Honors Calculus - 2

Number Systems and Basic Funtions

To prove $A=B$, we can prove $A \subset B$ and $B \subset A$

Number Sets

• Field
• Order Field

Interval Testing Method + Square Method

A set is bounded if it has both upper and lower bounds.

• supremum : $\beta = \sup E$
• infimum : $\alpha = \inf E$

least-upper-bound property or Completeness axiom :

for any nonempty subset $E$ of $S$, if $E$ is bounded above, then sup $E$ exists in $S$

Proof $\gamma=\sup E$

1. $\forall p\le \gamma$
2. $\exist q>\gamma - \varepsilon$ (use $\varepsilon$ to describe parameter)

So $R$ ais has least-upper-bound property, while $Q$ has not.

Also : An ordered set with the least-upper-bound property also has the greatest-lower-bound property.

Extended real number system : $\overline{\mathbb{R}}=\mathbb{R}\cup\{-\infty,\infty\}$ (It is not a field)

Complex Numbers

• $(a,b)\times (c,d)=(ac-bd,ad+bc)$
• Complex field
  • Not an ordered field
  • $\mathbb{R}$ is an subfield of it
• $i=(0,1)$ is a unit imaginary number
• $z=a+bi$, where $a$ is called real part, $b$ is called imaginary part. Let $a=\text{Re}(z),b=\text{Im}(z)$
• $\overline{z}=a-bi$ is the conjugate of $z$
• $|z|=(z\overline z)^{1/2}$ is the modulus, or absolute value
• Triangle inequality : $|z+w|\le|z|+|w|$

Very Important :

$$|y_1-y_3|\le|y_1-y_2|+|y_2-y_3|$$

$$|x|\le|y|+|x-y|$$

$$||x|-|y||\le |x-y|$$