IB Mathematics: Analysis & Approaches HL · 鼎睿学苑

Unit A5: Proof
& Algebraic Manipulation单元 A5：证明与代数运算

The closing sub-unit of Topic 1 (Number & Algebra). Master four proof techniques (direct, counter-example, contradiction, induction), the partial-fraction decomposition that prepares the integration toolkit, and the Gaussian elimination + RREF machinery that solves $3\times 3$ linear systems. The induction, partial fractions, and linear-systems pieces are HL-only — they appear repeatedly on Paper 1B and Paper 3.Topic 1（数与代数）的收官子单元。掌握四种证明（proof）方法：直接证明、反例、反证法、数学归纳法；掌握部分分式（partial fractions）分解（积分工具箱的入门）；掌握 Gauss 消元（Gaussian elimination）与 RREF 解 $3\times 3$ 线性方程组的全套机制。归纳法、部分分式、线性方程组都是 HL 专属 —— 反复出现在 Paper 1B 与 Paper 3。

IB AA HL · Topic 1.6 / 1.11 / 1.15 Papers 1B · 3 6 Concepts · HL-heavy6 个核心概念 · HL 重点
Read me first先读这里

How to use this guide本指南使用说明

A5 is the most proof-heavy unit in Topic A — and proof is one of the IB's high-leverage skills. Approach it differently than computational units like A2 and A3:A5 是 Topic A 中证明分量最重的单元 —— 而证明（proof）正是 IB 的高价值技能之一。不要把它当成 A2、A3 那类计算型单元来读：

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If you're cramming如果你在临阵磨枪

Memorise the 3-line induction template (base, IH, IS) and the partial-fraction set-up. Skim one worked example per section. Skip the "going deeper" details. For linear systems, practise one $3\times 3$ end-to-end on the GDC.

背熟归纳法三段式模板（基础步骤、归纳假设、归纳步骤）和部分分式列式。每节挑一道例题扫一眼，跳过"深入"部分。对线性方程组，在 GDC 上端到端做一道$3\times 3$ 题目。

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If you're going for a 7如果你目标是 7 分

Write out every induction proof you encounter by hand, with the three-step structure explicit. Practise partial-fraction decomposition until you can do irreducible-quadratic and repeated-linear cases in one pass. Solve $3\times 3$ systems both by elimination and by RREF on the GDC, and confirm they agree.

每道归纳题都动手手写，把"基础、假设、推导"三段写清。把部分分式练到能一遍处理不可约二次因式与重复线性因式。$3\times 3$ 方程组用消元和 GDC RREF 两种方法都做，验证一致。

HL flagHL 标记说明 Mathematical induction, partial fractions, and $3\times 3$ linear systems are all HL-only on the AA syllabus. Direct proof and proof by contradiction appear on both SL and HL papers but at different depths — HL routinely expects students to construct a proof from scratch.数学归纳法、部分分式、$3\times 3$ 线性方程组在 AA 大纲中均为 HL 专属。直接证明与反证法 SL、HL 两边都考，但 HL 要求更深 —— 通常要求学生从零构造一个完整证明。
Topic A5.1 · Direct ProofTopic A5.1 · 直接证明

Direct Proof & Counter-Example直接证明与反例 SL 1.6

Two reciprocal techniques.
  • Direct proof of "if $P$ then $Q$": assume $P$, apply algebra/definitions, conclude $Q$. Layout: Assume… Then… Therefore $Q$. $\square$
  • Counter-example to "for all $x$, $P(x)$": find one specific $x_{0}$ where $P(x_{0})$ fails. A single value is enough.
Use direct proof to verify universal statements; use counter-example to refute them.
两种互逆技巧。
  • 直接证明"若 $P$ 则 $Q$"：假设 $P$ 成立，应用代数与定义，推出 $Q$。书写模板：设…故…所以 $Q$。 $\square$
  • 反例（counter-example）反驳"对所有 $x$，$P(x)$ 成立"：找一个具体的 $x_{0}$ 使 $P(x_{0})$ 不成立即可。一个反例就够。
用直接证明验证全称命题；用反例反驳。
Worked Example A5.1a — Direct proofA5.1a 例题 —— 直接证明

Prove: the product of two even integers is even.证明：两个偶数的乘积是偶数。

Set up. Let $a, b \in \mathbb{Z}$ be even. By definition, $a = 2m$ and $b = 2n$ for some integers $m, n$.

