📚 Example 1.8.1: Algebraic Proof | 示例 1.8.1：代数证明
In AQA A-Level Mathematics, algebraic proof is a core skill that appears throughout the Pure specification. This article uses Example 1.8.1 to show how to construct rigorous, exam-ready proofs and avoid common pitfalls. The worked example is a classic proof about consecutive integers, but the methods extend to many other types of proof questions.
在 AQA A-Level 数学中，代数证明是贯穿 Pure 模块的核心技能。本文以例题 1.8.1 为例，演示如何构建严谨、符合考试要求的证明，并避免常见错误。该例题是一个关于连续整数的经典证明，但所用方法可推广到许多其他类型的证明题。
1. What Is Algebraic Proof? | 什么是代数证明？
Algebraic proof uses variables to represent arbitrary numbers, so a single argument covers all possible cases. Unlike numerical checking, a proof must show why a statement is always true. In AQA exams, you will be asked to prove results about integers, odd and even numbers, squares, and sums.
代数证明用变量表示任意数，因此一个论证可以覆盖所有可能情形。与数值验证不同，证明必须说明为什么命题总是成立。在 AQA 考试中，你会被要求证明关于整数、奇偶数、平方数和和式的结论。
For example, if you substitute n = 1, 2, 3 into a formula and it works, that only gives evidence. It does not prove the formula is true for every integer. A valid proof must use algebra to establish the result generally, without relying on specific examples.
例如，你把 n = 1、2、3 代入某个公式并且结果成立，这只能作为证据，并不能证明该公式对所有整数都成立。一个有效的证明必须使用代数一般性地建立结论，而不是依赖具体例子。
2. The Worked Example 1.8.1 | 例题 1.8.1 详解
The classic example asks you to prove that the sum of any three consecutive integers is divisible by 3. The cleanest approach is to let the first integer be n, so the three integers are n, n + 1 and n + 2. Their sum is n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1). Since n + 1 is an integer, the sum is 3 times an integer, so it is divisible by 3.
这个经典例题要求证明任意三个连续整数的和能被 3 整除。最简洁的方法是设第一个整数为 n，因此三个整数为 n、n + 1 和 n + 2。它们的和为 n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1)。因为 n + 1 是整数，所以和是某个整数的 3 倍，因此能被 3 整除。
Sum = n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1)
This proof is complete because it does not restrict n to any particular value. Letting n be any integer ensures that the argument covers all triples of consecutive integers, such as 1, 2, 3 or 100, 101, 102 or −7, −6, −5.
这个证明是完整的，因为它没有将 n 限制为任何特定值。设 n 为任意整数保证了该论证覆盖所有连续整数三元组，例如 1、2、3，或 100、101、102，或 −7、−6、−5。
3. Setting Up the Algebra | 建立代数表达式
The key step is to define variables clearly. Write ‘Let n be any integer’ before using it. For consecutive integers, n, n + 1, n + 2 is standard. For consecutive even numbers, use 2n, 2n + 2, 2n + 4. For consecutive odd numbers, use 2n + 1, 2n + 3, 2n + 5. Always state the domain of n, such as n ∈ Z (integer).
关键步骤是清晰地定义变量。在使用变量前写下“设 n 为任意整数”。对于连续整数，n、n + 1、n + 2 是标准设置。对于连续偶数，使用 2n、2n + 2、2n + 4。对于连续奇数，使用 2n + 1、2n + 3、2n + 5。始终说明 n 的取值范围，例如 n ∈ Z（整数）。
In AQA mark schemes, the first mark is often awarded for correctly defining variables. If you write ‘Let n be an integer’ and then use 2n for an even number, you have shown that all even numbers are of that form. Avoid using different letters for the same purpose, as this can make the proof unclear.
