📚 Proof-Writing Framework and Model Essays for Year 13 CIE Further Mathematics | Year 13 CIE 进阶数学论文写作框架与范文
In CIE Year 13 Further Mathematics, presenting a rigorous, well-structured argument is just as vital as reaching the correct result. Whether you are tackling induction, vectors, complex numbers or contradiction proofs, examiners reward clarity, logical flow and precise notation. This article provides a reusable writing framework together with annotated model essays that demonstrate how to earn full marks.
在 CIE 进阶数学中，呈现严谨、结构清晰的论证与得出正确结果同等重要。无论你面对的是归纳法、向量、复数还是反证法，考官看重的是清晰的思路、流畅的逻辑和准确的符号。本文将提供一个可复用的写作框架，并以带批注的范文展示如何拿到满分。
1. Understanding CIE Proof Mark Schemes | 理解 CIE 证明题评分标准
Before writing, it is essential to know that marks are allocated for stating the goal, setting up a valid method, showing each logical step, and arriving at a conclusion with a clear statement.
动笔之前，必须清楚：分数会分配给明确目标、设置有效方法、展示每一个逻辑步骤，以及用明确陈述得出结论。
Examiners do not just look at your final line; they follow your reasoning. A missing justification or an omitted algebraic simplification can cost method marks even if the answer is correct.
考官不只是看最后一行，他们会追踪你的推理过程。即使最终答案正确，缺失的理据或跳过的代数化简也可能导致方法分丢失。
2. The PREP Writing Framework | PREP 写作框架
Adopt the four-stage PREP model for every proof: Plan, Restate, Execute, Prove. This gives your solution a natural structure that is easy for examiners to follow.
每个证明都可套用四阶段 PREP 模型：P 计划、R 重述、E 执行、P 证明。这会让你的解答拥有一个考官容易跟上的自然结构。
- Plan: Identify the type of proof and sketch the key stages on scrap paper.
- 重述：明确写出已知条件和要证明的结论。
- Execute: Carry out the algebraic or geometric manipulation, showing every substitution and simplification.
- 证明：以陈述句收尾，例如 “Hence, by mathematical induction, the statement is true for all n ∈ ℕ.”
3. Opening Statements: Knowns and Goal | 开篇陈述：已知与目标
Begin every solution by translating the problem into crisp mathematical language. For an induction question, write "Let P(n) be the statement that …". For a vector proof, define points and vectors explicitly.
每一道题的解答都应从将问题转译为精炼的数学语言开始。遇到归纳法题目，写下 "设 P(n) 表示命题 …"；遇到向量证明，明确写出点和向量的定义。
This step secures the communication marks and prevents you from drifting away from what needs to be shown.
这一步骤能锁定表达分，并防止你在证明过程中偏离待证目标。
4. Logical Connectives and Smooth Transitions | 逻辑连接词与流畅过渡
Use words such as "Assume", "Then", "Hence", "Since", "Therefore", and "It follows that" to guide the reader. Avoid presenting a wall of equations without commentary.
使用 “Assume”、“Then”、“Hence”、“Since”、“Therefore”、“It follows that” 等词语引导读者。切忌只堆砌等式而没有任何文字说明。
In induction, the word "Assume" introduces the inductive hypothesis; "Consider n = k+1" signals the next phase. Clear signposting makes your reasoning transparent.
在归纳法中，“Assume” 引出归纳假设，“Consider n = k+1” 标志着下一阶段。清晰的标识符会使你的推理过程一目了然。
5. Maintaining Correct Notation | 保持正确符号
se bold or underlined letters for vectors, write complex numbers as z = a + bi, and employ “∀” or “∈” only when comfortable. Ensure indices, powers and brackets are unambiguous.
严格遵循 CIE 考试中的符号惯例：向量用粗体或下划线表示，复数写作 z = a + bi，在熟练掌握的情况下可使用 “∀”、“∈” 等符号。确保角标、幂次和括号没有歧义。
| Preferred | Avoid |
| ∑r=1n r² | sum of r squared from 1 to n (ambiguous) |
| (k+1)(k+2)(2k+3)/6 | missing brackets in the numerator |
6. Model Essay 1: Induction for Summation | 范文一：求和公式的数学归纳法
Problem: Prove that for all positive integers n, ∑r=1n r² = n(n+1)(2n+1)/6.
