Lecture A1

• Ordered Sets
• Fields
• The Real Field

Notation

• Set of integers:

$$\mathbb{Z}:=\{0,\pm 1,\pm 2,\pm 3,\cdots \}$$

$$\mathbb{N}:=\{1,2,3,4,\cdots \}$$

• Rationals

$$\mathbb{Q}:=\{p/q:p,\in \mathbb{Z},q\in \mathbb{N}\}$$

• Real numbers that are not rational

$${\mathbb{Q}}^c:=\mathbb{R}\backslash \mathbb{Q}$$

Lemma

The equation $x^2=2$ does not have any rational solutions.

Proof:

Suppose by contradiction that there is a rational solution. That is,

$$\exists \frac{p}{q}\in \mathbb{Q}$$

such that

$${\left(\frac{p}{q}\right)}^2=2$$

WLOG, we can assume that not both $p,q$ are even (otherwise you could reduce it WLOG). Consider

$${\left(\frac{p}{q}\right)}^2=2⟺\frac{p^2}{q^2}=2⟺p^2=2q^2$$

Which implies that $p^2$ is even (since we multiply by $2$) and thus $p$ is even (by contrapositive, $p\text{ odd}⟹p^2\text{ odd}$). So,

$$\begin{array}{rl}{p}^2=2q^2& ⟺(2k{)}^2=2q^2\\ & ⟺2k^2=q^2\\ & ⟹q^2\text{ is even}\\ & ⟹q\text{ is even}\end{array}$$

which is a contradiction.

Supremum

Ordered Sets