$$

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.

Tips

• 令 或 考慮 通常在說一個事實或是設定。
  Let
  $ϵ>0$.
  Let
  $\sqrt{2}$ be a rational number.
• 假設 通常是反證法的或是設定。
  Suppose
  $\sqrt{2}$ can be written as
  $\frac{a}{b}$ for some integers
  $a$ and
  $b$.
  Suppose
  $0$ is a rational number.
• 「if」是連接詞，不會單獨成一個句子
  If
  $x$ is a real number, then
  $x^2\ge 0$.
  If
  $x$ is a real number. Then
  $x^2\ge 0$.
• 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.

Tips

• 英文中的因為和所以不會同時出現。
  Since
  $x$ is even,
  $x+1$ is odd.
  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。
  Since
  $x$ is even,
  $x+1$ is odd.
  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）
  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

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.

Exercise 2 (一對一): Determine if the following functions are injective.

1. $f:\mathbb{R}\to \mathbb{R}$ defined by
   $f(x)=x^3$.
2. $g:\mathbb{R}\to \mathbb{R}$ defined by
   $f(x)=x^2$.
3. $h:[0,1]\to \mathbb{R}$ defined by
   $g(x)=x^2$.

Claim:

$f$ is injective.
_____

$a$ and

$b$ be numbers in

$\mathbb{R}$ such that

$a\ne b$. 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 distinct

$a$ 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.

Exercise 3 (映成): Determine if the following functions are surjective.

1. $f:\mathbb{R}\to \mathbb{R}$ defined by
   $f(x)=x^3$.
2. $g:\mathbb{R}\to \mathbb{R}$ defined by
   $f(x)=x^2$.
3. $h:[0,1]\to \mathbb{R}$ 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 not 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

$0\le x^2\le 1$ whenever

$0\le x\le 1$, _____ we may choose, for example,

$y=100$ so that

$y\ne h(x)$ for any

$x\in [0,1]$. _____,

$h$ is not surjective.

Exercise 4 (反證): Prove that

$\sqrt{2}$ is an irrational 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

$$2=\frac{a^2}{b^2},$$

_____

$2b^2=a^2$. _____,

$a$ 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.