as follows: x-y is x+(-y), x/y is defined as x⋅1/y, 1+1=2 and x⋅x=x²
In addition to the field axioms, the real numbers also satisfy 4 order axioms. These axioms identify the properties of the relation "<". As we all learned some many years ago, we can write y>x instead of x<y and x≤y is an abbreviation of "x<y or x=y". The notation "≥" is abbreviated analogously. A number is said to be nonnegative if x≥0 and positive if x>0. Here are the 4 order axioms
O1. For all x,y, ∈ R, exactly one of the relations x=y, x>y, or x<y holds (trichotomy law)
O2. For all x,y,z ∈ R, if x<y and y<z, then x<z.
O3. For all x,y,z ∈ R, if x<y, then x+z < y+z
O4. For all x,y,z ∈ R, if x<y and z>0, then xz < yz
The following theorem illustrates how the axioms may be used to derive some familiar algebraic properties
Let x, y, and z be real numbers
T1. If x+z = y+z, then x = y
T2. x ⋅ 0 = 0
T3. -0 = 0
T4. (-1) ⋅ x = -x
T5. xy = 0 iff x=0 or y=0
T6. x<y iff -y < -x
T7. If x<y and z<0, then xz > yz
Now, let's attempt to apply these to a few examples.
- 1) xy = 0 iff x = 0 or y = 0
Proof: If x = 0 or y = 0, then xy = 0 by M2. Conversely, suppose xy=0 and x=/=0. Then, y = 0. Since x=/=0, 1/x exists by M5. Thus,
0 = 1/x ⋅ 0 T2
0 = 1/x ⋅ (xy) b/c xy = 0
0 = (1/x ⋅x) ⋅ y by M3
0 = (x ⋅ 1/x) ⋅ y M2
0 = 1⋅y M5
0 = y M2 and M4
- 2) x < y and z<0, then xz > yz
Proof: Suppose x<y and z<0. Then, -z>-0 by T6. And -z > 0 by T3. Thus, x(-z) > y(-z) by O4. But
x(-z) = x[(-1)(z)] T4
= [x(-1)](z) M3
= [(-1)(x)](z) M2
= (-1)(xz) M M3
= -xz T4
Similarly, y(-z) = -yz. Thus, -xz < -yz. But then, yz < xz by T6
Any mathematical system that satisfies these 15 axioms are called an ordered field. Thus, the real numbers are an example of an ordered field. But there are other example, specifically with rational number Q are also an ordere pairs, because
Q = {m/n : m, n ∈ Z and n=/= 0}
In particular, we can use ordered fields in a more unorthodox way where we let F be the set of all rational numbers functions, where F is the set of all quotients of polynomials. We can define an order on F by saying a quotient is positive iff the quotient is greater than 0. For example
(3x^2 + 4x - 1)/(7x^5 + 5) >0
In this case, (3)(7) > 0. However, for the next example,
(4x^5 + 3x - 2)/(-7x^3 + 6x^2) < 0
Because (4)(-7) < 0
If p/q and f/g are rational functions, we say p/q > f/g iff p/q - f/g > 0
or (pg - fq)/qg >0
We can apply this to the next 2 problems:
- 1) Which is larger, x^2 or 3/x?
x^2 / 1 - 3/x = (x^3 - 3)/x > 0, so x^2 > 3/2
- 2) Which is larger, x/(x+2) or x/(x+1)?
x/(x+2) - x/(x+1) = -x/(x^2+3x+2)<0, so x/(x+2) < x/(x+1)
Lastly, we will denote the definition of an absolute value. If x∈R, then the absolute value of x, denoted by |x|, is defined by
|x| = { x if x≥0 or -x if x≤0 }
The basic properties of absolute value are summarized in the following theorem:
Let x,y∈R and let a≥0, Then
(a) |x| ≥ 0
(b) |x| ≤ a iff -a≤x≤a
(c) |xy| = |x| ⋅ |y|
(d) |x+y| ≤ |x| + |y|
