Cheat Sheet to Write Mathematics in English for Chinese
27 Sep 2015The following is by no means ‘the’ introduction on how to write mathematical proofs in English – English is my third tongue, therefore I do not master it very well myself. Just take the following as building blocks so that I can follow your proofs. After a while, you may improve by looking at proof arguments in professional English textbooks – I am not a reference on that topic. That’s also why I recommend you to read only textbooks in English! At the beginning I accept – even if I hate it – the use of logical symbols such as \(\Rightarrow\), \(\Leftrightarrow\), \(\forall\), \(\exists\), etc. in text. Try however to replace them by their English counterparts, it is nicer to read.*
Logic:
- Assertion: 断言
- Hypothesis: 假设
- True, False: 真，假
- \(A\Rightarrow B\): “\(A\) implies \(B\)” \(A\) 推出 \(B\) or “From \(A\) follows \(B\)” 若 \(A\) 则 \(B\), or “If \(A\) (is true), then (so is) \(B\).” 若 \(A\) 为真则 \(B\) 为真
- \(A\Leftrightarrow B\): “\(A\) is equivalent to \(B\)” \(A\) 等价于 \(B\) or “\(A\) holds if, and only if, \(B\) holds”. \(A\) 成立当且仅当 \(B\) 成立
- Element: 元素
- Set: 集
- Collection: 类
- \(x\in A\): “\(x\) is in \(A\)”. \(x\) 属于 \(A\)
- \(A\subseteq B\): “\(A\) is a subset of \(B\)”. \(A\) 包含于 \(B\)
- \(\forall x\): “For all \(x\)” (or “For every \(x\)”). 对任意 \(x\)
- \(\exists x\): “For some \(x\)”, or (there exists \(x\)). 存在 \(x\)
Basic math vocabulary:
- \(x\geq y\): \(x\) is greater than \(y\), or \(y\) is less than \(x\). \(x\) 不小于 \(y\)
- \(x\geq 0\): \(x\) is positive, and \(-x\) is negative. \(x\)不小于\(0\)
- \(x>y\) or \(x>0\): \(x\) is strictly greater than \(y\), or \(x\) is strictly positive. \(x\)严格大于\(y\) \(x\)严格大于\(0\)
- \(\mathbb{N}\): natural numbers (自然数) sometimes called strictly positive integers (正整数).
- \(\mathbb{Z}\): integers 整数
- \(\mathbb{Q}\): rational numbers 有理数
- \(\mathbb{R}\): real numbers 实数
- \(f:X\to Y\): function \(f\) from the domain \(X\) to the codomain \(Y\). \(f\) 是从定义域 \(X\) 映射到值域 \(Y\) 的函数
- \(f(A)\) and \(f^{-1}(B)\): image of \(A\subseteq X\) and preimage of \(B\subseteq Y\) under \(f\). \(A\) 在映射\(f\)下的像，\(B\)的原像
- \(f(x)\leq f(y)\) for \(x\leq y\): \(f\) is increasing. 单调不减的
- \(f(x)\geq f(y)\) for \(x\leq y\): \(f\) is decreasing. 单调不增的
- \(f\) either increasing or decreasing: \(f\) is monotone. \(f\) 是单调的
- Finite family 有限族
- Countable family 可数族
Writing proofs:
- We show that \(A\) and \(B\) implies \(C\). 我们说明A 和 B 推出C
- Under the assumption [of the theorem/proposition/exercise], it holds [this/that].根据定理的假设，我们可以得出
- Suppose/assume that [something] holds.假设…成立
- By contradiction, suppose that [something] holds. (show that it can not be true).由于矛盾，说明…成立
- If [something] holds, then it follows that … 如果…成立则说明…
- Since [something] holds, it follows that… 因为…成立，我们有…
- Hence,… 由于
- [something] yields [something] …表明…
- However, [something is true], therefore, [another thing] holds. …成立，因此，…成立