列式。设 $a, b \in \mathbb{Z}$ 均为偶数。由偶数定义，存在整数 $m, n$ 使 $a = 2m$、$b = 2n$。

Compute the product.

计算乘积。

$$ ab \;=\; (2m)(2n) \;=\; 4mn \;=\; 2(2mn). $$

Conclude. $2mn \in \mathbb{Z}$, so $ab = 2k$ for some integer $k = 2mn$, hence $ab$ is even. $\square$

得证。$2mn \in \mathbb{Z}$，所以 $ab = 2k$（取 $k = 2mn$），即 $ab$ 为偶数。 $\square$

Worked Example A5.1b — Counter-exampleA5.1b 例题 —— 反例

Disprove: for all $n \in \mathbb{Z}^{+}$, $n^{2} - n + 41$ is prime.反驳："对所有正整数 $n$，$n^{2} - n + 41$ 均为质数（prime）"。

Look for a structural cancellation. Try $n = 41$:

找结构性化简。试 $n = 41$：

$$ 41^{2} - 41 + 41 \;=\; 41^{2} \;=\; 1681 \;=\; 41 \cdot 41, $$

which is not prime (it's $41 \cdot 41$). So $n = 41$ is a counter-example, and the claim fails. $\square$

不是质数（等于 $41 \cdot 41$）。故 $n = 41$ 是一个反例，命题不成立。 $\square$

Why structure helps. The expression $n^{2} - n + 41$ factors as $n(n-1) + 41$. Plugging in $n = 41$ gives $41 \cdot 40 + 41 = 41(40 + 1) = 41^{2}$, immediately exposing the composite factor.

为什么"看结构"很有用。$n^{2} - n + 41 = n(n-1) + 41$。代入 $n = 41$ 得 $41 \cdot 40 + 41 = 41(40 + 1) = 41^{2}$，复合因子立即暴露。

Which of the following is a valid counter-example to "for all positive integers $n$, $n^{2} + n$ is even"?下列哪个是对"对所有正整数 $n$，$n^{2} + n$ 为偶数"的有效反例？
A5.1 · Q1
$n = 2$ (gives $6$)
$n = 3$ (gives $12$)
None — the statement is true没有 —— 命题本身为真
$n = 0$
$n^{2} + n = n(n+1)$ is the product of two consecutive integers, one of which is even, so the product is always even. No counter-example exists.$n^{2} + n = n(n+1)$ 是两个相邻整数之积，必有一个为偶数，所以乘积总是偶数。不存在反例。
$n(n+1)$ is always even (one of two consecutive integers is even). The statement is true; no counter-example exists.$n(n+1)$ 总为偶数（两个相邻整数中必有一个为偶数）。命题为真，不存在反例。
Topic A5.2 · ContradictionTopic A5.2 · 反证法

Proof by Contradiction反证法 SL 1.6

To prove $P$, assume $\neg P$ and derive a contradiction. The contradiction can be:
  • An algebraic absurdity ($0 = 1$, $\sqrt{2} \in \mathbb{Q}$ contradicting lowest-terms).
  • A contradiction with a known fact (e.g. an integer is both even and odd).
The contradiction shows $\neg P$ is impossible, so $P$ must hold. Layout: Suppose, for contradiction, that $\neg P$. Then… But this contradicts… Therefore $P$. $\square$
反证法（proof by contradiction）：欲证 $P$，先假设 $\neg P$ 并推出矛盾。矛盾可以是：
  • 代数上的荒谬（$0 = 1$、$\sqrt{2} \in \mathbb{Q}$ 与"最简分数"自相矛盾）；
  • 与已知事实冲突（如某整数既是偶数又是奇数）。
矛盾说明 $\neg P$ 不可能，故 $P$ 成立。书写模板：设若 $\neg P$ 成立。则…但这与…矛盾。故 $P$ 成立。 $\square$
Worked Example A5.2 — Irrationality of $\sqrt{2}$A5.2 例题 —— $\sqrt{2}$ 是无理数

