Mathematical Induction: Application to Divisibility | 数学归纳法：整除性应用

📚 Mathematical Induction: Application to Divisibility | 数学归纳法：整除性应用

In this section we explore how mathematical induction is used to prove divisibility statements, such as “n³ – n is divisible by 3 for all positive integers n”. These proofs are a classic AQA A-Level Mathematics topic.

本节探讨如何使用数学归纳法证明整除性命题，例如“对所有正整数 n，n³ – n 可被 3 整除”。这类证明是 AQA A-Level 数学的经典考点。

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1. What Does “Divisible” Mean? | “整除”的含义

An integer m is divisible by an integer k ≠ 0 if there exists an integer q such that m = kq. Equivalently, k divides m, written as k | m.

若存在整数 q 使得 m = kq，则称整数 m 能被整数 k（k ≠ 0）整除。等价地，k 整除 m，记作 k | m。

For example, 12 is divisible by 3 because 12 = 3 × 4.

例如，12 能被 3 整除，因为 12 = 3 × 4。

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2. The Principle of Mathematical Induction | 数学归纳法原理

Mathematical induction is a method for proving that a statement P(n) holds for every positive integer n. It requires two steps: the base case and the inductive step.

数学归纳法是一种证明命题 P(n) 对所有正整数 n 都成立的方法。它需要两步：基础步骤和归纳步骤。

If P(1) is true, and if P(k) being true implies P(k+1) is true, then P(n) is true for all n ∈ ℕ.

若 P(1) 为真，并且 P(k) 为真能推出 P(k+1) 为真，那么 P(n) 对所有 n ∈ ℕ 都为真。

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3. Structure of a Divisibility Proof | 整除性证明的结构

To prove that “for all n ≥ 1, expression E(n) is divisible by d”, we often write E(n) = d × A(n), where A(n) is an integer.

要证明“对所有 n ≥ 1，表达式 E(n) 能被 d 整除”，我们通常写成 E(n) = d × A(n)，其中 A(n) 是整数。

During the inductive step, a common tactic is to express E(k+1) in terms of E(k), then manipulate the algebra to factor out d.

在归纳步骤中，常用策略是将 E(k+1) 用 E(k) 表示，然后通过代数变形分解出因子 d。

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4. Step-by-Step Method | 分步方法

• Base case: Verify that the statement is true for the first allowed value, usually n = 1.

  基础步骤：验证命题对于第一个允许的值（通常是 n = 1）成立。

• Inductive hypothesis: Assume the statement is true for n = k.

  归纳假设：假设命题对 n = k 成立。

• Inductive step: Show that the statement is then true for n = k+1.

  归纳步骤：证明命题对 n = k+1 也成立。

• Conclusion: Since the base case and inductive step hold, the statement is true for all positive integers n.

  结论：由于基础步骤和归纳步骤成立，命题对所有正整数 n 都成立。

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5. Example 1: n³ – n is Divisible by 3 | 例1：n³ – n 能被 3 整除

Prove that n³ – n is divisible by 3 for all positive integers n.

证明：对所有正整数 n，n³ – n 能被 3 整除。

Base case: For n = 1, 1³ – 1 = 0, and 0 = 3 × 0, so the statement is true.

基础步骤：当 n = 1 时，1³ – 1 = 0，而 0 = 3 × 0，所以命题成立。

Inductive hypothesis: Suppose that k³ – k is divisible by 3. Then k³ – k = 3m for some integer m.

归纳假设：假设 k³ – k 能被 3 整除，即存在整数 m，使得 k³ – k = 3m。

Inductive step: Consider (k+1)³ – (k+1). Expand:

归纳步骤：考虑 (k+1)³ – (k+1)，展开得：

(k+1)³ – (k+1) = k³ + 3k² + 3k + 1 – k – 1 = k³ + 3k² + 2k

Now rewrite k³ + 3k² + 2k = (k³ – k) + 3k² + 3k = 3m + 3(k² + k) = 3(m + k² + k).

现在改写 k³ + 3k² + 2k = (k³ – k) + 3k² + 3k = 3m + 3(k² + k) = 3(m + k² + k)。

Since m + k² + k is an integer, (k+1)³ – (k+1) is divisible by 3. By induction, n³ – n is divisible by 3 for all n ≥ 1.

由于 m + k² + k 是整数，所以 (k+1)³ – (k+1) 能被 3 整除。由归纳法，对所有 n ≥ 1，n³ – n 都能被 3 整除。

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6. Example 2: 7ⁿ – 1 is Divisible by 6 | 例2：7ⁿ – 1 能被 6 整除

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