OCR数学 H240 证明 反证法 穷举 反例
OCR 的 A Level 数学（H240）把证明放在纯数学的第一节，编号 1.01。这一条下面的四小块，正好对应本篇文章要讲的四种方法。很多学生以为证明是天赋题，其实它是一套可以逐句检查的写法。下面按 spec 的顺序拆开讲，每一节给一个完整例子。
The Proof section of OCR's A Level Mathematics specification is coded 1.01, and its four sub-clauses line up with the four methods discussed here. Proofs are marked on the chain of reasoning rather than on inspiration, so the writing can be rehearsed. Each section below takes one method and works through an example.
证明在 H240 里的位置：1.01 的四条内容
打开 OCR 官网的 H240 spec，纯数学第一节就是 Proof。这一条下面分了四小块，分别是直接证明、逻辑联结词、反例否定与反证法。AS 阶段（H230）只要求前三块，反证法是 A Level 才加的内容。spec 写明，反证法题目只会出在考生已经熟悉的内容上，多数来自 GCSE 或 AS 学过的数与式。卷面上 H240 有三份两小时、各 100 分的卷子，每份各占三分之一。证明题常出现在纯数学部分，分值多在 4 到 6 分之间。很多学生以为证明是独立的大题，实际上它常常藏在函数题或数列题的第二问里。卷面结构上，Paper 1 只考纯数学，Paper 2 与 Paper 3 的 A 部分是纯数学，B 部分分别是统计与力学。所以证明不只出现在第一份卷子里。
The proof content sits at the start of the pure mathematics section, and the specification separates what AS candidates must know from what the full A Level adds; contradiction is the addition. OCR also states that contradiction questions are set on content candidates have already met, which keeps the algebra familiar. Proof marks are usually four to six and rarely arrive as a standalone question, so expect to justify something inside a functions or sequences part.
直接证明：从定义出发的链条
直接证明（proof by deduction）的写法，是从已知条件出发，每一步都用定义或已证的结论推到下一步。奇偶性是最常见的入口：设两个奇数分别是 2m+1 与 2n+1，其中 m、n 是整数。相加得 2m+2n+2，也就是 2(m+n+1)，因此这个和是偶数。这类题的关键不在算，而在把「存在整数 m 使这个数等于 2m+1」这句话写出来。很多学生以为举出 3 加 5 等于 8 就算证明，其实那只是两个例子。定义里的量词写清之后，后面每一步才有依据。阅卷时看的是这条链有没有断，而不是结论对不对。写完之后数一数用到了哪些条件，题目给出的条件通常在证明里都要用到。
A deduction proof starts from the given conditions and moves forward one justified step at a time. For parity, write the odd numbers as 2m+1 and 2n+1 with m and n declared as integers, add them to obtain 2(m+n+1), then say what that factor of two establishes. Examples such as 3+5=8 illustrate a claim but never prove it, and a missing declaration about m and n costs the mark even when the arithmetic is correct.
穷举法：情形有限才列得尽
穷举法（proof by exhaustion）适用于情形有限、能一一列尽的命题。整数按奇偶分成两类，是最省事的穷举：n 为偶数时，n² 与 n 都是偶数，相加仍是偶数。n 为奇数时，n² 是奇数，n 也是奇数，两个奇数相加得偶数。写法上要先说清分类的依据，再逐类给出结论。常见的错误是漏掉一类，比如按正负分类却忘了零。情形超过四五种时，穷举就变成体力活，这时候一般该换反证法。spec 只要求把穷举用在熟悉的内容上，不会出几百种情形的题。写答案时把每一类的结论写成完整句子，只留一个式子通常不算完成。如果两类情形能用一个通式覆盖，写出通式比逐个举例更快。
Exhaustion works when the cases are finite and can all be listed, and parity is the standard example; writing n as either 2k or 2k+1 covers every integer exactly once. State the basis of the split before working through each case, because a missing case invalidates the whole argument. Once the list of cases runs past about four, the question is usually steering you towards contradiction instead.