Prove that $\sqrt{2}$ is irrational.证明 $\sqrt{2}$ 是无理数（irrational）。

Suppose, for contradiction, that $\sqrt{2}$ is rational. Then $\sqrt{2} = \dfrac{a}{b}$ for some integers $a, b$ with $b \ne 0$ and $\gcd(a, b) = 1$ (i.e. the fraction is in lowest terms).

反设 $\sqrt{2}$ 是有理数（rational）。则存在互素整数 $a, b$（$b \ne 0$、$\gcd(a, b) = 1$）使 $\sqrt{2} = \dfrac{a}{b}$。

Square and rearrange.

两边平方并整理。

$$ 2 \;=\; \frac{a^{2}}{b^{2}} \;\Longrightarrow\; a^{2} \;=\; 2b^{2}. $$

So $a^{2}$ is even. Since the square of an odd integer is odd, $a$ itself must be even. Write $a = 2k$:

$a^{2}$ 为偶数。又因奇数的平方仍为奇数，故 $a$ 本身必为偶数。设 $a = 2k$：

$$ (2k)^{2} \;=\; 2b^{2} \;\Longrightarrow\; 4k^{2} \;=\; 2b^{2} \;\Longrightarrow\; b^{2} \;=\; 2k^{2}. $$

By the same argument, $b$ must be even.

同理，$b$ 也必为偶数。

Contradiction. Both $a$ and $b$ are even, so $\gcd(a, b) \ge 2$ — contradicting our choice $\gcd(a, b) = 1$. Therefore $\sqrt{2}$ cannot be rational. $\square$

矛盾。$a$、$b$ 同为偶数，故 $\gcd(a, b) \ge 2$，与"$\gcd(a, b) = 1$"矛盾。故 $\sqrt{2}$ 不是有理数。 $\square$

▸ Going deeper — what to assume and why▸ 深入 —— 假设什么、为什么这样设

The "lowest terms" stipulation $\gcd(a, b) = 1$ is what produces the contradiction. Without it, the proof stalls — there's no clash with "both even." Always include the lowest-terms clause when assuming a rational representation; it's the lever that breaks the assumption.

"最简分数"$\gcd(a, b) = 1$ 这一前提才是矛盾的来源。没有这一条，证明会卡住 —— "同为偶数"不会与任何条件冲突。一旦假设某数为有理数，务必加上"最简分数"前提；这是破假设的杠杆。

In a proof by contradiction of "$\sqrt{3}$ is irrational," the contradiction reached is:"$\sqrt{3}$ 是无理数"的反证法中，所得矛盾是：
A5.2 · Q1
$a^{2}$ is positive but $a$ is negative$a^{2}$ 为正而 $a$ 为负
$3 \mid a$ and $3 \mid b$, but we assumed $\gcd(a,b) = 1$$3 \mid a$ 且 $3 \mid b$，但已设 $\gcd(a,b) = 1$
$\sqrt{3}$ is greater than $1$$\sqrt{3} > 1$
$a^{2}/b^{2}$ is not equal to $3$$a^{2}/b^{2}$ 不等于 $3$
From $a^{2} = 3b^{2}$, the prime $3$ divides $a^{2}$, hence divides $a$. Substituting $a = 3k$ gives $b^{2} = 3k^{2}$, so $3 \mid b$. Both $3 \mid a$ and $3 \mid b$ contradicts $\gcd(a,b) = 1$.由 $a^{2} = 3b^{2}$，质数 $3$ 整除 $a^{2}$，故 $3 \mid a$。代入 $a = 3k$ 得 $b^{2} = 3k^{2}$，故 $3 \mid b$。两者同除 $3$，与 $\gcd(a,b) = 1$ 矛盾。
The proof mirrors the $\sqrt{2}$ case but with the prime $3$ playing the role of $2$. Both $a, b$ end up divisible by $3$, contradicting lowest terms.证明仿照 $\sqrt{2}$ 的情形，把 $2$ 换成质数 $3$。最终 $a$、$b$ 同被 $3$ 整除，与"最简分数"矛盾。
Topic A5.3 · Mathematical InductionTopic A5.3 · 数学归纳法

