NOTE: ordered field
Field is a nonempty set \(\mathbb{F}\) with two binary operations
- addition: \(+\)
- multiplication: \(\cdot\)
satisfying the following five axioms.
- Commutative Law: If \(a, b \in \mathbb{F}\), then \(a + b = b + a\) and \(a \cdot b = b \cdot a\)
- Distributive Law: If \(a, b, c \in \mathbb{F}\), then \(a \cdot (b + c) = a \cdot b + a \cdot c\)
- Associative Law: If \(a, b, c \in \mathbb{F}\), then \((a + b) + c = a + (b + c)\) and \((a \cdot b) \cdot c = a \cdot (b \cdot c)\)
- Identity Law: There are special elements \(0, 1 \in \mathbb{F}\), where \(a + 0 = a\) and \(a \cdot 1 = a\) for all \(a \in \mathbb{F}\)
- Inverse Law: For each \(a \in \mathbb{F}\), there is an element \(-a \in \mathbb{F}\) such that \(a + -a = 0\). If \(a \neq 0\), then there is also an element \(a^{-1} \in \mathbb{F}\) such that \(a \cdot a^{-1} = 1\).
0.1. examples
- Natural numbers \(\mathbb{N}\) didn't form a field, it didn't satisfy \(a + (- a) = 0\) since there has no \(-a \in \mathbb{N}\) for any \(a \in \mathbb{N} \gt 0\)
- Integers \(\mathbb{Z}\) is not a field, it didn't satisfy \(a \cdot a^{-1} = 1\), for example, where \(a = 2\), then \(a^{-1} = \frac{1}{2}\), but \(\frac{1}{2} \notin \mathbb{Z}\)
- \(\mathbb{Q}\) form a field
1. positive set
With an additional axiom, you will get an ordered field \(\mathbb{F}\)
1.0.1. Order Axiom
There is a nonempty subset \(P \subseteq \mathbb{F}\), called the positive elements, such that
- If \(a, b \in P\), then \(a + b \in P\) and \(a \cdot b \in P\)
- If \(a \in \mathbb{F}\) and \(a \neq 0\), then \(a \in P\) or \(-a \in P\)
2. Metric properties
Since positive elements are all you need to define the absolute value, it give metric properties(distance function). You can define inequality formally now.
2.0.1. Inequality in Ordered Field
If \(\mathbb{F}\) is an ordered field and \(a, b \in \mathbb{F}\), then we say \(a < b\) if \(b - a \in P\), where \(a \leq b\) is \(a = b\) or \(a < b\)
3. Standard properties of inequalities
- If \(a < b\), then \(a + c < b + c\)
- Transitivity: If \(a < b\) and \(b < c\), then \(a < c\)
- If \(a < b\), then \(ac < bc\) if \(c > 0\), else \(ac > bc\)
- If \(a \neq 0\), then \(a^2 > 0\)
3.1. Proofs
3.1.1. First: If \(a < b\), then \(a + c < b + c\)
By definition of inequality, we know the following conversion
\[ a + c < b + c \\ \to (b+c) - (a+c) \in P \\ \to b + c - a - c = b - a \]
And we already know \(b - a \in P\)
Q.E.D.
3.1.2. Second: If \(a < b\) and \(b < c\), then \(a < c\)
By definition of inequality, \(a < c\) mean \(c - a \in P\)
\[ c - a \\ \to (c - b) + (b - a) \]
We know \(c - b \in P\) and \(b - a \in P\), by first rule of Order Axiom, we know
If \(a, b \in P\), then \(a + b \in P\)
So \(c - a = (c - b) + (b - a) \in P\)
Q.E.D.
3.1.3. Third: If \(a < b\), then \(ac < bc\) if \(c > 0\), else \(ac > bc\)
By definition of inequality, we know
If \(c > 0\)
\[ ac < bc = bc - ac \in P \]
In this case, \(bc - ac = c (b - a)\), we know
- \(c \in P\)
- \(b - a \in P\) since \(a < b\)
Thus, by first rule of Order Axiom, \(c (b - a) \in P = ac < bc\)
If \(c < 0\)
\[ ac > bc = ac - bc \in P \]
In this case, \(ac - bc = c (a - b)\), we know
- \(c \notin P\), so \(-c \in P\), by second rule of Order Axiom
- \(a - b \notin P\) since \(a < b\), so \(-(a - b) \in P\)
Thus, by first rule of Order Axiom, \(-c \cdot -(a - b) \in P\)
And \(-c \cdot -(a - b) \in P = c (a - b) \in P = ac - bc \in P = ac > bc\)
Q.E.D.
3.1.4. Fourth: If \(a \neq 0\), then \(a^2 > 0\)
- If \(a \in P\), by first rule of Order Axiom, \(a \cdot a \in P\)
- If \(a \notin P\), by first rule of Order Axiom, \(-a \cdot -a \in P\), and \(-a \cdot -a = a \cdot a\)
Q.E.D.