在 AQA 评分标准中，第一个得分点通常是正确设定变量。如果你写下“设 n 为整数”，然后用 2n 表示偶数，你就已经展示了所有偶数都可以写成这种形式。避免用不同字母表示同一含义，这会让证明变得不清楚。
4. Proof by Deduction | 演绎证明
Deduction is the most common form of proof in AQA. You start from known facts, such as the definition of an integer, and use algebra to reach the conclusion. Each line must follow logically from the previous one. In Example 1.8.1, the deduction is: sum = 3(n + 1), and because n + 1 is an integer, the expression is a multiple of 3.
演绎是 AQA 中最常见的证明形式。你从已知事实出发，例如整数的定义，然后通过代数推导得出结论。每一行都必须从上一行逻辑推出。在例题 1.8.1 中，演绎过程为：和 = 3(n + 1)，而由于 n + 1 是整数，该表达式就是 3 的倍数。
A strong deductive proof has three parts: setting up the variables, simplifying the expression algebraically, and writing a conclusion that explains why the simplified expression proves the claim. Missing any of these parts can cost marks.
一个强有力的演绎证明包含三个部分：设定变量、对表达式进行代数化简，以及写出结论来解释为什么化简后的表达式证明了命题。缺少任何一部分都可能导致失分。
5. Proof by Exhaustion | 穷举证明
Exhaustion is used when there are only a finite number of cases to check. For example, prove that n² − n is even for n = 1, 2, 3, 4, 5, 6. You must check every case explicitly and state that the proof is complete because no other cases exist. In AQA, exhaustion often appears with small sets or modulo arithmetic.
穷举法用于只需检查有限多个情形的问题。例如证明 n² − n 在 n = 1、2、3、4、5、6 时为偶数。你必须逐一检查每个情形，并说明由于不存在其他情形，证明已经完成。在 AQA 中，穷举法常出现在小集合或模运算问题中。
To use exhaustion correctly, list all cases in a table or a sequence of calculations. For instance, checking n = 1 to 6 for n² − n gives 0, 2, 6, 12, 20, 30, all even. Then conclude: since every possible case has been checked, the statement is proved.
要正确使用穷举法，请将所有情形列成表格或一系列计算。例如，对 n = 1 到 6 检查 n² − n，得到 0、2、6、12、20、30，全部为偶数。然后下结论：由于所有可能情形都已检查，命题得证。
6. Proof by Counterexample | 反例证明
To disprove a statement, one counterexample is enough. For example, the statement ‘all prime numbers are odd’ is false because 2 is prime and even. Write the counterexample clearly and explain why it contradicts the statement. AQA questions may ask you to find a counterexample or to decide whether a statement is true or false.
要反驳一个命题，一个反例就足够了。例如“所有素数都是奇数”是假命题，因为 2 是素数且为偶数。清楚地写出反例，并解释它为何与命题矛盾。AQA 题目可能要求你找出反例，或判断命题真假。
When giving a counterexample, do not just state the number. Show that it satisfies the condition but fails the conclusion. For example, for the statement ‘n² + n + 41 is prime for all integers n’, testing n = 41 gives 41² + 41 + 41 = 41(41 + 1 + 1) = 41 × 43, which is composite.
给出反例时，不要只写出数字。要说明它满足条件但不满足结论。例如，对于命题“对所有整数 n，n² + n + 41 都是素数”，检验 n = 41 得到 41² + 41 + 41 = 41(41 + 1 + 1) = 41 × 43，这是合数。
7. Common Mistakes in Algebraic Proof | 代数证明常见错误
Students often substitute specific numbers, such as n = 1, 2, 3, and think they have proved the result. This only verifies examples, not the general case. Another error is forgetting to define variables, writing expressions like ‘odd + odd = even’ without proof. Also avoid mixing up notation: 2n is always even, 2n + 1 is always odd when n is an integer.
学生经常代入具体数字，例如 n = 1、2、3，就认为已经证明了结论。这只是验证了例子，并不是一般情形。另一个错误是忘记定义变量，写出“奇数+奇数=偶数”却没有证明。还要避免混淆记号：当 n 为整数时，2n 总是偶数，2n + 1 总是奇数。
| Common error | 常见错误 | Correct approach | 正确方法 |
|---|---|
| Checking n = 1, 2, 3 only | 只检查 n = 1,
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