题目：证明对所有正整数 n，∑r=1n r² = n(n+1)(2n+1)/6。
Step 1 – Define the proposition: Let P(n) be the statement ∑r=1n r² = n(n+1)(2n+1)/6.
步骤 1——定义命题：设 P(n) 表示命题 ∑r=1n r² = n(n+1)(2n+1)/6。
Step 2 – Base case: For n=1, LHS = 1² = 1, RHS = 1×2×3/6 = 1. Thus P(1) is true.
步骤 2——基础情况：当 n=1 时，左边 = 1² = 1，右边 = 1×2×3/6 = 1。因此 P(1) 成立。
Step 3 – Inductive hypothesis: Assume P(k) is true for some k ≥ 1, i.e. ∑r=1k r² = k(k+1)(2k+1)/6.
步骤 3——归纳假设：假设对某个 k ≥ 1，P(k) 成立，即 ∑r=1k r² = k(k+1)(2k+1)/6。
Step 4 – Inductive step: Consider n = k+1. Then ∑r=1k+1 r² = ∑r=1k r² + (k+1)². Substitute the hypothesis: = k(k+1)(2k+1)/6 + (k+1)².
步骤 4——归纳递推：考虑 n = k+1。此时 ∑r=1k+1 r² = ∑r=1k r² + (k+1)²。代入假设：= k(k+1)(2k+1)/6 + (k+1)²。
+1)[k(2k+1)/6 + (k+1)] = (k+1)[(2k²+k + 6k+6)/6] = (k+1)(2k²+7k+6)/6 = (k+1)(k+2)(2k+3)/6.
因式分解 (k+1)： k+1)[k(2k+1)/6 + (k+1)] = (k+1)[(2k²+k + 6k+6)/6] = (k+1)(2k²+7k+6)/6 = (k+1)(k+2)(2k+3)/6。
actly the right-hand side of P(k+1) since (k+1)((k+1)+1)(2(k+1)+1)/6 = (k+1)(k+2)(2k+3)/6.
这正好是 P(k+1) 的右边，因为 (k+1)((k+1)+1)(2(k+1)+1)/6 = (k+1)(k+2)(2k+3)/6。
Step 5 – Conclusion: If P(k) is true then P(k+1) is true. Since P(1) is true, by mathematical induction P(n) is true for all n ∈ ℕ.
步骤 5——结论：若 P(k) 成立则 P(k+1) 成立。因 P(1) 成立，由数学归纳法知 P(n) 对所有正整数 n 均成立。
7. Model Essay 2: Vector Proof of Collinearity | 范文二：向量共线证明
Problem: Show that the points A(1, 2, 3), B(3, 5, 7) and C(5, 8, 11) are collinear.
题目：证明点 A(1, 2, 3)、B(3, 5, 7) 和 C(5, 8, 11) 共线。
Step 1 – Define vectors: Choose a common origin. Let position vectors be a = i+2j+3k, b = 3i+5j+7k, c = 5i+8j+11k.
步骤 1——定义向量：选择公共原点，设位置向量 a = i+2j+3k，b = 3i+5j+7k，c = 5i+8j+11k。
Step 2 – Find direction vectors: i + (5-2)j + (7-3)k = 2i + 3j + 4k. AC = c - a = (5-1)i + (8-2)j + (11-3)k = 4i + 6j + 8k.
步骤 2——求方向向量：AB = b - a = 2i + 3j + 4k。AC = c - a = 4i + 6j + 8k。
Step 3 – Show one is a scalar multiple: Clearly AC = 2×(2i + 3j + 4k) = 2 AB. Hence AC is parallel to AB.
步骤 3——证明成比例：显然 AC = 2×(2i + 3j + 4k) = 2 AB。因此 AC 与 AB 平行。
Step 4 – Conclude collinearity: Since AC is parallel to AB and they share point A, the three points A, B and C lie on the same straight line.
步骤 4——得出共线结论：因为 AC 与 AB 平行且二者共享点 A，故三点 A、B、C 位于同一直线上。
8. Model Essay 3: Modulus of a Complex Product | 范文三：复数相乘的模
Problem: Prove that for any complex numbers z₁ and z₂, |z₁z₂| = |z₁||z₂|.
题目：证明对任意复数 z₁ 和 z₂，有 |z₁z₂| = |z₁||z₂|。
Step 1 – Express in algebraic form: Let z₁ = a + bi and z₂ = c + di, where a, b, c, d ∈ ℝ.
