Real Numbers

• set of infinite decimal expansions
• $a_0.a_1{a}_2{a}_3\dots$, where $a_0\in \mathbb{Z}$ and $0\le a_i\le 9$ for $i\ge 1$.
• ‘fills in number line’
• $0.\bar{9}=1$

Field Axioms §

On $\mathbb{R}$, there are 2 binary operations $+$ and $\cdot$ such that

[!math|{“type”:“axiom”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Commutativity”,“label”:“commutativity”,“_index”:0}] Axiom 1 (Commutativity). $x=y=y+x$, $xy=yx$.

[!math|{“type”:“axiom”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Associativity”,“label”:“associativity”,“_index”:1}] Axiom 2 (Associativity).

$$\begin{array}{c}x+(y+z)\ge (x+y)+z,\\ x(yz)=(xy)z\end{array}$$

[!math|{“type”:“axiom”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Distributivity”,“label”:“distributivity”,“_index”:2}] Axiom 3 (Distributivity).

$$x(y+z)=xy+xz$$

[!math|{“type”:“axiom”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Existence of $0,1$”,“label”:“existence-of-01”,“_index”:3}] Axiom 4 (Existence of $0,1$).

$$\begin{array}{c}0+x=0,\\ 1\cdot x=x\end{array}$$

[!math|{“type”:“axiom”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Existence of Negatives”,“label”:“existence-of-negatives”,“_index”:4}] Axiom 5 (Existence of Negatives).

$$\forall x\in \mathbb{R},\exists y\mid y+x=0$$

This is the definition of $-x$.

We may now define

[!math|{“type”:“definition”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Comparator”,“label”:“comparator”,“_index”:5}] Definition 6 (Comparator).

$$\begin{array}{c}x<y⟺y-x\in {\mathbb{R}}_{+}\\ x>y⟺y<x⟺x-y\in {\mathbb{R}}_{+}\end{array}$$

In particular, $x>0$ iff $x\in {\mathbb{R}}_{+}$.

[!math|{“type”:“lemma”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Transitivity of Comparator”,“label”:“transitivity-of-comparator”,“_index”:6}] Lemma 7 (Transitivity of Comparator). If $a<b$ and $b<c$, then $a<c$. \begin{proof} $a<b$ and $b<c$ mean that $b-a\in {\mathbb{R}}_{+}$ and $c-b\in {\mathbb{R}}_{+}$.

But, by ,

$$\begin{array}{c}(c-b)+(b-a)\in {\mathbb{R}}_{+}\\ ⟹c-a\in {\mathbb{R}}_{+}\\ ⟹a<c\end{array}$$

\end{proof}

[!math|{“type”:“lemma”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:7}] Lemma 8. If $a<b$ and $c>0$, then $ac<bc$. \begin{proof} $b-a>0$ and $c>0$ implies by \ref that $(b-a)c>0⟹bc-ac>0⟹ac<bc$. \end{proof}

[!math|{“type”:“lemma”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:8}] Lemma 9.

$$a\ne 0⟹a^2>0$$

\begin{proof} If $a>0$, then $a^2>0$ by previous lemma.

Assume $a^2\le 0$. Then, because $a\ne 0$, we must have $a<0$. However, that implies by axiom 8 that $-a>0⟹(-a{)}^2>0⟹a^2>0$, a contradiction. \end{proof}

[!math|{“type”:“definition”,“number”:“auto”,“setAsNoteMathLink”:false,“label”:“upper-bound”,“_index”:9,“title”:“Upper Bound”}] Definition 10 (Upper Bound). Suppose $S$ is a nonempty subset of $\mathbb{R}$ and $B\in \mathbb{R}$ is such that $x\le B$ for all $x\in S$. Then, we can say $B$ is an upperbound for $S$. If $B\in S$, we call B the maximum of $S$.

