---
description: math runway, jephian, nsysu, 林晉宏
tags: talk, learning-together, math-runway
---

$\newcommand{\trans}{^\top}
\newcommand{\adj}{^{\rm adj}}
\newcommand{\cof}{^{\rm cof}}
\newcommand{\inp}[2]{\left\langle#1,#2\right\rangle}
\newcommand{\dunion}{\mathbin{\dot\cup}}
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\newcommand{\bb}{\mathbf{b}}
\newcommand{\bc}{\mathbf{c}}
\newcommand{\bd}{\mathbf{d}}
\newcommand{\be}{\mathbf{e}}
\newcommand{\bh}{\mathbf{h}}
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\newcommand{\by}{\mathbf{y}}
\newcommand{\bz}{\mathbf{z}}
\newcommand{\bu}{\mathbf{u}}
\newcommand{\bv}{\mathbf{v}}
\newcommand{\bw}{\mathbf{w}}
\newcommand{\tr}{\operatorname{tr}}
\newcommand{\nul}{\operatorname{null}}
\newcommand{\rank}{\operatorname{rank}}
%\newcommand{\ker}{\operatorname{ker}}
\newcommand{\range}{\operatorname{range}}
\newcommand{\Col}{\operatorname{Col}}
\newcommand{\Row}{\operatorname{Row}}
\newcommand{\spec}{\operatorname{spec}}
\newcommand{\vspan}{\operatorname{span}}
\newcommand{\Vol}{\operatorname{Vol}}
\newcommand{\sgn}{\operatorname{sgn}}
\newcommand{\idmap}{\operatorname{id}}
\newcommand{\am}{\operatorname{am}}
\newcommand{\gm}{\operatorname{gm}}
\newcommand{\mult}{\operatorname{mult}}
\newcommand{\iner}{\operatorname{iner}}$

# How to write a good proof?

## 不同版本的證明

Prove the Cauchy–Schwarz inequality:
For any vectors $\bx$ and $\by$ in $\mathbb{R}^n$,

$$
\inp{\bx}{\by} \leq \|\bx\|\|\by\|.
$$

:::warning

**老師上課版** :male-teacher:

Case 1: $\bx = \bzero$ or $\by = \bzero$.
:male-teacher: Obvious, think about why.

Case 2: Both $\bx$ and $\by$ are nonzero vectors with length $1$.
:male-teacher: We have the following inequality.
$\|\bx - \by\|^2 \geq 0$
:male-teacher: Recall that the square of the norm is the inner product of the vector with itself, we have ...
$\inp{\bx - \by}{\bx - \by} \geq 0$
:male-teacher: This means ...
$\|\bx\|^2 + \|\by\|^2 - 2\inp{\bx}{\by} \geq 0$
:male-teacher: Note that we have ...
Note: $\|\bx\| = \|\by\| = 1$.
:male-teacher: so ...
$2 - 2\inp{\bx}{\by} \geq 0$
$\implies 2\inp{\bx}{\by} \leq 2$
$\implies \inp{\bx}{\by} \leq 1 = \|\bx\|\|\by\|$

Case 3: Both $\bx$ and $\by$ are nonzero vectors.
:male-teacher: You may replace $\bx$ with $\frac{\bx}{\|\bx\|}$ and $\by$ with $\frac{\by}{\|\by\|}$. Then apply the previous case. This is left as your homework.
:::

:::spoiler 優缺點
:+1: 屬於證明的導讀，著重證明的直覺。
:-1: 很多都只有講沒有寫下來。
:-1: 一堆細節都說 --- 這個回家當練習。
:::

:::danger

**寫在你筆記上的版本** :notebook:

Case 1: $\bx = \bzero$ or $\by = \bzero$.
Obvious.

Case 2: Both $\bx$ and $\by$ are nonzero vectors with length $1$.
$\|\bx - \by\|^2 \geq 0$
$\inp{\bx - \by}{\bx - \by} \geq 0$
$\|\bx\|^2 + \|\by\|^2 - 2\inp{\bx}{\by} \geq 0$
Note: $\|\bx\| = \|\by\| = 1$.
$2 - 2\inp{\bx}{\by} \geq 0$
$\implies 2\inp{\bx}{\by} \leq 2$
$\implies \inp{\bx}{\by} \leq 1 = \|\bx\|\|\by\|$

Case 3: Both $\bx$ and $\by$ are nonzero vectors.
Homework.
:::

---

:::success

**課本版** :book:

We first consider the case when either $\bx = \bzero$ or $\by = \bzero$. In this case, we have $\inp{\bx}{\by} = 0$ and $\|\bx\|\|\by\| = 0$, so the inequality is true.

