04 - Axioms for the usual ordering on ℝ

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

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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Results on inequalities

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

1. $0<1$
2. $a<b$ if $-b<-a$
   In particular, $a>0$ if and only if $-a<0$
3. If $a<b$ then $a+c<b+c$
4. If $a<b$ and $0<c$ then $ac<bc$
5. $a^2\ge 0$, with equality if and only if $a=0$
6. $a>0$ if and only if $\frac{1}{a}>0$
7. If $a,b>0$ and $a<b$ then $\frac{1}{b}<\frac{1}{a}$

Note that 1, 2, 3 hold when using $\le$ instead of $<$

Proof

1. By trichotomy then either $0<1$ or $0=1$ or $0>1$
   By Axioms for Arithmetic (To Avoid Total Collapse) then $0\ne 1$ so either

$$0<1\text{ or }1>0$$

Suppose for a contradiction that $0>1$ then

$$(-1)\in \mathbb{P}$$

Hence by $\cdot$ and ordering

$$(-1)\cdot (-1)\in \mathbb{P}$$

However $(-1)\cdot (-1)=1\in P$ (Properties of arithmetic (12))
So $0<1$ but this contradicts trichotomy hence

$$0<1$$

2. Using properties of addition

$$\begin{array}{rl}a<b& ⟺b-a\in \mathbb{P}\\ & ⟺(-a)-(-b)\in \mathbb{P}\\ & ⟺-a>-b\end{array}$$

3. Assume that $a<b$

$$(b+c)-(a+c)=b-a>0\text{so}a+c<b+c$$

4. Assume that $a<b$ and $0<c$

$$bc-ac=b(a-c)>0(\cdot \text{ and ordering})$$

5. Naturally $a^2=0⟺a=0$ by Properties of Arithmetic (10 & 13)
   If $a\ne 0$ then by trichotomy then either $a>0$ or $-a>0$
   Either way $a^2=a\cdot a=(-a)\cdot (-a)>0$($\cdot$ and ordering)
6. Suppose for a contradiction that $a>0$ and $\frac{1}{a}<0$, hence $-\frac{1}{a}>0$
   Then

$$-1=-\left(a\cdot \frac{1}{a}\right)=a\cdot \left(-\frac{1}{a}\right)>0$$

This contradicts $(1)$ hence $a>0$ if $\frac{1}{a}>0$
Similarly if $a<0$ and $\frac{1}{a}>0$ then $-\frac{1}{a}<0$

$$-1=-\left(a\cdot \frac{1}{a}\right)=a\cdot \left(-\frac{1}{a}\right)>0$$

This contradicts $(1)$ hence

$$a>0⟺\frac{1}{a}>0$$

7. Suppose that $a,b,>0$ and $a<b$
   Then $\frac{1}{a},\frac{1}{b}>0$ by $(6)$
   So

$$a\cdot \frac{1}{a}\cdot \frac{1}{b}<b\cdot \frac{1}{a}\cdot \frac{1}{b}(\text{by }(4))$$

Hence

$$\frac{1}{b}<\frac{1}{a}$$