--- description: math runway, jephian, nsysu, 林晉宏 tags: talk, learning-together, math-runway --- $\newcommand{\trans}{^\top} \newcommand{\adj}{^{\rm adj}} \newcommand{\cof}{^{\rm cof}} \newcommand{\inp}[2]{\left\langle#1,#2\right\rangle} \newcommand{\dunion}{\mathbin{\dot\cup}} \newcommand{\bzero}{\mathbf{0}} \newcommand{\bone}{\mathbf{1}} \newcommand{\ba}{\mathbf{a}} \newcommand{\bb}{\mathbf{b}} \newcommand{\bc}{\mathbf{c}} \newcommand{\bd}{\mathbf{d}} \newcommand{\be}{\mathbf{e}} \newcommand{\bh}{\mathbf{h}} \newcommand{\bp}{\mathbf{p}} \newcommand{\bq}{\mathbf{q}} \newcommand{\br}{\mathbf{r}} \newcommand{\bx}{\mathbf{x}} \newcommand{\by}{\mathbf{y}} \newcommand{\bz}{\mathbf{z}} \newcommand{\bu}{\mathbf{u}} \newcommand{\bv}{\mathbf{v}} \newcommand{\bw}{\mathbf{w}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\nul}{\operatorname{null}} \newcommand{\rank}{\operatorname{rank}} %\newcommand{\ker}{\operatorname{ker}} \newcommand{\range}{\operatorname{range}} \newcommand{\Col}{\operatorname{Col}} \newcommand{\Row}{\operatorname{Row}} \newcommand{\spec}{\operatorname{spec}} \newcommand{\vspan}{\operatorname{span}} \newcommand{\Vol}{\operatorname{Vol}} \newcommand{\sgn}{\operatorname{sgn}} \newcommand{\idmap}{\operatorname{id}} \newcommand{\am}{\operatorname{am}} \newcommand{\gm}{\operatorname{gm}} \newcommand{\mult}{\operatorname{mult}} \newcommand{\iner}{\operatorname{iner}}$ # Common English usage in Mathematics ## 課堂常用句法 ### 老師 - You are correct ... but your answer is not quite complete. Please consider ... - Can someone volunteer to summarize in your own words the main points we just discussed? - That is all for today---we will see you next class. - The results seem to demonstrate that ... - These findings prove that ... - To be honest, it seems to me that ... ### 學生 - I don't quite understand ... . I am fine with _something that you are okay_ but I got lost when we are talking about _something that you don't understand_. Could you give me some hints? - Could you go back to ... . Why _something_ is _something_? - Could you say _something_ again? - I need some help on _something_. ## 寫證明常用句法 ### 令、考慮、假設 - **Let** _condition_. **Then** _implication_. - **Consider** _condition_. **Then** _implication_. - **Suppose** _condition_. **Then** _implication_. - **Assume** _condition_. **Then** _implication_. - **If** _condition_, **then** _implication_. :::info Tips :bulb: - **令** 或 **考慮** 通常在說一個事實或是設定。 :o: Let $\epsilon > 0$. :x: Let $\sqrt{2}$ be a rational number. - **假設** 通常是反證法的或是設定。 :o: Suppose $\sqrt{2}$ can be written as $\frac{a}{b}$ for some integers $a$ and $b$. :x: Suppose $0$ is a rational number. - 「if」是連接詞,不會單獨成一個句子 :o: If $x$ is a real number, then $x^2 \geq 0$. :x: If $x$ is a real number. Then $x^2 \geq 0$. - 英文標點後空一格,標點前不空格 ::: ### 因為、所以 - **Since** _cause_, _effect_. - **Becuase** _cause_, _effect_. - **By** _cause_, _effect_. - _Cause_ **implies** _effect_. - _Cause_, **so** _effect_. - _Cause_. **Therefore**, _effect_. - _Cause_. **Consequently**, _effect_. - _Cause_. **Hence**, _effect_. - _Cause_. **Thus**, _effect_. :::info Tips :bulb: - 英文中的因為和所以不會同時出現。 :o: Since $x$ is even, $x+1$ is odd. :x: Since $x$ is even, so $x+1$ is odd. - 「because」或是「since」放句首的話子句結束要加逗點。 :o: Since $x$ is even, $x+1$ is odd. :o: We know $x+1$ is odd since $x$ is even. (通常不會讓數學式放句首,所以會塞一些不改變語意的文字) :x: Since $x$ is even $x+1$ is odd. ::: ### 論證手法概述 - **It is sufficient/enough to show** ... . - **Prove by induction on $n$**. - We will prove the **contrapositive statement**. - **Suppose, for the purpose of yielding a contradiction,** _hypothesis_. ### 練習 :::success **Exercise 1 (集合相等)**: Prove that $\{3k + 1: k\in\mathbb{Z}\} = \{3k - 2: k\in\mathbb{Z}\}$. ==_____== $X = \{3k + 1: k\in\mathbb{Z}\}$ and $Y = \{3k - 2: k\in\mathbb{Z}\}$. ==_____== to show that $x\in X$ implies $x\in Y$ and $y\in Y$ implies $y\in X$. ==_____== $x\in X$. ==_____== $x$ can be written as $x = 3k - 1$ for some $k$. ==_____==, $x = 3(k-1) + 2$ ==_____== $x\in Y$. On the other hand, ==_____== $y\in Y$. ==_Finish the other direction._== ::: :::success **Exercise 2 (一對一)**: Determine if the following functions are injective. 1. $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = x^3$. 2. $g: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = x^2$. 3. $h: [0,1] \rightarrow \mathbb{R}$ defined by $g(x) = x^2$. **Claim**: $f$ is injective. ==_____== $a$ and $b$ be numbers in $\mathbb{R}$ such that $a \neq b$. We may ==_____== $a < b$. This ==_____== that $f(a) = a^3 < b^3 = f(b)$, ==_____== $f$ is injective. **Claim**: $g$ is not injective. It is ==_____== to find distince $a$ and $b$ in $\mathbb{R}$ such that $g(a) = g(b)$. For example, $g(-1) = g(1)$, ==_____== $g$ is not injective. **Claim**: $h$ is injective. ==_Finish this case._== ::: :::success **Exercise 3 (映成)**: Determine if the following functions are surjective. 1. $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = x^3$. 2. $g: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = x^2$. 3. $h: [0,1] \rightarrow \mathbb{R}$ defined by $g(x) = x^2$. **Claim**: $f$ is surjective. ==_____== $y$ be a number in $\mathbb{R}$. ==_____== we may find $x = \sqrt[3]{y}$ such that $f(x) = y$. ==_____==, $f$ is surjective. **Claim**: $g$ is surjective. ==_Finish this case._== **Claim**: $h$ is not surjective. It is ==_____== to find some $y$ in $\mathbb{R}$ such that $y$ cannot be written as $h(x)$ for any $x\in [0,1]$. Observe that $0 \leq x^2 \leq 1$ whenever $0 \leq x \leq 1$, ==_____== we may choose, for example, $y = 100$ so that $y \neq h(x)$ for any $x \in [0,1]$. ==_____==, $h$ is not surjective. ::: :::success **Exercise 4 (反證)**: Prove that $\sqrt{2}$ is a rational number. ==_____== $\sqrt{2}$ is a rational number. ==_____== it can be written as $\sqrt{2} = \frac{a}{b}$ for some integers $a$ and $b$ with $\gcd(a,b) = 1$. By taking the square on both sides, we have $$ 2 = \frac{a^2}{b^2}, $$ ==_____== $2 b^2 = a^2$. ==_____==, $a$ has to be an even number and we may write it as $a = 2k$. ==_____==, $2b^2 = 4k^2$ and $b^2 = 2k^2$. This again ==_____== $b$ is an even number. However, the fact that $a$ and $b$ are both even numbers violates our assumption $\gcd(a,b) = 1$, ==_____== $\sqrt{2}$ is not a rational number. :::