📂Analysis

Three Axioms of Analysis: The Second Order Axiom

Axioms¹

For real numbers $ a,b,c \in \mathbb{R}$, the following properties are accepted:

Trichotomy: For any given $a,b$, it must be that $a<b$ or $a>b$ or $a=b$

Transitivity: If $a<b$ and $b<c$ then $a<c$

Additivity: If $a<b$ and $c\in \mathbb{R}$ then $a+ c< b + c$

Multiplicativity: If $a<b$ and $c>0$ then $ac< bc$, or if $c<0$ then $ac> bc$

Description

Although the terms are quite old-fashioned, they deal with such basic truths that understanding them should pose no problem. While field axioms concern numbers and operations, order axioms deal with the magnitude relationships between numbers.

In a way, we might be more familiar with order axioms than with field axioms. In analysis, they are the second set of axioms, but we learn inequalities (order axioms) in elementary school and irrational numbers (field axioms) in middle school.

Field axioms alone don’t allow for much, but with the addition of order axioms, new possibilities arise. For example, the following theorem can be proved.

Theorem

The power of a non-zero real number is always positive.

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It might seem pointless to prove such things. However, historically, scholars did not first create axioms to develop proofs from the ground up; rather, they tried to prove everything and those they could not prove, they grouped together and called axioms.

That is, axioms were not created to facilitate these proofs, but were found necessary in the process of attempting to prove these concepts. Meanwhile, precisely proving and rigorously verifying everything is, from a mathematician’s standpoint, a matter of course. So, don’t feel too aggrieved, and try to study joyfully, imagining you’re laying the groundwork step by step.

Proof

By trichotomy, if $a \ne 0$, then there are only two possibilities: $a>0$ or $a<0$.

• Case 1.

  Multiplying both sides of $a>0$ by $a$, by multiplicativity

  $$ a^2>a \cdot 0 $$

• Case 2.

  Multiplying both sides of $a<0$ by $a$, by multiplicativity

  $$ a^2>a \cdot 0 $$

Given that multiplying any real number by $0$ yields $0$, in either case

$$ a^2 > a \cdot 0=0 $$

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