📚 A-Level Maths: A Summary of Common Proof Techniques | A-Level 数学：常见数学证明方法归纳
Proof is the heart of mathematics. In A-Level mathematics, you are not only required to perform calculations but also to justify why a statement is true. This article summarises the most common proof techniques you will encounter, with worked examples and exam-style guidance.
证明是数学的核心。在 A-Level 数学中，你不仅需要完成计算，更要说明某个命题为何成立。本文归纳了考试中最高频的几种证明方法，配以典型例题与应试要点，帮助你系统掌握证明题的思路。
1. Direct Proof | 直接证明法
Direct proof starts from known facts, definitions, or previously proven theorems and uses logical deduction to reach the desired conclusion. It is the most straightforward and commonly used method.
直接证明法从已知事实、定义或已证定理出发，通过逻辑推理直接推出结论。这是最基础、最常用的一种证明方式。
For example, to prove that the sum of two even numbers is even: let the two even numbers be 2m and 2n, where m and n are integers. Their sum is 2m + 2n = 2(m + n), which is clearly even.
例如，证明两个偶数之和为偶数：设两个偶数分别为 2m 和 2n，其中 m、n 为整数。它们的和为 2m + 2n = 2(m + n)，显然是一个偶数。
2m + 2n = 2(m + n) ⇒ even
Key tip: always clearly state your assumptions before starting the deduction.
关键提示：开始推理之前，务必清楚写出你的假设条件。
2. Proof by Exhaustion | 穷举证明法
Proof by exhaustion is used when a statement can be split into a finite number of cases, and each case can be checked individually. This is common in number theory problems involving modular arithmetic or inequalities.
穷举证明法适用于命题可以被拆分为有限多种情形、且每种情形都可以逐一验证的情况。在涉及模运算或不等式的数论题中尤为常见。
For instance, prove that n² + n is even for all integers n. Consider two cases: if n is even, then n² is even and n is even, so their sum is even. If n is odd, then n² is odd and n is odd, and odd + odd = even. Hence in both cases, n² + n is even.
例如，证明对所有整数 n，n² + n 为偶数。分两种情形：若 n 为偶数，则 n² 为偶数、n 为偶数，二者之和为偶数；若 n 为奇数，则 n² 为奇数、n 为奇数，而奇数 + 奇数 = 偶数。因此两种情形下 n² + n 均为偶数。
Case 1: n even ⇒ n² + n = even; Case 2: n odd ⇒ n² + n = even
Key tip: when there are too many cases, consider whether another method (such as contradiction or induction) is more efficient.
关键提示：当情形过多时，请考虑是否换用反证法或数学归纳法会更高效。
3. Proof by Contradiction | 反证法
Proof by contradiction assumes the opposite of what you want to prove, then shows that this assumption leads to an impossible or absurd consequence. This means the original statement must be true.
反证法先假设要证明的命题不成立（即假设结论的反面成立），再从这个假设推出矛盾或荒谬的结论，从而说明原命题为真。
A classic A-Level example is proving that √2 is irrational. Assume √2 is rational, so √2 = p/q in lowest terms, where p and q are coprime integers. Squaring gives p² = 2q², so p² is even, which implies p is even. Let p = 2k. Then 4k² = 2q², so q² = 2k², meaning q is also even. This contradicts the assumption that p and q are coprime. Therefore √2 is irrational.
A-Level 经典例子是证明 √2 为无理数。假设 √2 是有理数，即 √2 = p/q 为最简分数，其中 p、q 互质。两边平方得 p² = 2q²，所以 p² 为偶数，进而 p 为偶数。设 p = 2k，则 4k² = 2q²，即 q² = 2k²，所以 q 也为偶数。这与 p、q 互质的假设矛盾。因此 √2 是无理数。
√2 = p/q ⇒ p even ⇒ q even ⇒ contradiction with gcd(p,q) = 1
Key tip: the contradiction must be explicit. State clearly which assumption is violated.
关键提示：矛盾必须清晰指出，明确说明它违背了哪一条假设。
4. Proof by Induction | 数学归纳法
Mathematical induction is a powerful method for proving statements that depend on a positive integer n. It consists of two steps: the base case and the inductive step. If both hold, the statement is true for all positive integers n.
