📚 A-Level Mathematics: Basic Methods of Mathematical Proof | A-Level 数学：数学证明的基本方法
In A-Level mathematics, proof is the foundation that distinguishes mathematics from mere calculation. A proof is a logical argument that demonstrates the truth of a statement beyond any doubt, using axioms, definitions, and previously established results.
在 A-Level 数学中，证明是区分数学与单纯计算的基础。证明是一种逻辑论证，利用公理、定义和先前确立的结论，毫无疑问地展示一个命题的真实性。
1. Direct Proof | 直接证明
Direct proof begins with known facts or assumptions and applies deductive reasoning step by step to reach the desired conclusion. It is the most straightforward and commonly used method in algebra and number theory.
直接证明从已知事实或假设出发，通过一步一步的演绎推理，推出所要的结论。它是最直接、最常用的方法，广泛应用于代数和数论中。
Example: Prove that the sum of any two even integers is even.
例如：证明任意两个偶数的和仍然是偶数。
Let the two even integers be m = 2a and n = 2b, where a and b are integers. Then their sum is m + n = 2a + 2b = 2(a + b). Since a + b is an integer, m + n is divisible by 2, hence even.
设两个偶数为 m = 2a 和 n = 2b，其中 a、b 是整数。则它们的和为 m + n = 2a + 2b = 2(a + b)。因为 a + b 是整数，所以 m + n 能被 2 整除，即为偶数。
If P is true, and P ⇒ Q by logical steps, then Q is true.
如果 P 为真，且能通过逻辑步骤得到 P ⇒ Q，则 Q 为真。
2. Proof by Contradiction | 反证法
Proof by contradiction assumes that the statement we want to prove is false. If this assumption leads to an impossible or self-contradictory result, then the original statement must be true.
反证法假设我们想要证明的命题为假。如果这个假设导致一个不可能或自相矛盾的结果，那么原命题必然为真。
Classic example: Prove that √2 is irrational.
经典例子：证明 √2 是无理数。
Suppose √2 is rational. Then √2 = p/q, where p and q are integers with no common factor, q ≠ 0, and the fraction is in lowest terms. Squaring both sides gives 2 = p²/q², so p² = 2q². This means p² is even, so p must be even. Let p = 2k.
假设 √2 是有理数，则 √2 = p/q，其中 p、q 为互质整数，q ≠ 0，且该分数已化为最简形式。两边平方得 2 = p²/q²，即 p² = 2q²。这说明 p² 是偶数，因此 p 必然是偶数。令 p = 2k。
Substituting p = 2k into p² = 2q² gives 4k² = 2q², so q² = 2k². Thus q² is even, so q is also even. But then p and q share the common factor 2, contradicting the assumption that p/q is in lowest terms. Therefore √2 cannot be rational; it must be irrational.
将 p = 2k 代入 p² = 2q²，得到 4k² = 2q²，即 q² = 2k²。因此 q² 是偶数，q 也是偶数。但这样一来 p 和 q 就有公因数 2，与 p/q 是最简分数的假设矛盾。因此 √2 不可能是有理数，它必然是无理数。
To prove Q: assume ¬Q and derive a contradiction (e.g. A ∧ ¬A). Therefore Q.
要证 Q：假设 ¬Q 并推出矛盾（如 A ∧ ¬A），因此 Q 成立。
3. Proof by Induction | 数学归纳法
Mathematical induction is used to prove that a statement holds for all positive integers. It consists of two essential steps: the base case and the inductive step.
数学归纳法用于证明一个命题对所有正整数都成立。它包含两个关键步骤：基础步骤和归纳步骤。
Example: Prove that 1 + 2 + 3 + … + n = n(n + 1)/2 for all positive integers n.
例如：证明对所有正整数 n，有 1 + 2 + 3 + … + n = n(n + 1)/2。
Base case: For n = 1, the left side is 1 and the right side is 1 × 2 / 2 = 1. So the statement is true for n = 1.
基础步骤：当 n = 1 时，左边为 1，右边为 1 × 2 / 2 = 1。因此命题对 n = 1 成立。
Inductive step: Assume the statement is true for some positive integer k. That is, 1 + 2 + … + k = k(k + 1)/2. We must show it is true for k + 1.
归纳步骤：假设命题对某个正整数 k 成立，即 1 + 2 + … + k = k(k + 1)/2。我们需要证明它对 k + 1 也成立。
Add (k + 1) to both sides: 1 + 2 + … + k + (k + 1) = k(k + 1)/2 + (k + 1) = (k + 1)(k/2 + 1) = (k + 1)(k + 2)/2. This is exactly the formula for n = k + 1. By induction, the formula holds for all positive integers.
