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Math4111
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Math4111 L1

Lecture 1

Introduction

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

Chapter 1: The real number and complex number systems

  • Natural numbers: N={1,2,3,4....}\mathbb{N}=\{1,2,3,4....\} note by some conventions, 00 is also a natural number

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

  • Rational numbers: Q={mn:m,nZ and n0}\mathbb{Q}=\{\frac{m}{n}:m,n\in\mathbb{Z}\ and\ n\neq 0\}

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

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

Theorem (2\sqrt{2} is irrational)

pQ,p2=2\exist p\in \mathbb{Q},p^2=2 is false.

pQ,p2=2\equiv\cancel{\exist} p\in \mathbb{Q}, p^2=2

pQ,p22\equiv p\in \mathbb{Q},p^2\neq 2

Proof

Suppose for contradiction, pQ\exist p\in \mathbb{Q} such that p2=Qp^2=\mathbb{Q}.

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

p2=2p^2=2 and p=mnp=\frac{m}{n}, so m2=2n2m^2=2n^2, so m2m^2 is even, mm is even.

So m2m^2 is divisible by 4, 2n22n^2 is divisible by 4.

So n2n^2 is even. but they are not both even.

EOP

Theorem (No closest rational for a irrational number)

Let A={pq,p>0 and p22}A=\{p\in \mathbb{q}, p>0\ and\ p^2\leq 2\}, Then AA does not have a largest element.

i.e. pA\exist p\in A such that qA,qp\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 pA,pQp\in A,p\in \mathbb{Q}, p>0,p2<2p>0, p^2<2.

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

(p+δ)2<2p2+2pδ+δ2<2δ(2p+δ)<2p2δ<2p22pδ\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+δ)2<2(p+\delta)^2<2, we know δ<2\delta<2 (this is a crude bound, δ<2\delta<\sqrt{2}).

So one choice can be δ=2p22p+2\delta=\frac{2-p^2}{2p+2}

Proof

pA\forall p\in A, we can find a δ=2p22p+2\delta=\frac{2-p^2}{2p+2} which is greater than zero (p2<2,2p2>0,2p+2>0,δ>0p^2<2,2-p^2>0,2p+2>0,\delta>0) and construct a new number (p+δ)2(p+\delta)^2 such that p2<(p+δ)2<2p^2<(p+\delta)^2<2.

Here we construct a formula for approximate 2=limip0=1,pi+1=pi+2pi22pi+2\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 δ\delta to δ<2p\delta< 2-p, since (p+δ)2<2,p+δ<2(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

ΠR\Pi\in \mathbb{R}

use ,\subset,\subsetneq in this class.

  • ABA\subset B, xA,xB\forall x\in A, x\in B
  • A=BA=B, ABA\subset B and BAB\subset A
  • AA\subsetneq means ABA\subset B and ABA\neq B