步骤 1——代数形式表示：设 z₁ = a + bi，z₂ = c + di，其中 a, b, c, d 为实数。
Step 2 – Compute the product: z₁z₂ = (a+bi)(c+di) = (ac - bd) + (ad + bc)i.
步骤 2——计算乘积：z₁z₂ = (a+bi)(c+di) = (ac - bd) + (ad + bc)i。
Step 3 – Square the modulus: |z₁z₂|² = (ac - bd)² + (ad + bc)².
步骤 3——模的平方：|z₁z₂|² = (ac - bd)² + (ad + bc)²。
Expand: = a²c² - 2abcd + b²d² + a²d² + 2abcd + b²c² = a²c² + b²d² + a²d² + b²c².
展开：= a²c² - 2abcd + b²d² + a²d² + 2abcd + b²c² = a²c² + b²d² + a²d² + b²c²。
Factor by grouping: = a²(c²+d²) + b²(c²+d²) = (a²+b²)(c²+d²).
分组因式分解：= a²(c²+d²) + b²(c²+d²) = (a²+b²)(c²+d²)。
Step 4 – Relate to individual moduli: |z₁|² = a²+b² and |z₂|² = c²+d², so |z₁z₂|² = |z₁|²|z₂|². Taking the positive square root gives |z₁z₂| = |z₁||z₂|.
步骤 4——联系各自模：|z₁|² = a²+b²，|z₂|² = c²+d²，因此 |z₁z₂|² = |z₁|²|z₂|²。取正平方根即得 |z₁z₂| = |z₁||z₂|。
9. Model Essay 4: Proof by Contradiction – Irrationality of √2 | 范文四：反证法证明 √2 为无理数
Problem: Prove that √2 is irrational.
题目：证明 √2 是无理数。
Step 1 – Assume the opposite: Suppose, for a contradiction, that √2 is rational. Then it can be written as √2 = p/q, where p, q ∈ ℤ, q ≠ 0, and p/q is in lowest terms (coprime).
步骤 1——反设：假设 √2 是有理数，以追求矛盾。那么可写作 √2 = p/q，其中 p, q ∈ ℤ，q ≠ 0，且 p/q 已化为最简分数（互质）。
Step 2 – Manipulate: Squaring both sides gives 2 = p²/q² ⇒ p² = 2q². Hence p² is even, which forces p to be even. Write p = 2k for some integer k.
步骤 2——变形：两边平方得 2 = p²/q² ⇒ p² = 2q²。因此 p² 为偶数，于是 p 必为偶数。令 p = 2k。
Step 3 – Substitute back: (2k)² = 2q² ⇒ 4k² = 2q² ⇒ q² = 2k². Thus q² is even, so q is also even.
步骤 3——回代：(2k)² = 2q² ⇒ 4k² = 2q² ⇒ q² = 2k²。因此 q² 为偶数，故 q 亦为偶数。
Step 4 – Reach contradiction: Both p and q are even, contradicting the assumption that p/q was in lowest terms. Therefore our original supposition is false, and √2 cannot be rational; it is irrational.
步骤 4——推出矛盾：p 与 q 皆偶，这与 p/q 为最简分数相矛盾。故原假设不成立，√2 不能为有理数，它必为无理数。
10. Avoiding Common Pitfalls | 常见错误规避
Pitfall 1: ng the base case in induction or assuming P(k) without stating it explicitly. Always write 'Assume true for n = k'. 错误 1： 归纳法跳过基础步骤，或未明确写出归纳假设。务必写出 “假设 n = k 时成立”。
Pitfall 2: nce, writing AB = 2 without vector symbol leads to confusion between length and direction. 错误 2： 向量符号使用不当。例如写 AB = 2 而不带向量标记，容易混淆长度与方向。
Pitfall 3: In complex proofs, forgetting to specify that a, b, c, d are real numbers. Without this, the algebraic expansion may not be justified. 错误 3： 复数证明中忘记指明 a, b, c, d 为实数。缺少此点，代数展开就可能失去依据。
Pitfall 4: Ending a contradiction proof without stating the contradiction clearly. Always finish with a sentence like 'This contradicts the initial assumption; hence the statement holds.' 错误 4： 反证法没有清晰点明矛盾所在。结尾务必用一句话如 “这与原假设矛盾，故命题成立”。
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