两边同时加上 (k + 1)：1 + 2 + … + k + (k + 1) = k(k + 1)/2 + (k + 1) = (k + 1)(k/2 + 1) = (k + 1)(k + 2)/2。这正是 n = k + 1 时的公式。由数学归纳法，公式对所有正整数成立。
Base case: P(1) is true. Inductive step: P(k) ⇒ P(k + 1) for all k ≥ 1. Therefore P(n) is true for all n ∈ ℤ⁺.
基础步骤：P(1) 为真。归纳步骤：对任意 k ≥ 1，P(k) ⇒ P(k + 1)。因此 P(n) 对所有 n ∈ ℤ⁺ 为真。
4. Proof by Exhaustion | 穷举证明
Proof by exhaustion, also called proof by cases, involves splitting the situation into a finite number of possible cases and verifying each one individually. If every case works, the statement is true.
穷举证明又称分情况证明，它将情况分成有限个可能的类别，并逐一验证每个类别。如果每一种情况都成立，则命题为真。
Example: Prove that n² + n is even for all integers n.
例如：证明对所有整数 n，n² + n 都是偶数。
Case 1: n is even. Let n = 2k. Then n² + n = (2k)² + 2k = 4k² + 2k = 2(2k² + k). This is a multiple of 2, so it is even.
情况一：n 为偶数。令 n = 2k，则 n² + n = (2k)² + 2k = 4k² + 2k = 2(2k² + k)。这是 2 的倍数，因此是偶数。
Case 2: n is odd. Let n = 2k + 1. Then n² + n = (2k + 1)² + (2k + 1) = 4k² + 4k + 1 + 2k + 1 = 4k² + 6k + 2 = 2(2k² + 3k + 1). This is also a multiple of 2, so it is even.
情况二：n 为奇数。令 n = 2k + 1，则 n² + n = (2k + 1)² + (2k + 1) = 4k² + 4k + 1 + 2k + 1 = 4k² + 6k + 2 = 2(2k² + 3k + 1)。这也是 2 的倍数，因此是偶数。
Since both possible cases are covered, n² + n is even for every integer n.
由于所有可能的情况都已覆盖，所以对每个整数 n，n² + n 都是偶数。
5. Disproof by Counterexample | 反例证伪
A single counterexample is enough to disprove a universal statement, such as “all … are …”. To show such a statement is false, find one value that satisfies the conditions but violates the conclusion.
一个反例就足以推翻一个全称命题，例如“所有……都是……”。要说明这样的命题为假，只需找到一个满足条件但不符合结论的值。
Example: Disprove the statement “Every prime number is odd.”
例如：证明“所有素数都是奇数”是假命题。
The number 2 is a prime number, but it is even. Since 2 is a prime that is not odd, the statement is false.
数字 2 是一个素数，但它是偶数。因为 2 是素数却不满足“为奇数”，所以该命题为假。
To disprove ∀x ∈ S, P(x): find one a ∈ S such that P(a) is false.
要推翻 ∀x ∈ S, P(x)：只需找到一个 a ∈ S，使得 P(a) 为假。
6. Constructive Proof | 构造性证明
A constructive proof establishes the existence of an object by explicitly constructing it. Instead of arguing indirectly, it provides a concrete example or a recipe for building the object.
构造性证明通过明确地构造出某个对象来证明它的存在性。它不是间接论证，而是给出一个具体的例子或构造方法。
Example: Prove that there exists an irrational number between 0 and 1.
例如：证明存在一个介于 0 和 1 之间的无理数。
Consider the number x = √2 / 2. We know √2 is irrational. Dividing an irrational number by a non-zero rational number 2 leaves it irrational, so x is irrational. Also, 0 < √2 < 2, so 0 < √2 / 2 < 1. Hence x = √2 / 2 is an irrational number between 0 and 1.
考虑数 x = √2 / 2。我们知道 √2 是无理数，将一个无理数除以非零有理数 2 仍是无理数，所以 x 是无理数。又因为 0 < √2 < 2，所以 0 < √2 / 2 < 1。因此 x = √2 / 2 是一个介于 0 和 1 之间的无理数。
To prove “there exists x with property P”, explicitly define one such x and verify P(x).
要证明“存在具有性质 P 的 x”，只需具体定义一个这样的 x 并验证 P(x)。
7. Proof by Contrapositive | 逆否证明
The contrapositive of “if P then Q” is “if not Q then not P”. A conditional statement and its contrapositive are logically equivalent, so proving the contrapositive proves the original statement.
