Order Axioms

• Trichotomy: $\forall a,b \in \mathbb{R}$ exactly one of these holds: $a > b$, $a = b$, $a<b$
• Transitivity: $\forall a,b,c \in \mathbb{R}; a<b \land b<c \implies a<c$
• Operation with addition: $\forall a,b \in \mathbb{R}; a<b \implies a+c<b+c$
• Operation with mutliplication: $\forall a,b,c \in \mathbb{R}; a<b \land 0<c \implies ac<bc$

Definitions

• $a < b \equiv b > a$
• $a \le b \equiv a<b\lor a=b$
• $a\not =b\equiv a<b\lor a>b$
• $\lvert{x}\rvert=\begin{cases}x & \text{if } x \ge 0, \\-x & \text{if } x < 0\end{cases}$

Triangular inequalities

$$\lvert{a}\rvert - \lvert{b}\rvert \le \lvert{a+b}\rvert \le \lvert{a}\rvert + \lvert{b}\rvert$$ $$\Big\lvert{ \lvert{a}\rvert - \lvert{b}\rvert } \Big\rvert \le \lvert{a+b}\rvert$$

Proof Hint

For first:

• Use $-|a| \le a \le |a|$

For second:

• Use the below substitutions in first conclusion
  • $a=a-b \;\;\land\;\; b = b$
  • $a=b-a \;\;\land\;\; b=a$

Required proofs

• $\forall a,b,c \in \mathbb{R}; a<b \land c<0 \implies ac>bc$
• $1 > 0$
• $-\lvert{a}\rvert\le a\le\lvert{a}\rvert$
• Triangular inequalities

Theorems

• $\exists{a}\;\forall{\epsilon>0},\,a<\epsilon\implies{a}\le{0}$
• $\exists{a}\;\forall{\epsilon>0},\,0\le{a}<\epsilon\implies{a}={0}$
• $\forall\epsilon>0\;\exists a,a<\epsilon\centernot\implies a\le 0$

Caution

$\forall{\epsilon>0}\;\exists{a},\,a<\epsilon\implies{a}\le{0}$ is not valid.$$