99 lines
2.6 KiB
Markdown
99 lines
2.6 KiB
Markdown
# Lecture 1
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## Introduction
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Reading is not recommended before class, it's hard.
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## Chapter 1: The real number and complex number systems
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* Natural numbers: $\mathbb{N}=\{1,2,3,4....\}$ note by some conventions, $0$ is also a natural number
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* IntegersL $\mathbb{Z}=\{...,-2,-1,0,1,2,...\}$
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* Rational numbers: $\mathbb{Q}=\{\frac{m}{n}:m,n\in\mathbb{Z}\ and\ n\neq 0\}$
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* Real numbers: $\mathbb{R}$ the topic of chapter
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* Complex numbers: $\mathbb{C}=\{a+bi:a,b\in \mathbb{R}\}$
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### Theorem ($\sqrt{2}$ is irrational)
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$\exist p\in \mathbb{Q},p^2=2$ is false.
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$\equiv\cancel{\exist} p\in \mathbb{Q}, p^2=2$
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$\equiv p\in \mathbb{Q},p^2\neq 2$
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<details>
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<summary>Proof</summary>
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Suppose for contradiction, $\exist p\in \mathbb{Q}$ such that $p^2=\mathbb{Q}$.
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Let $p=\frac{m}{n}$, where $m,n \in \mathbb{Z}$ are not both even. (reduced form)
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$p^2=2$ and $p=\frac{m}{n}$, so $m^2=2n^2$, so $m^2$ is even, $m$ is even.
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So $m^2$ is divisible by 4, $2n^2$ is divisible by 4.
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So $n^2$ is even. but they are not both even.
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</details>
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### Theorem (No closest rational for a irrational number)
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Let $A=\{p\in \mathbb{q}, p>0\ and\ p^2\leq 2\}$, Then $A$ does not have a largest element.
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i.e. $\exist p\in A$ such that $\forall q\in A, q\leq p$ is false.
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> 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...)
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#### Thought process
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Let $p\in A,p\in \mathbb{Q}$, $p>0, p^2<2$.
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We want a $\delta\in\mathbb{Q}$ such that $\delta>0$ and $(p+\delta)^2<2$.
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$$
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\begin{aligned}
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(p+\delta)^2&<2\\
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p^2+2p\delta+\delta^2&<2\\
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\delta(2p+\delta)&< 2-p^2\\
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\delta&<\frac{2-p^2}{2p-\delta}
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\end{aligned}
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$$
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From $(p+\delta)^2<2$, we know $\delta<2$ (this is a crude bound, $\delta<\sqrt{2}$).
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So one choice can be $\delta=\frac{2-p^2}{2p+2}$
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#### Proof
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$\forall p\in A$, 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 $p^2<(p+\delta)^2<2$.
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_Here we construct a formula for approximate $\sqrt{2}=\lim_{i\to \infty}p_0=1,p_{i+1}=p_i+\frac{2-p_i^2}{2p_i+2}$_
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Interesting...
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We can also further optimize the formula by changing the bound of $\delta$ to $\delta< 2-p$, since $(p+\delta)^2<2,p+\delta<2$
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```python
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def sqrt_2(acc):
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if acc==0: return 1
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c=sqrt_2(n-1)
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return c+((2-c**2)/(2*c+2))
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```
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### Definition and notations for sets
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Some set notation
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$\Pi\in \mathbb{R}$
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use $\subset,\subsetneq$ in this class.
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* $A\subset B$, $\forall x\in A, x\in B$
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* $A=B$, $A\subset B$ and $B\subset A$
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* $A\subsetneq$ means $A\subset B$ and $A\neq B$
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