OCR数学 ALevel 证明 演绎 反例 反证法 穷举
OCR A Level Mathematics A 的代码是 H240，纯数内容的第一块就是 1.01 Proof，而且它不单独占一张卷：三张卷都考，任何一道代数、三角或积分题都可能顺带要求你说明理由。这篇文章按 OCR 官方 spec 给出的四个方法展开，把演绎、穷举、反例、反证法各自适用的场景、写法的固定结构，以及考生最常见的丢分点讲清楚，最后给出 H240 的卷面口径与三周复习安排。文中引用的卷面结构、命令词释义与 AO 权重来自 OCR 官方 spec（H240，2017 年 9 月起施教）与 OCR 官网 1.01 Proof 的教学指南。写证明不是把题目给的式子化简一遍，而是从已知条件一步步推到结论。
Proof appears at the start of the OCR A Level Mathematics A specification (H240) as section 1.01, and it is examined across all three papers rather than in a paper of its own; algebraic, trigonometric and integration questions can all demand a justification. This article follows the four methods OCR names in the specification, namely deduction, exhaustion, counter-example and contradiction, and sets out where each one fits, the fixed structure examiners expect, and the marks candidates drop most often. A proof is a chain of reasoning from given conditions to a conclusion, not a restatement of the expression you were handed.
H240 三张卷里证明题的位置与分值
OCR A Level Mathematics A 的三张卷分别是 H240/01 Pure Mathematics、H240/02 Pure Mathematics and Statistics 和 H240/03 Pure Mathematics and Mechanics，每张卷 2 小时、满分 100 分、各占总成绩的三分之一，整门课合计 300 分。纯数内容在三张卷上都会考，OCR 官方 spec 写明任何纯数主题都可能出现在任何一张卷上。以 H240/03 为例，官方 Summer 2024 examiner report 写明这张卷分成两段：Section A 纯数 51 分，Section B 力学 49 分；H240/02 同样是纯数加统计的两段结构，统计题还假定你熟悉 pre-release large data set。AS 与 A Level 的分界也落在证明这一块，AS 只要求证明的语言与结构加三个方法，反证法是 A Level 才引入的。证明题本身通常只有 3 到 5 分，但它挂在 AO2 上：OCR spec 给出的 AO2 权重是 23% 到 27%，H240/01 一张卷里 AO2 占 25 到 29 分。丢掉一道 4 分的证明，看起来是丢了一个小题，实际是在三分之一的卷面上少拿一块，因为三张卷各占三分之一。
The qualification is H240, and its three papers are H240/01 Pure Mathematics, H240/02 Pure Mathematics and Statistics, and H240/03 Pure Mathematics and Mechanics; each paper lasts 2 hours, carries 100 marks and contributes a third of the final grade, giving 300 marks in total. Pure mathematics is examined on all three papers, and the OCR examiner report for H240/03 states that Section A carries 51 marks of pure mathematics against 49 marks of mechanics in Section B. Proof questions rarely carry more than 3 to 5 marks, yet they belong to Assessment Objective 2, which the OCR specification weights at 23% to 27% of the qualification; on H240/01 alone that is 25 to 29 marks.
证明的语言：定义变量、写出假设与结论
写证明的第一个动作是定义变量，OCR spec 对命令词 Prove 的释义里明确要求清楚定义变量，也就是在开头写出 let n be an integer 这类句子。变量没有定义，后面的代数再漂亮也缺一个给分点。接着要写清假设了什么、要证明什么，A Level 的证明是从已知出发推向结论，不是把题面重抄一遍。符号上，等号用于两个相等的东西，恒等号用于在整个定义域内都成立的等式；必要与充分条件要分清，A 是 B 的充分条件说的是 A 成立就足以推出 B，此时 B 是 A 的必要条件。很多学生以为把表达式化简成某个漂亮形式就算证完，其实还差最后那句结论。结论句要明确写出被证明的命题成立，例如 so the difference is a multiple of 8 for all integers n，这一句在 3 到 4 分的题里往往就是 1 分。
Start by defining the variables, because the OCR specification's definition of the command word Prove asks for variables to be defined clearly, so a line such as let n be an integer earns its mark before any algebra appears. State what is assumed and what has to be shown, since a proof runs from given conditions to a conclusion rather than repeating the question. Use an equals sign for equal quantities and an identity sign for an equation that holds throughout a domain, keep necessary and sufficient conditions apart, and finish by naming the result, as in so the difference is a multiple of 8 for all integers n.
