Methods of Proof | 证明方法

📚 Methods of Proof | 证明方法

In A-Level Mathematics, proof is not just ‘showing a calculation’ – it is a rigorous chain of reasoning that establishes a statement for all possible cases. Edexcel expects you to know direct proof, proof by exhaustion, proof by contradiction, proof by contrapositive, and disproof by counterexample.

在 A-Level 数学中，证明不只是“展示计算过程”，而是严格的推理链条，能够确立一个命题对所有可能情形成立。Edexcel 要求你掌握直接证明、穷举证明、反证法、逆否命题证明以及反例反驳。

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1. What is a Mathematical Proof? | 什么是数学证明？

A proof is a logical argument that starts from axioms, definitions and known theorems, and uses valid rules of inference to reach the statement being proved. A valid proof leaves no room for exceptions.

证明是从公理、定义和已知定理出发，使用有效推理规则得出结论的逻辑论证。有效证明不容许例外。

In Edexcel exams, you may be asked to prove a given result, to complete a partially written proof, or to identify an error in an argument. You must therefore understand both the structure and the language of proof.

在 Edexcel 考试中，题目可能要求证明给定结论、补全部分证明，或找出论证中的错误。因此你既要理解证明的结构，也要掌握证明的语言。

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2. Direct Proof | 直接证明（演绎法）

Direct proof, also called deduction, starts from known facts and applies algebraic manipulation, definitions or logical steps to arrive directly at the conclusion.

直接证明又称演绎法，从已知事实出发，通过代数变形、定义或逻辑步骤直接推出结论。

For example, prove that the sum of two even numbers is even. Let a = 2m and b = 2n, where m and n are integers.

例如，证明两个偶数之和为偶数。设 a = 2m，b = 2n，其中 m 和 n 为整数。

a + b = 2m + 2n = 2(m + n)

Since m + n is an integer, a + b is a multiple of 2, so it is even. This direct argument proves the result for all even numbers.

因为 m + n 是整数，所以 a + b 是 2 的倍数，故为偶数。这个直接论证证明了对所有偶数都成立。

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3. Proof by Exhaustion | 穷举证明

Proof by exhaustion works when a statement only involves a finite number of cases. You check each case separately, and if every case satisfies the statement, the proof is complete.

当命题只涉及有限多种情形时，可以用穷举证明。逐一检查每种情形，若全部满足，则证明完成。

A useful example is proving that no square number ends in 7. Consider the last digit d of any integer n. The last digit of n² depends only on d, so it is enough to check d = 0, 1, 2, …, 9.

一个有用的例子是证明平方数的末位不可能是 7。考虑整数 n 的末位数字 d。n² 的末位只取决于 d，因此只需检查 d = 0, 1, 2, …, 9。

d	0	1	2	3	4	5	6	7	8	9
last digit of d²	0	1	4	9	6	5	6	9	4	1

None of the final digits is 7, so no square number ends in 7. This is a complete exhaustion because every possible last digit has been checked.

这些末位数字中没有 7，因此平方数的末位不可能是 7。这是一次完整的穷举，因为所有可能的末位数字都被检查了。

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4. Disproof by Counterexample | 反例反驳

A universal claim can be disproved by one counterexample. This is not a proof of truth; it shows that the statement is false for at least one case.

一个全称命题只需一个反例即可被反驳。这不是证明命题为真，而是说明该命题至少有一个情形不成立。

For example, consider the claim: ‘For every integer n, n² + n + 41 is prime.’ If n = 40, then:

例如，考虑命题：“对于所有整数 n，n² + n + 41 都是质数。”当 n = 40 时：

n² + n + 41 = 40² + 40 + 41 = 1600 + 40 + 41 = 1681 = 41²

Since 1681 = 41², it is not prime. Therefore the original claim is false. A single counterexample is enough to disprove a universal statement.

因为 1681 = 41²，它不是质数。因此原命题为假。一个反例就足以反驳一个全称命题。

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5. Proof by Contradiction | 反证法（矛盾证明）

In proof by contradiction, you assume the opposite of the statement you want to prove. If this assumption leads to a logical contradiction, the original statement must be true.

反证法中，先假设要证明的命题的反面成立。如果该假设推出逻辑矛盾，则原命题必为真。

The usual steps are: identify the statement P, assume P is false, use valid reasoning to derive an impossible result such as 1 = 0 or a number being both even and odd, then conclude P is true.

通常步骤为：确定命题 P，假设 P 不成立，通过有效推理推出不可能的结果，例如 1 = 0 或某数既为偶数又为奇数，然后得出 P 成立。

Proof by contradiction is especially useful for negative statements, such as ‘there is no greatest integer’ or ‘√2 is irrational’.

反证法特别适用于否定性命题，例如“不存在最大整数”或“√2 是无理数”。

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6. Worked Example: Proving √2 is Irrational | 典型例题：证明 √2 是无理数

The classic Edexcel contradiction proof is that √2 is irrational. Suppose, for contradiction, that √2 is rational.

Edexcel 经典反证法例题是证明 √2 为无理数。假设，为了导出矛盾，√2 是有理数。

Then √2 = p/q, where p and q are integers, q ≠ 0, and the fraction is in lowest terms. Squaring gives:

则 √2 = p/q，其中 p 和 q 为整数，q ≠ 0，且分数为最简形式。两边平方得：

√2 = p/q → 2 = p²/q² → p² = 2q²

From p² = 2q², p² is even, so p must be even. Write p = 2k for some integer k. Then p² = 4k² = 2q², so q² = 2k².

由 p² = 2q² 可知 p² 为偶数，因此 p 必为偶数。设 p = 2k，其中 k 为整数。则 p² = 4k² = 2q²，故 q² = 2k²。

Hence q² is even, so q is even. Both p and q are even, which contradicts the assumption that p/q is in lowest terms.

于是 q² 为偶数，所以 q 也为偶数。p 和 q 均为偶数，这与 p/q 为最简分数的假设矛盾。

Therefore

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