数学归纳法用于证明与正整数 n 相关的命题。它包含两个步骤：基础情形（base case）和归纳步骤（inductive step）。若两者成立，则命题对所有正整数 n 成立。
To prove the formula 1 + 2 + 3 + … + n = n(n + 1)/2 for all n ≥ 1:
证明公式 1 + 2 + 3 + … + n = n(n + 1)/2 对所有 n ≥ 1 成立：
Base case: for n = 1, the formula gives 1 = 1(1 + 1)/2 = 1, which is true. Inductive step: assume the formula holds for n = k, so 1 + 2 + … + k = k(k + 1)/2. Then for n = k + 1, adding (k + 1) to both sides gives 1 + 2 + … + k + (k + 1) = k(k + 1)/2 + (k + 1) = (k + 1)(k + 2)/2. This is exactly the formula for n = k + 1.
基础情形：当 n = 1 时，公式给出 1 = 1(1 + 1)/2 = 1，成立。归纳步骤：假设公式对 n = k 成立，即 1 + 2 + … + k = k(k + 1)/2。那么当 n = k + 1 时，两边同时加上 (k + 1)，得到 1 + 2 + … + k + (k + 1) = k(k + 1)/2 + (k + 1) = (k + 1)(k + 2)/2，这正是公式在 n = k + 1 时的形式。
Base: n = 1 ✓; Assumption: n = k ✓; Conclusion: n = k + 1 ✓
Key tip: never forget the base case. Also, explicitly write the phrase ‘by induction, the statement is true for all n ≥ 1’.
关键提示：不要遗漏基础情形。另外，务必写出「由归纳法可知，命题对所有 n ≥ 1 成立」这一句。
5. Disproof by Counter-Example | 反例法
Sometimes a statement is false. The quickest way to show this is to provide one counter-example. A single counter-example is enough to disprove a universal statement.
有时命题本身是假的。最快的方式是举出一个反例。对于全称命题，一个反例足以推翻它。
For example, consider the statement: ‘All prime numbers are odd.’ The number 2 is prime, but 2 is even. Hence the statement is false.
例如，考虑命题「所有素数都是奇数」。数字 2 是素数，但 2 是偶数。因此该命题不成立。
Counter-example: n = 2 (prime but not odd) ⇒ statement false
Key tip: when testing a possible counter-example, always verify it satisfies all the premise conditions.
关键提示：检验反例时，务必确认该对象满足命题的所有前提条件。
6. Proof Involving Inequalities | 不等式证明
Inequality proofs are common in A-Level mathematics. One typical technique is to work backwards from the desired inequality to an obviously true statement, provided every step is reversible. Another is to use the fact that the square of a real number is non-negative.
不等式证明在 A-Level 数学中非常常见。一种典型技巧是从目标不等式倒推，直到得到一个显然成立的事实，且保证每一步都可逆。另一种是利用任意实数的平方非负这一性质。
For example, prove that for all real numbers a and b, a² + b² ≥ 2ab. Rearranging gives a² + b² – 2ab = (a – b)² ≥ 0, which is always true because the square of any real number is non-negative.
例如，证明对任意实数 a 和 b，有 a² + b² ≥ 2ab。移项得 a² + b² – 2ab = (a – b)² ≥ 0。由于任何实数的平方都非负，因此该式恒成立。
a² + b² – 2ab = (a – b)² ≥ 0 ⇒ a² + b² ≥ 2ab
Another useful method for inequalities is ‘proof by contradiction’: assume the inequality is false, then derive a contradiction.
另一种有效方法是反证法：假设不等式不成立，推出矛盾。
Key tip: always check whether the steps you performed would still be valid for negative numbers.
关键提示：始终检查你所做的步骤对负数情形是否依然有效。
7. Proof by Contrapositive | 逆否命题法
The contrapositive of ‘if P then Q’ is ‘if not Q then not P’. A statement and its contrapositive are logically equivalent, so proving the contrapositive proves the original statement. This method is useful when P is difficult to work with directly.
「若 P 则 Q」的逆否命题是「若非 Q 则非 P」。原命题与其逆否命题在逻辑上等价，因此证明逆否命题成立即可证明原命题。当 P 条件难以直接处理时，这一方法非常有效。
For example, prove that if n² is odd, then n is odd. The contrapositive is: if n is even, then n² is even. Let n = 2k. Then n² = 4k² = 2(2k²), which is even. The contrapositive is proved, so the original statement is true.