四个方法的分工：演绎、穷举、反例与反证法
OCR 官方 blog 把四个方法的分工写得很清楚：AS 的内容是证明的语言与结构、proof by deduction、proof by exhaustion 和 disproof by counter-example，到了 A Level 才引入 proof by contradiction。反证法在 AS 阶段不需要，但 A Level 卷上完全可能考。OCR 官网同时提醒，每年都有考生用数学归纳法去做 A Level 的证明题，induction 属于 Further Mathematics，正确解答照样给分，但比用考纲内的方法更费时间，也不会额外加分。四个方法怎么选看题目给的信息：要证一个对所有整数成立的等式，通常用演绎；命题被限制在有限多种情形里，用穷举；遇到对所有都成立的断言，先想反例；要求证明某个对象不存在或某个结论不可能，反证法是首选。很多学生以为反证法与反例是一回事，OCR 官方 blog 把这一点直接点名为 common misconception：反例需要一个反面的具体例子，反证法是假设命题不成立，再推出自相矛盾。方法选错，整道题基本拿不到分。
The OCR blog sets out the split clearly: AS content covers the language and structure of proof, proof by deduction, proof by exhaustion and disproof by counter-example, while proof by contradiction is introduced at A Level. Proof by induction belongs to Further Mathematics, and although a correct induction argument would be credited on an A Level paper it brings no advantage over the specification methods and usually costs more time. A single numerical example never proves a general statement, whereas one counter-example is enough to destroy it, and the OCR blog files the confusion between contradiction and counter-example as a misconception that can cost a candidate the whole question.
演绎证明：连续奇数的平方差是 8 的倍数
演绎证明的模板可以用一道经典题看清楚：证明任意两个连续奇数的平方差是 8 的倍数。第一步定义变量，写 let the two consecutive odd numbers be 2n - 1 and 2n + 1 where n is an integer。第二步展开，注意中间项的符号：(2n + 1)² - (2n - 1)² = (4n² + 4n + 1) - (4n² - 4n + 1) = 8n。第三步写结论：8n 是 8 的倍数，所以两个连续奇数的平方差是 8 的倍数。整段只有三行，给分点却分布在定义变量、展开正确、结论完整三处，缺一处就少吃一分。同一个套路用来证明 n² - n 对所有整数都是偶数：写成 n(n - 1)，连续两个整数里必有一个是偶数，乘积就是偶数。学生最常把变量范围写掉，n 是不是整数决定了结论成不成立，写成实数整段推理立刻失效。官方口径很直接，代入几个数值不算证明，只有当范围有限、能把所有情形列完时，数值检查才有意义。
A standard deduction question asks you to prove that the difference of the squares of two consecutive odd numbers is a multiple of 8. Write the numbers as 2n - 1 and 2n + 1 for integer n, expand with attention to the signs so that (2n + 1)² - (2n - 1)² collapses to 8n, and close with the sentence that 8n is a multiple of 8; the marks sit on the variable definition, the expansion and the conclusion. An answer that declares real numbers instead of integers collapses before the algebra has even been assessed, which is why the domain is worth writing down first.
配方与不等式：证明对所有实数成立
配方是不等式证明里最稳的一条路，OCR 的题面常写成 prove that x² - 6x + 10 > 0 for all real x。做法是把它写成完全平方加常数：x² - 6x + 10 = (x - 3)² + 1。取值范围随后直接读出来，(x - 3)² 对所有实数都大于等于 0，所以整个式子大于等于 1，自然大于 0。常见的错误是只写到 (x - 3)² + 1 就停笔，没写为什么这就完成了证明，那一步的说明在 mark scheme 里是独立的一分。另一条路是判别式：把 x² - 6x + 10 = 0 当二次方程看，判别式 36 - 40 = -4 小于 0，开口向上，所以图像与 x 轴没有交点，函数恒为正。题目若换成 prove that x² + 2x + 5 is always positive，同样的配方得到 (x + 1)² + 4，最小值 4 在 x = -1 处取到。两条路都合法，配方在解释性上更好写，也更容易在 3 分的题里一次拿满。
Completing the square turns an inequality proof into a one-line reading of a range, since writing x² - 6x + 10 as (x - 3)² + 1 shows at once that the expression is at least 1 for every real x. Candidates who stop after the completed-square form lose the mark that depends on saying why a square is never negative and why that settles the claim. The discriminant offers an alternative route: setting the expression to zero gives 36 - 40, a negative discriminant with a positive leading coefficient, so the curve never meets the x-axis and stays above it.