反例否定：一个反例就够
否定一个全称命题只需要一个反例，这条规则在 spec 1.01c 里写得很清楚。命题写成「若 P 成立则 Q 成立」时，找一个让 P 真而 Q 假的 x 就够了。常见例子是「所有素数都是奇数」，x 取 2 就推翻它。另一个经典例子是「n²+n+41 对所有正整数都是素数」，n 取 40 得 1681，而 1681 等于 41 乘 41。反例题要把 x 的值写出来，并且验证 P 真、Q 假这两个条件。只写「取 2 不成立」这种句子，通常拿不到全分。反方向也要记住：再多的例子也证不出全称命题。
One counterexample destroys a universal statement, and OCR spells the rule out in 1.01c. For a claim of the form if P then Q, you need a single value where P holds and Q fails, such as the prime 2 for the assertion that all primes are odd. The polynomial n²+n+41 is another standard case: at n = 40 it gives 1681, which is 41 squared, so a bare assertion without checking both conditions rarely scores.
反证法的三步骨架
反证法的骨架只有三步：先假设结论不成立，再从假设出发做正确的推理，最后指出推出来的东西与已知事实矛盾。矛盾必须写明是跟什么矛盾，比如与「a、b 互质」矛盾，或者与「任何大于 1 的整数都有素因数」矛盾。常见的错误是推着推着就宣布「所以矛盾」，读者看不出矛盾在哪里，这一步的分就没了。spec 1.01d 明确说反证法用在你已熟悉的内容上，所以题目给的往往是一个数的性质，而不是新定义。写答案时把假设句单独成行，后面每句都标出依据。结尾那句要具体，把被违反的条件点名。还有一个常被忽略的细节：反证法收尾时，要把原命题再写一遍作为结论，别停在矛盾上。
Contradiction has three moves: assume the negation, reason correctly from that assumption, then name the fact you have contradicted. Naming it matters, because a bare claim of contradiction without saying what it clashes with usually earns nothing. OCR keeps these questions on familiar ground, so the assumption is nearly always about a number property you have already worked with.
根号 2 无理性的完整链条
根号 2 是无理数这条证明，要练到能默写。设根号 2 等于 a 除以 b，其中 a、b 是互质的正整数。两边平方得 a² 等于 2b²，因此 a² 是偶数，于是 a 是偶数，写成 a 等于 2k。代回得 4k² 等于 2b²，约掉 2 得 b² 等于 2k²，同样推出 b 是偶数。a 与 b 都是偶数，就与「互质」矛盾，假设不成立。整条链的引擎是「若 a² 是偶数则 a 是偶数」，这一句要用逆否命题补一句：a 为奇数时 a² 也是奇数。
The irrationality of root two runs on one engine step, namely that an even square forces an even root, justified through the contrapositive: an odd number squares to an odd number. Write root two as a over b in lowest terms, square both sides, deduce that a and then b are even, and point out that a shared factor of two contradicts lowest terms. Skipping the contrapositive justification is the usual reason this proof loses a mark.
素数无穷：欧几里得的构造
素数无穷多的证明用的是欧几里得的构造。假设素数只有有限个，把它们全部列出来，记作 p1 到 pk。令 N 等于这些素数的乘积再加 1。N 除以列表中任何一个素数都余 1，所以列表里的素数都除不尽 N。可是任何大于 1 的整数都有素因数，于是存在一个不在列表里的素数。这与「列表已列出全部素数」矛盾，所以素数有无限多个。这里有个容易忽略的点：N 本身不一定是素数。用 2、3、5、7、11、13 构造得 30031，它等于 59 乘 509，两个因数都不在原列表里。
Euclid's argument assumes a complete finite list, multiplies the primes together and adds one, so that every prime on the list leaves remainder one. The constructed number still has a prime factor, and that factor cannot be on the list, which is the contradiction. A subtlety many miss is that the number need not be prime: the first six primes give 30031, which factorises as 59 times 509.