- Thus,… 那么
- This completes the proof. (or CQFD.) 证明完成
Exemplary sentences of the lecture:
- Let \(x \in \mathbb{R}\) be such that… 令x属于实数集则有
- Let \((x_n)\) be a sequence of elements in \(A\) such that… 令\((x_n)\)是A中的一列元素
- Let \(f:X\to Y\) be a function such that… 令f是从X映射到Y的函数，那么
- Let \((A_i)\) be a family of subsets of \(\Omega\) such that… 令\((A_i)\)是\((\Omega)\)中的子集族
- Let \((A_n)\) be a countable family of subsets of \(\Omega\) such that…令\((A_i)\)是\((\Omega)\)中的可数子集族
- Since \(\mathcal{F}\) is a $\sigma$-algebra, it follows that \(\cup A_n \in \mathcal{F}\) for every countable family \((A_n)\) of elements in \(\mathcal{F}\). 因为 \(\mathcal{F}\) 是一个 $\sigma$-代数, 对任意由 \(\mathcal{F}\)中的元素\((A_n)\)组成的可数集族可以得到 \(\cup A_n \in \mathcal{F}\)
- Let \((A_k)_{k\leq n}\) be a finite family of subsets of \(\Omega\) such that…令\((A_k)_{k\leq n}\)是\(\Omega\)的可数子集族，那么
- Suppose that \((A_n)\) is a countable family of subsets of \(\Omega\) such that…假设\((A_n)\)是\(\Omega\)的一个可数子集族
- Since the random variable $X$ is measurable, it follows that \(\{X\leq a\}\) is measurable for every \(a \in \mathbb{R}\).由于随机变量X可测，可以得到\(\{X\leq a\}\)对任意\(a \in \mathbb{R}\)可测
- Suppose that the random variable \(X\) is bounded, then it follows in particular that \(X\) is integrable. 假设随机变量X有界，那么特别地，X是可积的
Example
(推理)
Let \(n\) and \(m\) be two integers.Suppose that \(n\) is even and \(m\) is even.
Then, \(nm\) is also even.
令 \(n\) 和 \(m\) 是两个整数. 假设 \(n\) 是偶数 \(m\) 是偶数. 那么 \(nm\) 也是偶数.
(not nice but okay at the beginning) 证明（不完美的写法）
\(n,m \in \mathbb{Z}\) are even \(\Rightarrow\) \(\exists p,q \in \mathbb{Z}\) such that \(n=2p\) and \(m=2q\) \(\Rightarrow nm=2p2q=2(2pq)=2r\) where \(r:=2pq\in \mathbb{Z}\) \(\Rightarrow\) \(nm\) is even. CQFD.
\(n,m \in \mathbb{Z}\) 是偶数 \(\Rightarrow\) \(\exists p,q \in \mathbb{Z}\) 使得 n=2p 且 m=2q \(\Rightarrow nm=2p2q=2(2pq)=2r其中r=2pq\in \mathbb{Z}\) \(\Rightarrow\) \(mn\) 也是偶数. 证明完成.
(证明)
Let $n$ and $m$ be two even integers.
By definition, it follows that $n$ and $m$ are divisible by two, that is, \(n=2p\) and \(m=2q\) for some integers \(p\) and \(q\).
Hence, \(nm=2p2q=4pq=2(2pq)\).
It follows that \(nm=2r\) where \(r=2pq\) is an integer. Thus, \(nm\) is even, which completes the proof.
令 \(n\) 和 \(m\) 是两个偶数. 根据定义，\(m\) 和 \(n\) 可以被 \(2\) 整除，这说明 \(n=2p\) 且 \(m=2q\) 对某些整数 \(p,q\) 成立. 那么 \(mn=2r\) 其中 \(r=2pq\) 也是整数. 那么 \(nm\) 是偶数，证明完成.
Comments
- As French, I never use the US/UK terminology “non-decreasing”, “non-increasing”, “non-negative” and “non-positive” for “increasing”, “decreasing”, “positive” and “negative” as defined above. Indeed, I find this way of defining something by negation unnatural and actually only makes sense for total order. So it is a bad habit, and I want you to use the natural definitions as above. If, however, you decide to go for the Yankee way, then stick to it and do not mix.
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