穷举证明：有限范围内一个不漏
穷举证明适用于情形数有限、并且能说清已覆盖全部情形的题目。OCR 官方 blog 给过一个例子：证明任意整数的立方要么是 9 的倍数，要么比 9 的倍数大 1 或小 1。关键的一句是把所有整数写成 3a - 1、3a 或 3a + 1 三种形式：n = 3a - 1 时 n³ = 9(3a³ - 3a² + a) - 1，n = 3a 时 n³ = 9(3a³)，n = 3a + 1 时 n³ = 9(3a³ + 3a² + a) + 1，三种情形都落在命题说的范围内。另一类常见题是证明三个连续整数的平方和不能被 3 整除：n² + (n + 1)² + (n + 2)² = 3(n + 1)² + 2，余数是 2，除不尽；代入 n = 1 得 1 + 4 + 9 = 14，14 除以 3 余 2，与公式一致。平方数的个位数也可以用穷举处理，平方对 10 取余只有 0、1、4、5、6、9 六种结果，所以不存在个位为 2 的平方数，把 0 到 9 逐一检查就覆盖了全部整数。穷举的字面要求是逐项全覆盖，漏掉一种情形，结论就只是猜想，阅卷时最常见的问题正是列出了情形却没有说明为什么这几种够用。
Proof by exhaustion works when the cases are finite and every one of them can be listed. The OCR blog example notes that every integer can be written as 3a - 1, 3a or 3a + 1, so the cube of any integer is a multiple of 9 or one more or one less than a multiple of 9; the same style handles the sum of the squares of three consecutive integers, which is 3(n + 1)² + 2 and therefore never divisible by 3. Squares modulo 10 give only 0, 1, 4, 5, 6 and 9, which is why no square ends in 2, and what examiners look for is the sentence explaining that the list covers every case.
反例证伪：一个例子就够
反例证伪的规则只有一条：一个反例就能推翻一个全称命题。OCR 官方 blog 用的例子是 all square numbers are even 可以被 3² = 9 推翻，一个奇数平方就足够。更值得练的是 n² - n + 41 这道经典题，它在前 40 个正整数上都给出素数，所以只试几个值很危险；取 n = 41，得到 41² - 41 + 41 = 1681 = 41 × 41，不是素数，命题被推翻。写反例时要说清两件事：这个值确实落在命题的讨论范围内，它的结果确实与命题相矛盾。错得最多的是只写一个数字就收笔，没有那句 therefore the statement is false 的说明，或者拿一个不在定义域里的值当反例，例如用负数去推翻一个关于正整数的命题。反例题的分值通常不高，但它是唯一不需要完整推理链条的证明类题，先把这几分拿稳比较划算。
One counter-example is enough to disprove a universal claim, and the OCR blog's own illustration is that the statement all square numbers are even fails at 3² = 9. The harder classic is n² - n + 41, prime for the first 40 positive integers yet equal to 1681, that is 41 × 41, when n is 41, which is exactly why testing a handful of values proves nothing. Marks disappear when a candidate offers a number without stating that it lies inside the domain of the claim and that it contradicts the statement being tested.
反证法：无理数与素数无穷多
反证法的固定结构是三步：假设命题不成立，推出与已知事实或定义矛盾的结果，于是原命题成立。OCR 官方 blog 用 √7 做示范：假设 √7 是有理数，写成 p/q，其中 p 与 q 是互质的整数；两边平方得 7q² = p²，说明 p 是 7 的倍数，写成 p = 7k；代回得 7q² = 49k²，即 q² = 7k²，于是 q 也是 7 的倍数；p 与 q 同时是 7 的倍数，与互质的假设矛盾。同一套写法把 7 换成 2 就是 √2 无理的经典证明，H240 的真题里两类都出现过。素数有无限多个也用反证法：假设素数只有有限个，把它们全部相乘再加 1，得到的数除以名单里的任何一个素数都余 1，所以它要么本身是新的素数，要么能被一个不在名单里的素数整除。很多学生以为反证法的终点是那个矛盾，其实最后还要写一句结论，说明原假设不成立、命题得证。矛盾从哪来要写具体，与互质矛盾、与整除性矛盾，还是与已知定理矛盾，含糊的一句 contradiction 拿不到那 1 分。
Proof by contradiction has three moves: assume the opposite of the statement, derive something impossible, and conclude that the original statement must be true. For the irrationality of the square root of 7, assume it equals p over q in lowest terms, square both sides to get 7q² = p², deduce that both p and q are multiples of 7, and point out that this clashes with the assumption that they share no common factor; the same argument with 2 in place of 7 gives the classic proof for the square root of 2. The proof that there are infinitely many primes follows the same shape, because multiplying a supposed complete list of primes and adding 1 produces a number divisible by none of them.
