Lecture 2

Quotients

Constructing Quotients

Have equivalence relation on $\mathbb{Z}\times (\mathbb{Z}\setminus \{0\})$ given $(a,b)\sim (c,d)$ when $ad=bc$.

Then $\frac{\mathbb{Z}\times \mathbb{Z}\setminus \{0\}}{\sim }$ is the set of rational numbers $\mathbb{Q}$ with addition and multiplication defined in an obvious way.
$\frac{a}{b}=\frac{c}{d}$

Real Numbers

Constructing Real Numbers

Problem with $\mathbb{Q}$:
For example Pythagoras’ Theorem with sides 1 and 1 gives $\sqrt{2}$

$$\sqrt{2}\notin \mathbb{Q}$$ $$\sqrt{2}=\frac{a}{b},a,b\in \mathbb{Z},b\ne 0.$$

Which is a contradiction

Yet we can find a sequence $\{X_n\}$ of rational numbers such that $X_n^2\to 2$ as $n\to \infty$.
$x_1=1,x_2=1.4,x_3=1.41$
$\sqrt{2}=1.4142\dots$
By a sequence $\{x_n\}$ in a set $X$ we mean a function $f:\mathbb{N}\to X$ with $x_n\equiv f(n)$, and that a function or map $X\to Y$ between two sets ascribes one member of $Y$ to each member of $X$.

A sequence $\{x_n\}$ in $\mathbb{Q}$ converges to $x\in \mathbb{Q}$, written ${\lim }_{x_n\to \infty }=x$, if $\forall k\in \mathbb{N}$ $\exists N_k\in \mathbb{N}$ such that $|x-x_n|<\frac{1}{k},\forall n<N_k$.

So what is $\sqrt{2}$?

Real numbers are certain equivalence classes of Rational Cauchy Sequences

Two such sequences $\{x_n\}$, $\{y_n\}$ are equivalent if the “distance” ${\lim }_{|x_n-y_n|\to \infty }=0$ between them vanishes.
$\{x_n\}\in X\sim \{y_n\}\in X$
Their set of equivalence classes is the set of real numbers $\mathbb{R}$.
$X\setminus \sim =\mathbb{R}\ni [\{x_n\}]$
$[\{x_n\}]+[\{y_n\}]=[\{x_n+y_n\}]$
Get natural algebraic operations on $\mathbb{R}$ from $\mathbb{Q}$; check well-definedness. Then $\mathbb{Q}\subset \mathbb{R}$ as the classes containing constant sequences. An order $>$ on $\mathbb{R}$ is defined by declaring as positive those classes having sequences with only positive rational numbers.
$x>y⟹x-y>0$

Convergence of Real Numbers

A sequence of real numbers is said to converge to a real number if the “distance” between their representatives tend to zero.