[!math|{“type”:“definition”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Least Upper Bound”,“label”:“least-upper-bound”,“_index”:10}] Definition 11 (Least Upper Bound). $B\in \mathbb{R}$ is a least upper bound for a nonempty set $S$ if

1. $B$ is an upper bound for $S$.
2. No number less than $B$ is an upper bound for $S$

[!math|{“type”:“theorem”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Unique upper bound”,“label”:“unique-upper-bound”,“_index”:11}] Theorem 12 (Unique upper bound). The least upper bound for a set $S$ is unique \begin{proof} If there are 2 least upper bounds, $B_1$ and $B_2$, then $B_1\le B_2$ and $B_2\le B_1$ $⟹B_1=B_2$. \end{proof}

[!math|{“type”:“definition”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Supremum and Infimum”,“label”:“supremum-and-infimum”,“_index”:12}] Definition 13 (Supremum and Infimum). If it exists, the least upper bound of a subset $S$ of set $P$ is called the supremum of $S$, denoted $\text{sup}(S)$. The greatest lower bound is the infimum, denoted $\text{inf}(S)$.

[!math|{“type”:“axiom”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Continuitiy Axiom”,“label”:“continuitiy-axiom”,“_index”:13}] Axiom 14 (Continuitiy Axiom). Every nonempty set $S$ of $\mathbb{R}$ with an upper bound has a least upper bound or supremum.

[!math|{“type”:“theorem”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:14}] Theorem 15. The natural number $\mathbb{N}$ are unbounded above. \begin{proof}Suppose instead that $\mathbb{N}$ is bounded above. Then, by continuity axiom, it has a supremum $B$. So $B-1$ is not an upper bound, so there is some $n\in \mathbb{N}\mid n>B-1$. But then $n+1\in \mathbb{N}$ has $n+1>B$, contradicting that $B$ is an upper bound.\end{proof}

[!math|{“type”:“theorem”,“number”:“auto”,“setAsNoteMathLink”:false,“label”:“archimedean-property”,“_index”:15,“title”:“Archimedean Property”}] Theorem 16 (Archimedean Property). If $x>0$ any $y\in \mathbb{R}$, $\exists n\mid nx>y$. \begin{proof}If not, then $\frac{y}{x}\ge n$ for all $n\in \mathbb{N}$, so $\mathbb{N}$ would be bounded above.\end{proof}

[!math|{“type”:“theorem”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:”$sqrt2$”,“label”:“sqrt”,“_index”:16}] Theorem 17 ($\sqrt{2}$). 2 has a positive square root. \begin{proof}Let $S=\{x\in \mathbb{R}\mid x^2\le 2\}$. Consider $\text{sup}(S)$. We claim that $\text{sup}(S{)}^2=2$, so we need to disprove that $\text{sup}(S{)}^2>2$ and $\text{sup}(S{)}^2<2$. But first, we need to show that $\text{sup}(S)$ exists. To see that $S$ is non-empty, note that $0^2\le 2$. To see that $S$ is bounded above, note $2^2=4>2$. Thus, by Continuity Axiom, $\text{sup}(S)$ does exist.
Let $B=\text{sup}(S)$. Case A: $B^2>2$. Let $C=\frac{1}{2}(B+\frac{2}{B})$. Then, $C<B$. We can prove this by starting with $\frac{1}{2}\left(B+\frac{2}{B}\right)<B⟺\frac{1}{B}<\frac{B}{2}⟺2<B^2$. But also, $C^2=\frac{1}{4}(B^2+4+\frac{4}{B^2}=2+\cdots >2$. Therefore, $C$ is a smaller upper bound for $S$, contradicting the definition of $B=\text{sup}(S)$. Case B: $B^2<2$. Choose $C\mid C<B$, $C<\frac{2-B^2}{3B},C>0$. Then, $(B+C{)}^2=B^2+2BC+C^2$. That is $B^2+C(2B+C)<B^2+3BC<B^2+(2-B^2)=2$, contradicting that $B=\text{sup}(S)$. \end{proof}