逻辑联结词：¬ 与 ∧ 与 ∨ 以及 iff
spec 1.01b 要求看懂三个符号，也要求熟悉 congruence、if then 与 iff 这套说法。¬ 表示否定，∧ 表示并且，∨ 表示或者。写成「若 P 则 Q」时，只证了一个方向，就不能写成 P 当且仅当 Q。当且仅当意味着两个方向都成立，答案里要出现两段推理。很多学生以为把两个方向写在同一段里就算完成，实际上阅卷时会分开给分。符号只是缩写，句子之间的逻辑关系才是被考的东西。遇到 hence 或 show that 这类指令词，先确认它是否也要求逆命题。把命题改写成 P 蕴含 Q 的形式，往往能更快判断该用哪种方法。
Three connectives appear in 1.01b, and the exam expects you to read them as sentences rather than decoration. If and only if is the one that costs marks, because it asserts both directions and needs two separate arguments. Whenever a question says hence or show that, check whether the reverse implication is also being claimed before writing a double arrow.
记号与书写：结构分从哪里来
证明题的书写比计算题更依赖格式。开头写清设什么，中间用 therefore 或 hence 连接，结尾用一句「这与假设矛盾」收束。箭头可以表示推出，但它替代不了文字说明，尤其当一步用到两个已知条件时。阅卷时先找假设句，再找矛盾句，两者都在才给结构分。答案要避免整段没有标点，因为看不出一条链分了几步。把关键等式单独成行，检查的人一眼就能跟上。写完之后回读一遍，专门确认每个「所以」前面都有理由。答案里出现「显然」时，先问自己这一句能不能写出依据。
Presentation carries marks here: open with what you are assuming, link the middle with words rather than arrows alone, and close by stating the contradiction. An arrow shows direction but not the reason, so any step that uses two known facts needs a phrase saying so. Put the key equations on their own lines, since a solid block of prose hides the structure an examiner is looking for.
常用工具：代数变形与数的结构
证明题的素材基本来自几件熟悉工具。代数变形包括展开、因式分解与配方，平方非负是推出不等式的常用依据。判别式可以说明二次方程没有实根，奇偶与整除说明整数的结构。有理数的定义是能写成两个整数之比，无理数就是不能这样写。做题时先判断命题属于哪一类，再挑工具，比直接展开省时间。遇到「对所有整数」的命题，先想想能不能分类，再想想能不能反设。遇到「不存在」的命题，反证法通常比正面构造容易。工具选对之后，剩下的就是耐心。配方之后要说明平方非负，这条理由很短，却是整道题的支点。
Most proof questions lean on a small toolkit: algebraic manipulation, completing the square, the fact that squares are non-negative, the discriminant for real roots, and parity or divisibility for integer claims. Rational means a ratio of two integers, so an irrational argument always starts from that definition. Decide which tool the statement needs before expanding anything, and read a claim about all integers as an invitation to split into cases.
与进阶数学的边界：归纳法不在 H240
H240 的证明只考四条：直接证明、穷举、反例否定、反证法，归纳法不在其中。归纳法属于进阶数学的内容，写进 H240 的答案不会加分，还可能挤掉该写的那几步。另一个边界是反证法与逆否命题，两者结论相同但过程不同，题目要求哪种就写哪种。理解这条边界对复习很有用，证明题的范围是可以穷尽的，不必去翻大学教材。真正需要练的是把每一步的理由说清楚，而不是见更多题型。做完一道题之后，把方法名写在旁边，慢慢就能看出题目的暗示。
Induction is not part of H240; it belongs to Further Mathematics, and importing it into an A Level Mathematics answer earns nothing. Contrapositive and contradiction reach the same conclusion by different routes, so follow the wording of the question. The scope of proof here is finite, which is good news for revision: four methods, with two named standard results attached to the last of them.
备考与考场顺序：证明题怎么写满
复习时先按四种方法各找两三道真题，把标准答案的骨架抄一遍，再自己默写。考场上遇到证明题不要留空，先写假设句和要证的结论，结构分往往能拿到一部分。时间分配上，证明题的分值通常与计算题相当，但文字更多，留出五六分钟比较稳。常见的错误是把结论当条件用，一边证一边假设结论成立。检查时只看两件事：假设有没有写，矛盾有没有点名。这两件事做到，多数 4 分题能拿到 3 分以上。考前一周把四种方法的骨架各默写一遍，比刷十道新题更稳。
Build a skeleton for each of the four methods from past papers, then reproduce it from memory, because recognising which method a question wants is most of the work. In the exam, never leave a proof blank, as the assumption line and the target statement often carry marks on their own. Check two things at the end: that the assumption is stated and that the contradiction is named.
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