结论句与符号：把逻辑链闭合
证明的最后一步是把逻辑链闭合，这一步在 OCR 的 mark scheme 里通常对应最后一个 A1。结论句要直接写出题目要求的命题，例如 state that the difference of the squares of two consecutive odd numbers is a multiple of 8，而不是笼统写 done 或者只画一个方框符号。符号上也有几条硬规矩：等号连接两个相等的东西，恒等号表示在整个定义域内成立，if and only if 意味着两个方向都真，写它之前要确认反向也成立。必要与充分条件在题面上常以 needing 与 sufficient 这类措辞出现，审题时把这两个词圈出来，能省掉一次重写。阅卷时缺结论句的卷面很常见，考生把该做的代数都做了，最后一分却没拿到。另一种丢分是把待证的结论当成条件用，例如从 x² - 6x + 10 > 0 出发去推 (x - 3)² + 1 > 0，方向反了。落笔前先问一句：这个结论是我推出来的，还是我一开始就假设进来的。
The final line of a proof has to name the result rather than merely signal that the working is finished, and that line usually carries the last accuracy mark. Symbols need to stay honest as well: an equals sign joins equal quantities, an identity sign marks an equation true throughout a domain, and if and only if commits you to both directions, so it should not appear unless the converse has been shown. Reading the question for words such as necessary and sufficient prevents the most damaging error of all, which is working backwards and treating the statement you were asked to prove as though it were given.
阅卷口径：Prove Show that Determine Verify 的差别
OCR 的卷面命令词决定了答案要写到什么程度，官方 classroom guidance 对这几个词有明确定义。Prove 要求给出形式化的数学论证，变量定义清楚、代数变形正确、结论简洁；Show that 是已经把结果给了你，答案要覆盖每一步，任何一步缺失都会被视作论证不完整；Determine 允许先推理再给结论，过程同样要保留；Verify 只要求代入给定的值并说明它满足条件，不需要从头证一遍。Detailed reasoning 常出现在较长题目的某一问里，通常对应多分的推理链，写几个数字很难拿到分。很多学生以为 Show that 可以把印出来的结果当理由，先写 if x = 2 then 4 = 4 就当成证明；题目既然把结果印出来了，它评价的是你的过程，不是你的结论。这几种命令词在 H240/01 到 H240/03 三张卷上口径一致，官方 examiner report 明确写过，统计卷的纯数部分也按同样严格的尺度给分。
Command words decide how much working is required, and OCR publishes classroom guidance on them. Prove demands a formal argument with defined variables, correct algebra and a concise conclusion; Show that hands you the result and assesses the route, so a skipped step counts as incomplete; Determine expects reasoning before the answer; and Verify asks only for a substitution that confirms the given value. Candidates who treat Show that as an invitation to quote the printed result miss the point of the question, and the same standard applies to the pure section of every H240 paper.
真题与复习安排：三周一循环
复习证明的节奏可以按三周安排，重点是手写而不是重读。用 OCR 官网的 past papers，从 2018 年首次考试起的 H240/01、H240/02、H240/03 都可以拿来练，每张卷前几道题里通常就有一道证明。每周固定做 3 道，限时 6 分钟，做完立刻与 mark scheme 逐句对照，把少写的那一句记在本子上：变量没有定义、结论句缺失、或者穷举没有说明覆盖范围。六类题型值得各整理一张卡片：平方差与整除、配方与不等式、穷举的余数分类、反例、无理数的反证、素数无穷多的反证，每张卡片只写结构，不写题目原文。考前一周把卡片对着空白纸重写一遍，比再读一遍答案有用。这一步最容易出的问题是只看不写，以为自己会了，考场上却拼不出完整的链条。
A three-week cycle is enough to make proof routine if the practice is written rather than read. Work through OCR past papers from the 2018 series onwards, taking three proof questions a week under a six-minute limit and comparing each attempt line by line with the mark scheme; note every missing definition, conclusion or statement of coverage. Six cards cover the common shapes, namely difference of squares and divisibility, completing the square, exhaustion by remainders, counter-examples, irrationality by contradiction, and the infinitude of primes, and rewriting them from memory in the final week beats rereading solutions.
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