Methods of Mathematical Proof | 数学证明的常用方法汇总

📚 Methods of Mathematical Proof | 数学证明的常用方法汇总

Mathematical proof is the process of using known definitions, axioms and previously proven results to establish a statement beyond all doubt. In examination courses such as A-level mathematics, proof questions test logical reasoning, clear communication and careful choice of method. This article summarises the most common proof methods that you are likely to need.

数学证明是运用已知定义、公理以及已被证明的结论，使一个命题无可置疑地成立的过程。在 A-level 数学等考试课程中，证明题考查逻辑推理、清晰表达以及方法的正确选择。本文将汇总你考试中最常用的证明方法。

1. Direct Proof | 直接证明

Direct proof starts from the assumptions and uses logical steps to reach the required conclusion. This is the first approach you should try for almost any proof question. The key is to translate the assumption into an algebraic or definition-based form, then manipulate it until the conclusion appears.

直接证明从假设出发，通过逻辑推理逐步得到所需结论。面对证明题，这几乎是应该首先尝试的方法。关键在于把已知条件转化为代数形式或定义形式，然后逐步变化，直到结论显现。

For example, prove: If n is an even integer, then n² is even. Since n is even, we may write n = 2k for some integer k. Then n² = (2k)² = 4k² = 2(2k²). Because 2k² is an integer, n² must be even.

例如，证明：如果 n 是偶数，那么 n² 是偶数。因为 n 是偶数，可写 n = 2k，其中 k 为整数。于是 n² = (2k)² = 4k² = 2(2k²)。由于 2k² 是整数，所以 n² 一定是偶数。

n = 2k ⇒ n² = 4k² = 2(2k²)

Similarly, the sum of two even integers is even. If m = 2r and n = 2s, then m + n = 2r + 2s = 2(r + s), which is an even number.

类似地，两个偶数之和仍为偶数。若 m = 2r，n = 2s，则 m + n = 2r + 2s = 2(r + s)，结果仍是偶数。

2. Proof by Contradiction | 反证法

Proof by contradiction assumes that the statement you want to prove is false, then uses correct reasoning to arrive at an impossible conclusion. If that impossible conclusion is reached, the original assumption must have been false, so the original statement must be true. This method is particularly effective for statements beginning with “there is no” or for proving that a number is irrational.

反证法先假设要证明的命题为假，然后通过严谨推理导出一个不可能的矛盾。既然矛盾出现，就说明原假设不成立，原命题必然为真。这一方法尤其适合“不存在……”类命题的证明，也常用于证明某个数是无理数。

The classic A-level example is proving that √2 is irrational. Suppose, for contradiction, that √2 is rational. Then √2 = a/b, where a and b are integers with no common factor and b ≠ 0. Squaring both sides gives a² = 2b², so a² is even. Therefore a must be even, so write a = 2k. Substituting gives 4k² = 2b², which simplifies to b² = 2k², so b² is even and hence b is even. But a and b cannot both be even if they have no common factor. This contradiction proves that √2 is irrational.

A-level 中的经典例子是证明 √2 是无理数。假设相反，√2 是有理数，则 √2 = a/b，其中 a、b 是互质的整数且 b ≠ 0。两边平方得 a² = 2b²，所以 a² 是偶数，从而 a 是偶数。令 a = 2k，代入得 4k² = 2b²，即 b² = 2k²，于是 b² 也是偶数，b 也为偶数。但若 a 和 b 互质，它们不可能同为偶数。这一矛盾说明 √2 不是有理数，因此 √2 是无理数。

√2 = a/b ⇒ a² = 2b² ⇒ a = 2k ⇒ b² = 2k²

3. Proof by Contrapositive | 逆否命题证明

To prove a conditional statement p ⇒ q, we can instead prove its contrapositive ¬q ⇒ ¬p, because the two statements are logically equivalent. This is useful when the original statement is awkward to handle but the contrapositive is easier to prove directly.

要证明条件命题 p ⇒ q，可以转而证明它的逆否命题 ¬q ⇒ ¬p，因为二者在逻辑上等价。当原命题难以直接下手时，逆否命题往往更容易证明。

For example, prove: If n² is odd, then n is odd. The contrapositive is: If n is even, then n² is even. We already know how to prove this directly: if n = 2k, then n² = 4k² = 2(2k²), so n² is even. Since the contrapositive is true, the original statement is also true.

例如，证明：如果 n² 是奇数，那么 n 是奇数。它的逆否命题是：如果 n 是偶数，那么 n² 是偶数。我们已经知道如何直接证明：若 n = 2k，则 n² = 4k² = 2(2k²)，所以 n² 是偶数。逆否命题成立，因此原命题也成立。

p ⇒ q is equivalent to ¬q ⇒ ¬p

4. Proof by Induction | 数学归纳法

Proof by induction is ideal for statements that involve positive integers and claims that hold for every positive integer n. The method has three parts: a base case, an inductive hypothesis and an inductive step. If the base case is true and the inductive step works, then the statement is true for all positive integers.

数学归纳法适用于涉及正整数且对所有正整数 n 都成立的命题。这一方法包含三个部分：基础情形、归纳假设和归纳递推。如果基础情形成立，且归纳递推成立，那么命题对所有正整数都成立。

Prove that 1 + 2 + 3 + … + n = n(n + 1)/2 for all positive integers n.

证明：对所有正整数 n，1 + 2 + 3 + … + n = n(n + 1)/2。

Base case: when n = 1, the left-hand side is 1 and the right-hand side is 1(1 + 1)/2 = 1, so the statement is true for n = 1.

基础情形：当 n = 1 时，左边为 1，右边为 1(1 + 1)/2 = 1，所以命题对 n = 1 成立。

Inductive hypothesis: assume the statement is true for n = k, so 1 + 2 + … + k = k(k + 1)/2.

归纳假设：假设命题对 n = k 成立，即 1 + 2 + … + k = k(k + 1)/2。

Inductive step: add k + 1 to both sides. Then 1 + 2 + … + k + (k + 1) = k(k + 1)/2 + (k + 1) = (k

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