例如，证明「若 n² 为奇数，则 n 为奇数」。其逆否命题是「若 n 为偶数，则 n² 为偶数」。设 n = 2k，则 n² = 4k² = 2(2k²)，显然是偶数。逆否命题得证，因此原命题成立。
Original: n² odd ⇒ n odd; Contrapositive: n even ⇒ n² even ✓
Key tip: clearly identify both the contrapositive and why it is equivalent before starting the proof.
关键提示：在开始证明前，先明确写出逆否命题，并说明它与原命题等价。
8. Proof of Trigonometric Identities | 三角恒等式证明
Trigonometric identity proofs require you to transform one side of the equation into the other using standard identities, such as sin²θ + cos²θ = 1, tanθ = sinθ/cosθ, or the addition formulas.
三角恒等式证明要求你利用基本恒等式，如 sin²θ + cos²θ = 1、tanθ = sinθ/cosθ 或和角公式，将等式的一边变形为另一边。
For example, prove that (1 – cos²θ)/sinθ = sinθ. Since sin²θ + cos²θ = 1, we have 1 – cos²θ = sin²θ. Therefore (1 – cos²θ)/sinθ = sin²θ/sinθ = sinθ, as required.
例如，证明 (1 – cos²θ)/sinθ = sinθ。因为 sin²θ + cos²θ = 1，所以 1 – cos²θ = sin²θ。于是 (1 – cos²θ)/sinθ = sin²θ/sinθ = sinθ，得证。
1 – cos²θ = sin²θ ⇒ (1 – cos²θ)/sinθ = sinθ
Key tip: when working on both sides of an equation, do not perform operations that assume the equality is true. Keep the sides separate until they match.
关键提示：当对等式两边进行变形时，不要执行任何隐含「等式成立」的运算。两侧应保持独立，直到它们完全相同。
9. Proof of Sequences and Series Results | 数列与级数证明
Questions in this area often require you to prove a closed-form formula for the sum of a series or a term of a sequence. Methods include induction, algebraically manipulating partial sums, or using known summation formulas.
此类问题通常要求证明级数和的通项公式或数列的第 n 项公式。常用方法包括数学归纳法、对部分和进行代数变形，以及利用已知的求和公式。
For example, prove that 1² + 2² + 3² + … + n² = n(n + 1)(2n + 1)/6. One way is induction. Another way is to use the identity (k + 1)³ – k³ = 3k² + 3k + 1, sum from k = 1 to n, and rearrange.
例如，证明 1² + 2² + 3² + … + n² = n(n + 1)(2n + 1)/6。一种方法是数学归纳法；另一种是利用恒等式 (k + 1)³ – k³ = 3k² + 3k + 1，对 k = 1 到 n 求和后重新整理。
Σk² = n(n + 1)(2n + 1)/6, where Σ is from k = 1 to n
Key tip: if a proof seems algebraically complicated, induction is often the safest fallback for sequence and series formulas.
关键提示：如果证明过程显得代数上很繁复，归纳法往往是证明数列与级数公式最稳妥的方法。
10. Exam Strategy and Common Mistakes | 应试策略与常见错误
In A-Level exams, proof questions are often embedded in longer problems. You should read the question carefully, identify which method is being requested, and write your steps in a logical order with clear justifications.
在 A-Level 考试中，证明题通常嵌入在较长的题目中。你应仔细审题，判断题目要求使用哪种方法，并以清晰的逻辑顺序写出每一步骤和理由。
The most common mistakes are: skipping the base case in induction, omitting ‘by contradiction’ statements, failing to check the validity of algebraic manipulations, and using the result to prove itself.
最常见的错误包括：在归纳法中遗漏基础情形；未写出「反证」声明；未检查代数变形是否合法；以及用待证结论去证明自身（循环论证）。
- Always write down what is given and what needs to be proved.
- 在动笔前明确写出已知条件与待证结论。
- Do not skip steps in logical deduction.
- 不要在逻辑推导中跳步。
- If using contradiction, make the contradiction explicit.
- 使用反证法时，必须明确指出矛盾所在。
- If using induction, label the base case, assumption and inductive step clearly.
- 使用归纳法时，清晰标注基础情形、归纳假设与归纳步骤。
Key tip: practice proof-writing by explaining each line of your solution as if teaching a peer.
关键提示：练习证明题时，请像教同学一样解释每一行的依据，这能显著提升表达的严谨性。
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