0 - Ordered Sets, Suprema, Infima

The field of rational numbers are problematic.
As for construction, one can do it by defining equivalence classes on $Z^{2}$ , where $(a, b) \equiv (p, q)$ if $a q = b p$ . If we let $[a, b]$ denote the equivalence class of $(a, b)$ under the above relation, addition is defined as $[a, b] + [p, q] = [a q + b p, b q]$ , multiplication as $[a, b] [p, q] = [a p, b q]$ (both stand on addition and multiplication of integers), then It's not difficult to check that this is the familiar field of the rational numbers.

But we won't do this rational numbers as equivalence classes of $Z^{2}$ nonsense here.

\sqrt{2}

is irrational

There is no $p \in Q$ such that $p^{2} = 2$ .

$* p r o o f * :$
For the sake of contradiction, suppose there is a rational number $p$ for which $p^{2} = 2$ . Now, $p = \frac{a}{b}$ where $G C D (a, b) = 1$ . Then, we have $a^{2} = 2 b^{2}$ , which means $2 | a^{2}$ hence $2 | a$ . writing $a$ as $2 m$ , we have $2 m^{2} = b^{2}$ hence $2 | b$ as well. Therefore, $G C D (a, b) \geq 2$ . Arriving at the desired contradiction.

$◻$

The gap at

\sqrt{2}

in the rationals

Let $A = {p \in Q : p^{2} < 2}$ , $B = {p \in Q : p^{2} > 2}$ . Then $A$ has no largest element, and $B$ has no smallest element.

$* p r o o f * :$ Allow $q = p - \frac{(p^{2} - 2)}{p + 2} = \frac{2 p + 2}{p + 2}$ . Then, $q^{2} - 2 = \frac{2 (p^{2} - 2)}{(p + 2)^{2}}$ .

Now Suppose $p$ is the maximal element of $A$ . Using the rightmost equation, given that $p^{2} < 2$ , we see that $q^{2} - 2 < 0$ Hence $q \in A$ . However, using, the left most equation, we have that $q > p$ , hence we arrive at a contradiction.

Now suppose $p$ is the minimal element of $B$ . Again, using the rightmost equation, given that $p^{2} > 2$ , indeed, $q^{2} - 2 > 0$ Hence $q \in B$ .
However, using the leftmost equation, we have $q < p$ , arriving at a contradiction.

$◻$

Holes in the rational numbers

Even though the rational numbers are "dense" in the sense that between any two $a, b \in Q$ , we have $\frac{a + b}{2} \in Q$ , it seems that there are holes, certain equations have no solutions in $Q$ , moreover, it seems like we can get arbitrarily close to these holes from the left and from the right, as illustrated for the hole at " $\sqrt{2}$ ".

Ordered sets

A set $S$ with a relation $<$ on $S$ is called an orderd set if:

For each $x, y \in S$ exactly one of $x < y, x = y, y < x$ is true. This is often called the Trichotomy law.
If $x < y$ and $y < z$ for some $x, y, z \in S$ , then $x < z$ , meaning that order is transitive.
Sometimes, we write $x > y$ instead of $y < x$ , and $x \leq y$ is used to denote $x < y$ or $x = y$ .

Upper and Lower bounds

For the following discussion, let $S$ denote an ordered set.

A subset $E$ of $S$ is said to be bounded above if there exists $b \in S$ such that for all $e \in E, e < b$ . In such a case $b$ is called an upper bound for $E$ .

If $γ$ is an upper bound for $E$ such that for any upper bound $b$ of $E$ , $γ < b$ , then $γ$ is known as the least upper bound or supremum of $E$ . Often denoted $γ = sup E$ .

A subset $F$ of $S$ is said to be bounded below, if there exits $c \in S$ such that for all $f \in F, f > c$ . In such a case, $c$ is called a lower bound of $F$ .

if $β$ is a lower bound of $F$ such that for any lower bound $c$ of $F$ , $β > c$ , then $β$ is known as the greatest lower bound or infimum of $F$ . Often denoted $β = inf F$ .

A subset $P$ of $S$ is said to be bounded, if it is both: bounded above and below.