Then we consider a special case when both $\bx$ and $\by$ are unit vectors. Observe that $\|\bx - \by\|^2 \geq 0$. The left hand side can be written as the inner product

$$
\begin{aligned}
\|\bx - \by\|^2 &= \|\bx\|^2 + \|\by\|^2 - 2\inp{\bx}{\by} \\
&= 2 - 2\inp{\bx}{\by}.
\end{aligned}
$$

Note the second equality is because $\|\bx\| = \|\by\| = 1$. Thus, we have $2 - 2\inp{\bx}{\by} \geq 0$, which implies $\inp{\bx}{\by}\leq 1 = \|\bx\|\|\by\|$, so the inquality holds for this case.

Finally, when $\bx$ and $\by$ are both nonzero vector, we may apply the second case the the normalized vectors $\frac{\bx}{\|\bx\|}$ and $\frac{\by}{\|\by\|}$, which are unit vectors. This gives us

$$
\inp{\frac{\bx}{\|\bx\|}}{\frac{\by}{\|\by\|}} \leq \left\|\frac{\bx}{\|\bx\|}\right\|\left\|\frac{\by}{\|\by\|}\right\| = 1.
$$

Since

$$
\inp{\frac{\bx}{\|\bx\|}}{\frac{\by}{\|\by\|}} = \frac{1}{\|\bx\|\|\by\|}\inp{\bx}{\by},
$$

this leads to the desired inequality $\inp{\bx}{\by} \leq \|\bx\|\|\by\|$. This completes the proof.
:::

:::spoiler 優缺點
:+1: 非常完整。
:+1: 正式寫作規格。
:-1: 好長。
:-1: 英文字好多。
:::

---

## 寫一個好證明的重點

1. 想清楚每一個細節，看著你的筆記腦補完整的證明。
2. 懂了 $=$ ==能夠寫下來== $+$ 解釋給別人聽。
3. 分清楚直覺或是邏輯，想一下定義是什麼。
:x: Since $\inp{\bx}{\by} = \|\bx\|\|\by\|\cos\theta$ and $|\cos\theta| \leq 1$, we have $\inp{\bx}{\by} = \|\bx\|\|\by\|\cos\theta$. （為什麼內積會是 $\|\bx\|\|\by\|\cos\theta$？高維度的 $\theta$ 又是什麼？）
3. 句字要完整，連接用語跟數學式等同重要。

[Common English usage in Mathematics](https://hackmd.io/@jephianlin/eng-in-math): 一些基本證明的範例。
[風格指引](https://sagelabtw.github.io/LA-Tea/style.html): 正式寫作的一些建議。

## 練習

Play the [Kahoot!](https://create.kahoot.it/details/26e8163e-ebfb-441a-93a5-12622721c8dc) game if you wish. :smiley:

:::success
**Exercise 1 (集合寫法)**: Write down the the set of all multiple of $6$.

1. $6k$
2. $6k$, $k\in\mathbb{Z}$
3. $\{6k: k\in\mathbb{Z}\}$
4. $(6, 12, 18, \ldots)$

:::

:::success
**Exercise 2 (集合寫法)**: Which one is a set of vectors?

1. $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$
2. $\begin{bmatrix} 1 \\ 4 \end{bmatrix}$, $\begin{bmatrix} 2 \\ 5 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 6 \end{bmatrix}$
3. $c_1\begin{bmatrix} 1 \\ 4 \end{bmatrix} + c_2\begin{bmatrix} 2 \\ 5 \end{bmatrix} + c_3\begin{bmatrix} 3 \\ 6 \end{bmatrix}$ for some $c_1,c_2,c_3\in\mathbb{R}$
4. $\left\{\begin{bmatrix} 1 \\ 4 \end{bmatrix}, \begin{bmatrix} 2 \\ 5 \end{bmatrix}, \begin{bmatrix} 3 \\ 6 \end{bmatrix}\right\}$

:::

:::success
**Exercise 3 (集合包含)**: Let $A$ and $B$ be two sets. Which statement is equivalent to $A\subseteq B$.

