04. Ordering the real numbers

Axioms for the usual ordering on $\mathbb{R}$

There is a subset $\mathbb{P}$ of $\mathbb{R}$ such that for $a,b\in \mathbb{R}$

1. $+$ and ordering: If $a,b\in \mathbb{P}$ then $a+b\in \mathbb{P}$
2. $\cdot$ and ordering: If $a,b\in P$ then $a\cdot b\in \mathbb{P}$
3. positive, negative or $0$: Exactly one of $a\in \mathbb{P},a=0$ and $-a\in \mathbb{P}$ holds

Note that the elements of $\mathbb{P}$ are the positive numbers

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Inequalities with axioms of usual ordering on $\mathbb{R}$

Note that elements of $\mathbb{P}\cup \{0\}$ are the non-negative integers

1. $a<b$ or $b>a$ exactly when $b-a\in \mathbb{P}$
2. $a\le b$ or $b\ge a$ exactly when $b-a\in P\cup \{0\}$

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Properties of axioms of ordering

Take $a,b,c\in \mathbb{R}$ thhen

1. Reflexivity: $a\le a$

2. Antisymmetry: If $a\le b$ and $b\le a$ then $a=b$

3. Transitivity: If $a\le b$ and $b\le c$ then $a\le c$
   Works with $<$ so, If $a<b$ and $b<c$ then $a<c$

4. Trichotomy: Exactly one of $a<b,a=b$ and $a>b$ holds

Proof

1. We have $a-a=0\in P\cup \{0\}$ (additive inverse)
2. Suppose that $a\le b$ and $b\le a$
   If $a-b=0$ or $b-a=0$ then $a=b$ (properties of $+$) then that is what we needed
   If not, then $b-a\in P$ and $a-b\in P$
   But $b-a=-(a-b)$ (properties of $+$)
   So $a-b\in \mathbb{P}$ and $-(a-b)\in \mathbb{P}$, contradicting the positive, negative or $0$ ($3^{\text{rd}}$ axiom)
3. Note that $c-a=c+(-a)=c+0+(-a)=c+(-b)+b+(-a)=(c-b)+(b-a)$
   Using properties of $+$
   So if $a<b$ and $b<c$ then $a<c$ ($+$ and ordering)
   If $a=b$ and or $b=c$ then it is straightforward and the results for $\le$ follows
4. Follows from positive, negative or 0 axiom ($3^{\text{rd}}$ axiom)

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