Real Analysis – Basics II

— 2. Order Axioms —

Having studied the 5 previous axioms and seeing that they’re enough to prove all the results people are familiar with in common arithmetic were are now entitled to study a new set of axioms. The next group of axioms is known as the Order Axioms.

Axiom 6 The set of the positive real numbers is a subset of the real numbers which is closed for the { +} and { \cdot} operations.

\displaystyle   \mathbb{R}^{+} \subset \mathbb{R}: \forall x,y \in \mathbb{R}^{+} x+y \in \mathbb{R}^{+} \land x\cdot y \in \mathbb{R}^{+} \ \ \ \ \ (3)

We can now define negative numbers as the opposite of a positive real number.

Definition 3 There exists a set of numbers which will call negative numbers: { \mathbb{R}^{-}}. A number is said to be negative whenever its opposite is positive.

\displaystyle   x\in \mathbb{R}^{-} \Leftrightarrow -x \in \mathbb{R}^+ \ \ \ \ \ (4)

Axiom 7 Any real number different than { 0\,} either is positive or negative.

\displaystyle   x \in \mathbb{R} \setminus \{0\} \Rightarrow x \in \mathbb{R}^{+} \lor \mathbb{R}^{-} \ \ \ \ \ (5)

In an equivalent formulation one can also say: No real number is positive and negative at the same time { \mathbb{R}^{+} \cap \mathbb{R}^{-}=\emptyset }

A formal definition of the symbols { > } and { < } is now possible.

Definition 4 Let { x,y\in\mathbb{R}}. We say that { x} is lesser than { y} if and only if { y-x \in \mathbb{R}^{+}} and write it as { x < y} or { y > x}.
Theorem 6 (Trichotomy) { \forall x,y \in \mathbb{R}} one and only one of the following conditions is always verified: { x > y}, { x < y}, { x=y}. Proof: Omitted. \Box
Theorem 7 (Transitivity) { \forall x,y,z \in \mathbb{R}\quad x < y;\, y < z \Rightarrow x < z} Proof: Omitted. \Box

Any set where the the trichotomy and the transitivity properties exist is said to be an ordered set. So, we now know that { \mathbb{R}} is an ordered field.

Theorem 8 Let {x,y \in \mathbb{R}} and {x < y}. Then:

  • { z \in \mathbb{R} \Rightarrow x+z < y+z}
  • { u,v \in \mathbb{R}:\quad u < v \Rightarrow x+u < y+v}
  • { z > 0 \Rightarrow x\cdot z < y\cdot z} and { z < 0 \Rightarrow x\cdot z > y\cdot z}

Proof: Omitted. \Box

In axiom 4 it was said that { 0\,} and { 1} are distinct elements. By the law of trichotomy we must have { 0 < 1 } or { 0 > 1}. But which one of these two options is the valid one?

Let’s assume that { 1 < 0} is the valid proposition. In that case multiplying both sides of the inequality by { 1} we would have:

{ 1\cdot 1 > 1\cdot 0 \Rightarrow 1 > 0} which is contrary to what we initially assumed. Since all logical steps after the assumption are valid we have to conclude that the initial assumption was at fault and that {1 > 0}.

Usually the set of the natural numbers is denoted by { \mathbb{N}}, the integer numbers by { \mathbb{Z}}, and the rational numbers by { \mathbb{Q}}. And the following inclusions are valid: { \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}}.

Not wanting to be too rigorous about it we can see that not all integer numbers are natural numbers. That not all rational numbers are integer numbers. But is there any difference between the set { \mathbb{Q}} and the set { \mathbb{R}}? Yes there is!

For instance { x=\sqrt{2} \in \mathbb{R}} but { x=\sqrt{2}\notin \mathbb{Q}}! Mathematicians knew of this fact for a long time, but they just couldn’t formally differentiate the two sets. All of this changed when the Completeness Axiom came along. This new insight allowed mathematicians to put the set of the real numbers on a firm ground.

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