MATH 3150 Class Notes 01-23-2025

notereal-analysis

Today we will focus on $\mathbb{Z}$, $\mathbb{Q}$. We will introduce $\frac{n}{m}$ for $n,m\in \mathbb{Z}$ in an axiomatic way.

Properties (Axioms)

1. Addition
   1. $(x+y)+z=x+(y+z)$ associativity for all $x,y,z$.
   2. $x+y=y+x$ commutativity for all $x,y$.
   3. $\exists 0$ such that $x+0=0+x=x$ for all $x$.
   4. $\forall x,\exists (-x)$ such that $x+(-x)=(-x)+x=0$.
2. Multiplication
   1. $(x\cdot y)\cdot z=x\cdot (y\cdot z)$.
   2. $x\cdot y=y\cdot x$.
   3. $\exists 1$ such that $x\cdot 1=1\cdot x=x$ for all $x$.
   4. $\forall x\ne 0$, $\exists x^{-1}$ such that $x\cdot x^{-1}=x^{-1}\cdot x=1$.
3. Distributivity
   1. $x\cdot (y+z)=x\cdot y+x\cdot z$ for all $x,y,z$.

Remark

All known properties of rational (or real) numbers follow from these 9 axioms.

Ordering "$\le$"

1. $\forall x,y$ either $x\le y$ or $y\le x$.
2. $\forall x,y$ if $x\le y$ and $y\le x$ then $x=y$.
3. If $x\le y$ and $y\le z$, then $x\le z$.
4. If $x\le y$ then $x+z\le y+z$.
5. If $x\le y$ and $0\le z$, then $x\cdot z\le y\cdot z$.

Example

$0<1$ follows from Ordering axioms 1-5 (O1-O5).

Proof

$0\ne 1\stackrel{O1}{⟹}$ either $0\le 1$ or $1\le 0$.

Assume for the sake of contradiction that $1\le 0$, then by $O4$, $1+(-1)\le 0+(-1)\le -1$. From that we have that $0\le -1$, by which we can multiply both sides by $-1$, since we have that $0\le -1$, we have that $(0)\cdot (-1)\le (-1)\cdot (-1)$, or $0\le 1$, a contradiction.