Lecture 2: The real number system, $\mathbb{R}$

Why is it true that $\mathbb{Q}$ is a subfield of $\mathbb{R}$?

Consider the properties of integers

• Every $x \in \mathbb{Q}$ can be written $ x = \frac{p}{q}$, where p and q are integers and at least $p$ or $q$ is odd.
• If $n^2$ is even for $n \in \mathbb{N}$, then $n$ is even.

Proof: There is no rational whose square equals 2.

This proof will be by contradiction, ie suppose $x \in \mathbb{Q}$ and $x^2=2$

So $x^2 = 2 \iff \frac{p^2}{q^2} = 2 \iff p^2 = 2q^2 \implies p^2$ is even

If $p^2$ is even for $p \in \mathbb{N}$, then $p$ is even.

$p=2p'$ for $p' \in \mathbb{Z}$.

$p^2 = 2q^2 \iff (2p')^2 = 2q^2 \iff 2p'^2 = q^2 \implies q^2$ is even.

If $q^2$ is even for $q \in \mathbb{N}$, then $q$ is even.

Contradiction! $p$ or $q$ must be odd. The assertion is true. $\blacksquare$

Definition: A set is an aggregate/collection of (for our purposes, real) numbers.

Examples:

• $S = \{1, 2, 3, \frac{4}{5} \}$
• $S = \mathbb{N}$
• $S = \{x \in \mathbb{Q}$ s.t. $x + 7 = 9 \} $
• $S = \{x \in \mathbb{Q}$ s.t. $x > 7 \} $
• $S = \emptyset$ (empty set)

Here are some different ways to say the same thing:

• $x$ is in $S$
• $x$ is an element of $S$
• $x$ is a member of $S$
• $S$ contains $x$
• $x \in S$
• $S \ni x$

Definition: $M \in \mathbb{R}$ is an upper bound for for S (ie $S$ is bounded above by $M$) iff $\forall x \in S$, $x \subseteq M$. (Another way to say the last part: every element of S is less than or equal to M)

Example: $S = \{x \in \mathbb{Q}$ s.t. $ x \leq 4\}$. The upper bound $M = 4$, or any number greater than 4. $5, 6 ,7, \frac{100}{11}$ are all upper bounds.

Definition: $S$ has a least upper bound (also called L.U.B. or supremum) $L$ iff:

1. $L$ is an upper bound for $S$.
2. if $M$ is another upper bound for $S$, then $L \leq M$.

Here are some different ways to say the same thing:

• $L$ is the least upper bound of $S$
• $L$ is the supremum of $S$
• $L = L.U.B.(S)$
• $L = sup(S)$

Claim:If a supremum $L$ exists, then it is unique.

Proof:

Suppose $L$ and $L'$ satisfy the definition of supremum.

$L$ is an upper bound for $S$

$L'$ satisfies point 2, so $L' \leq L$.

Reverse the roles, and $L \leq L'$.

This implies that $L=L'$ so $L=sup(S)$ is unique. $\blacksquare$

Definition: If $M$ is an upper bound for $S$ and $M \in S$, then $M = max S$. (e.g. $M$ is a maximal element of $S$.)

Claim: if $M = max S$ exists, then it is unique.

Example: $S = \{1 - \frac{1}{n}$ s.t. $n \in N\}$

Recognize that $S$ is bounded above by 1, but $1 \notin S$.

In fact, $sup(S) = 1$, but $max(S)$ does not exist.

Example: A set of rationals may be bounded above such that its supremum (if it exists) is not rational. $S = \{x \in \mathbb{Q}$ such that $x^2 \leq 2\}$. $sup(S) = \sqrt(2)$ but $\sqrt(2) \notin \mathbb{Q}$.

Completeness axiom: If $S$ is a nonempty subset of $\mathbb{R}$ that is bounded above, then $sup(S)$ exists in $\mathbb{R}$.

Note: We will show that the completeness axiom allows to define $\sqrt(2)$ as a real number.

Proposition: Let $S$ be a set, $S \subset \mathbb{R}$, then $L$ is the supremum of $S \iff \forall x \in S$, $x \leq L$

$\forall \epsilon > 0$, $\exists x \in S$ such that $L - \epsilon < x \iff L < x + \epsilon$.

Proof: By contradiction, suppose $L$ = sup $S$

$L$ is an upper bound so (1) is satisfied.

By contradiction, if (2) is not satisfied, then $\exists \epsilon > 0$, $\forall x \in S$, $L = \epsilon \geq x \implies L - \epsilon$ is an upper bound for $S$.

Since $L$ = sup$S$, $L \leq L - \epsilon \iff \epsilon \leq 0$, but this is a contradiction! (2) holds true.

Suppose (1), (2) are true. (1) implies that $L$ is an upper bound.

We want to show that if M is an upper bound for $S$, then $L \leq M$.

Let $M$ be another upper bound for $S$, and by contradiction, assume that $L > M$, set $\epsilon = \frac{L - M}{2} > 0$.

By (2), $\exists x \in S$, $L - \epsilon < x$. $L \iff L - \frac{L}{2} + \frac{M}{2} = \frac{L + M}{2}$.

Because $M$ is an upper bound, $x \leq M$.

We get $\frac{L +M}{2} < x \leq M \implies \frac{L + M}{2} \leq M \implies L + M \leq 2 \implies L \leq M$

And we have another contradiction! So $L$ = sup$S$

Define $\sqrt(2)$: Let $S = \{x \in \mathbb{R}$ s.t. $x > 0$ and $x^2 \leq 2\}$. That is, $S$ is bounded above by 2. $x^2 \leq 2 \leq 4 = 2^2$ so $x^2 - 2^2 \leq 0$.

Since $x + 2 > 0$, then $x -2 \leq 0 \implies x \leq 2$

By the completeness axiom, sup $S$ exists. Call it $C$. Can prove that $c^2 = 2$ and $c > 0$. This defines $c$ as "$\sqrt(2)$"

More generally, for any $a > 0$, $a \in \mathbb{R}$, we can define:

• $\sqrt(2)$ = sup$\{s \in \mathbb{R}$ such that $x > 0, x^2 \leq a\}$
• $\sqrt[p]{a}$ = sup$\{s \in \mathbb{R}$ such that $x > 0, x^p \leq a\}$ for $p \in \mathbb{N}$

The construction of these only depends on field operators. The completeness axiom allows us to define algebraic roots as real numbers.