“若 P 则 Q”的逆否命题是“若非 Q 则非 P”。一个条件命题与其逆否命题在逻辑上等价，因此证明逆否命题即可证明原命题。
Example: Prove that if n² is odd, then n is odd.
例如：证明如果 n² 是奇数，则 n 是奇数。
Instead of proving this directly, we prove its contrapositive: if n is even, then n² is even. Let n = 2k. Then n² = (2k)² = 4k² = 2(2k²), which is even. Since the contrapositive is true, the original statement is also true.
我们不直接证明，而是证明它的逆否命题：如果 n 是偶数，则 n² 是偶数。令 n = 2k，则 n² = (2k)² = 4k² = 2(2k²)，它是偶数。因为逆否命题为真，所以原命题也为真。
P ⇒ Q is equivalent to ¬Q ⇒ ¬P.
P ⇒ Q 等价于 ¬Q ⇒ ¬P。
8. Proving Equivalence | 等价性证明
To prove the equivalence “P if and only if Q”, we must prove two directions: P ⇒ Q and Q ⇒ P. Both directions are necessary for a biconditional statement.
要证明等价命题“P 当且仅当 Q”，我们需要证明两个方向：P ⇒ Q 和 Q ⇒ P。双条件命题的两个方向都是必要的。
Example: Prove that n is even if and only if n² is even.
例如：证明 n 是偶数当且仅当 n² 是偶数。
First, prove n even ⇒ n² even. If n = 2k, then n² = 4k² = 2(2k²), so n² is even.
首先证明 n 为偶数 ⇒ n² 为偶数。若 n = 2k，则 n² = 4k² = 2(2k²)，所以 n² 是偶数。
Second, prove n² even ⇒ n even. We use contrapositive: if n is odd, then n² is odd. Let n = 2k + 1. Then n² = (2k + 1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1, which is odd. Therefore, if n² is even, n cannot be odd, so n is even.
其次证明 n² 为偶数 ⇒ n 为偶数。我们使用逆否命题：如果 n 为奇数，则 n² 为奇数。令 n = 2k + 1，则 n² = (2k + 1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1，它是奇数。因此，若 n² 为偶数，则 n 不可能为奇数，所以 n 为偶数。
Both directions are proved, so the equivalence holds.
两个方向都已证明，因此等价关系成立。
P ⇔ Q means (P ⇒ Q) ∧ (Q ⇒ P).
P ⇔ Q 表示 (P ⇒ Q) ∧ (Q ⇒ P)。
9. Common Mistakes in Proofs | 证明中的常见错误
One common mistake is circular reasoning, where the conclusion is used as an assumption in its own proof. This does not prove anything.
一个常见错误是循环论证，即把待证明的结论本身当作假设来使用。这样无法证明任何命题。
Another mistake is assuming the converse of a statement is true. For example, from P ⇒ Q, you cannot conclude Q ⇒ P unless it has been proved separately.
另一个错误是默认原命题的逆命题成立。例如，从 P ⇒ Q 不能推出 Q ⇒ P，除非单独证明过。
In induction, forgetting the base case makes the argument incomplete. The inductive step alone proves only a conditional relationship, not the statement itself.
在数学归纳法中，遗忘基础步骤会使论证不完整。仅归纳步骤本身只能证明条件关系，不能证明命题本身。
Finally, when using proof by exhaustion, missing one possible case means the proof is invalid. Every case must be covered carefully.
最后，在使用穷举法时，遗漏任何一种情况都会导致证明无效。必须仔细覆盖所有情况。
Always check: assumptions are legitimate, steps are reversible if needed, and no cases are omitted.
始终检查：假设是否合法、步骤是否在需要时可逆、是否遗漏了任何情况。
10. Conclusion | 总结
Mathematical proof is a rigorous tool that guarantees the truth of a statement. Direct proof, contradiction, induction, exhaustion, and the other methods each serve different purposes, and choosing the right one comes with practice.
数学证明是保证命题真实性的严谨工具。直接证明、反证法、归纳法、穷举法以及其他方法各有其用途，选择合适的方法需要不断练习。
For A-Level examinations, it is essential to recognise the structure of each proof method and to write clear, logical steps without gaps. By mastering these fundamental techniques, you will be able to tackle a wide range of problems confidently.
对于 A-Level 考试，重要是识别每种证明方法的结构，并写出清晰、无跳跃的逻辑步骤。掌握这些基本技巧后，你将能够自信地应对各种问题。
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