NoteNextra-origin/pages/Math4111/Math4111_L1.md

Lecture 1

Introduction

Reading is not recommended before class, it's hard.

Chapter 1: The real number and complex number systems

• Natural numbers: \mathbb{N}=\{1,2,3,4....\} note by some conventions, 0 is also a natural number

• IntegersL \mathbb{Z}=\{...,-2,-1,0,1,2,...\}

• Rational numbers: \mathbb{Q}=\{\frac{m}{n}:m,n\in\mathbb{Z}\ and\ n\neq 0\}

• Real numbers: \mathbb{R} the topic of chapter

• Complex numbers: \mathbb{C}=\{a+bi:a,b\in \mathbb{R}\}

Theorem (\sqrt{2} is irrational)

\exist p\in \mathbb{Q},p^2=2 is false.

\equiv\cancel{\exist} p\in \mathbb{Q}, p^2=2

\equiv p\in \mathbb{Q},p^2\neq 2

Proof

Suppose for contradiction, \exist p\in \mathbb{Q} such that p^2=\mathbb{Q}.

Let p=\frac{m}{n}, where m,n \in \mathbb{Z} are not both even. (reduced form)

p^2=2 and p=\frac{m}{n}, so m^2=2n^2, so m^2 is even, m is even.

So m^2 is divisible by 4, 2n^2 is divisible by 4.

So n^2 is even. but they are not both even.

EOP

Theorem (No closest rational for a irrational number)

Let A=\{p\in \mathbb{q}, p>0\ and\ p^2\leq 2\}, Then A does not have a largest element.

i.e. \exist p\in A such that \forall q\in A, q\leq p is false.

Remark: The book give a very slick proof trying to lean from these kinds of proofs takes some effort. (It is perfectly fine to write that solution this way...)

Thought process

Let p\in A,p\in \mathbb{Q}, p>0, p^2<2.

We want a \delta\in\mathbb{Q} such that \delta>0 and (p+\delta)^2<2.

\begin{aligned}
    
(p+\delta)^2&<2\\
p^2+2p\delta+\delta^2&<2\\
\delta(2p+\delta)&< 2-p^2\\
\delta&<\frac{2-p^2}{2p-\delta}
\end{aligned}

From (p+\delta)^2<2, we know \delta<2 (this is a crude bound, \delta<\sqrt{2}).

So one choice can be \delta=\frac{2-p^2}{2p+2}

Proof

\forall p\in A, we can find a \delta=\frac{2-p^2}{2p+2} which is greater than zero (p^2<2,2-p^2>0,2p+2>0,\delta>0) and construct a new number (p+\delta)^2 such that p^2<(p+\delta)^2<2.

Here we construct a formula for approximate $\sqrt{2}=\lim{i\to \infty}p_0=1,p_{i+1}=p_i+\frac{2-p_i^2}{2p_i+2}$_

Interesting...

We can also further optimize the formula by changing the bound of \delta to \delta< 2-p, since (p+\delta)^2<2,p+\delta<2

def sqrt_2(acc):
    if acc==0: return 1
    c=sqrt_2(n-1)
    return c+((2-c**2)/(2*c+2))

Definition and notations for sets

Some set notation

\Pi\in \mathbb{R}

use \subset,\subsetneq in this class.

• A\subset B, \forall x\in A, x\in B
• A=B, A\subset B and B\subset A
• A\subsetneq means A\subset B and A\neq B