1. For any $x\in A$, we have $x\in B$.
2. For any $x\in B$, we have $x\in A$.
3. For some $x\in A$, we have $x\in B$.
4. For some $x\in B$, we have $x\in A$.

:::

:::success
**Exercise 4 (一對一)**: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by $f(x) = x^3$. Show that $f$ is injective.

1. $x\neq y$ and $f(x)\neq f(y)$
2. $x\neq y$, $f(x) \neq f(y)$
3. We know $x \neq y$ implies $f(x) \neq f(y)$.
4. Suppose $x \neq y$. .... Then $f(x) \neq f(y)$.

:::

:::success
**Exercise 5 (映成)**: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by $f(x) = x^3$. Show that $f$ is surjective.

1. Let $y\in\mathbb{R}$. ... We found some $x$ such that $f(x) = y$.
2. For some $x$ and $y$, we have $f(x) = y$.
3. For some $y\in\mathbb{R}$, we can find an $x$ such that $f(x) = y$.
4. We know that $f(x) = y$.

:::

:::success
**Exercise 6 (線性獨立)**: Let $\bv_1,\ldots,\bv_n$ be vectors. How to show "If $c_1\bv_1 + \ldots + c_n\bv_n = \bzero$ for some $c_1,\ldots, c_n\in\mathbb{R}$, then $c_1 = \cdots = c_n = 0$."?

1. Suppose $c_1\bv_1 + \ldots + c_n\bv_n = \bzero$ for some $c_1,\ldots, c_n\in\mathbb{R}$. ... Therefore, $c_1 = \cdots = c_n = 0$.
2. Let $c_1 = \cdots = c_n = 0$. ...
3. If $c_1 = \cdots = c_n = 0$, then $c_1\bv_1 + \ldots + c_n\bv_n = \bzero$.
4. $c_1\bv_1 + \ldots + c_n\bv_n = \bzero$, $c_1 = \cdots = c_n = 0$

:::

:::success
**Exercise 7 (極限)**: How to show "For any $\epsilon > 0$, there is $\delta > 0$ such that XXX."?

1. Let $\epsilon = 0.00001$. We may find $\delta$ ...
2. $\epsilon > 0$, $\delta > 0$, XXX
3. Let $\epsilon > 0$. Suppose $\delta > 0$ ...
4. Let $\epsilon > 0$. We may find $\delta$ ...

:::

## 你在公 :shark: 小系列

:::danger

:bulb: 寫證明不是在應付論文比對

$\bx = \bzero$ or $\by = \bzero$ is obvious.

We know $\bx$ and $\by$ has length $1$.
$\|\bx - \by\|^2 \geq 0$
$\bx^2 + \by^2 - 2\bx\by \geq 0$
$\implies \inp{\bx}{\by} \leq 1 = \|\bx\|\|\by\|$

:::

:::danger

:bulb: 不要只有數學式

$\bx = \bzero$
$\by = \bzero$
Obvious.

$\|\bx\| = \|\by\| = 1$
$\|\bx - \by\|^2 \geq 0$
$\inp{\bx - \by}{\bx - \by} \geq 0$
$\|\bx\|^2 + \|\by\|^2 - 2\inp{\bx}{\by} \geq 0$
$2 - 2\inp{\bx}{\by} \geq 0$
$\implies 2\inp{\bx}{\by} \leq 2$
$\implies \inp{\bx}{\by} \leq 1 = \|\bx\|\|\by\|$

:::

:::danger

:bulb: 舉例 $\neq$ 證明

Let $\bx = (1,1,1,1)$ and $\by = (1,5,1,3)$.
$\inp{\bx}{\by} = 1 + 5 + 1 + 3$
$\|\bx\|\|\by\| = 2\cdot 6$
so $\inp{\bx}{\by} \leq \|\bx\|\|\by\|$.

:::

:::danger

:bulb: 感受一下題目的重點

Suppose $\bx = (x_1, \ldots, x_n)$ and $\by = (y_1, \ldots, y_n)$.
Then we would like to show

$$
(x_1y_1 + \cdots + x_ny_n)^2 \leq (x_1^2 + \cdots + x_n^2)(y_1^2 + \cdots + y_n^2).
$$

We know this inequality is true in high school, so $\inp{\bx}{\by} \leq \|\bx\|\|\by\|$.

:::