Mathematical Induction数学归纳法 HL AHL 1.6

Three-line template for proving "$P(n)$ holds for all $n \ge n_{0}$":
  1. Base case. Verify $P(n_{0})$ directly (usually $n_{0} = 1$).
  2. Inductive hypothesis (IH). Assume $P(k)$ is true for some $k \ge n_{0}$.
  3. Inductive step. Show $P(k) \Rightarrow P(k+1)$, using the IH explicitly.
Conclude: "By the principle of mathematical induction, $P(n)$ is true for all $n \ge n_{0}$." $\square$
证明"$P(n)$ 对所有 $n \ge n_{0}$ 成立"的三段式模板：
  1. 基础步骤（base case）。直接验证 $P(n_{0})$（通常 $n_{0} = 1$）。
  2. 归纳假设（inductive hypothesis，IH）。设某个 $k \ge n_{0}$ 时 $P(k)$ 成立。
  3. 归纳步骤（inductive step）。显式使用 IH，证明 $P(k) \Rightarrow P(k+1)$。
结论："由数学归纳法（principle of mathematical induction），$P(n)$ 对所有 $n \ge n_{0}$ 成立。" $\square$
Worked Example A5.3a — Sum of first $n$ integersA5.3a 例题 —— 前 $n$ 个正整数之和

Prove by induction: $1 + 2 + 3 + \cdots + n = \dfrac{n(n+1)}{2}$ for all $n \in \mathbb{Z}^{+}$.用归纳法证明：$1 + 2 + 3 + \cdots + n = \dfrac{n(n+1)}{2}$ 对所有正整数 $n$ 成立。

Base case ($n = 1$). LHS $= 1$. RHS $= \tfrac{1 \cdot 2}{2} = 1$. So $P(1)$ holds. ✓

基础步骤（$n = 1$）。LHS $= 1$，RHS $= \tfrac{1 \cdot 2}{2} = 1$。故 $P(1)$ 成立。 ✓

Inductive hypothesis. Assume that for some $k \ge 1$,

归纳假设。设某个 $k \ge 1$ 时

$$ 1 + 2 + 3 + \cdots + k \;=\; \tfrac{k(k+1)}{2}. \tag{IH} $$

Inductive step. Show that the formula then holds for $n = k+1$:

归纳步骤。证明公式在 $n = k+1$ 时也成立：

$$ \underbrace{1 + 2 + \cdots + k}_{= k(k+1)/2 \text{ by IH}} + (k+1) \;=\; \tfrac{k(k+1)}{2} + (k+1) \;=\; \tfrac{k(k+1) + 2(k+1)}{2} \;=\; \tfrac{(k+1)(k+2)}{2}. $$

This is exactly the formula with $n = k+1$. So $P(k) \Rightarrow P(k+1)$.

这恰是 $n = k+1$ 时的公式。故 $P(k) \Rightarrow P(k+1)$。

Conclusion. By the principle of mathematical induction, $P(n)$ holds for all $n \ge 1$. $\square$

结论。由数学归纳法，$P(n)$ 对一切 $n \ge 1$ 成立。 $\square$

Worked Example A5.3b — DivisibilityA5.3b 例题 —— 整除性

Prove that $7^{n} - 1$ is divisible by $6$ for every $n \in \mathbb{Z}^{+}$.证明对每个正整数 $n$，$7^{n} - 1$ 能被 $6$ 整除。

Base case ($n = 1$). $7^{1} - 1 = 6 = 6 \cdot 1$. ✓

基础步骤（$n = 1$）。$7^{1} - 1 = 6 = 6 \cdot 1$。 ✓

Inductive hypothesis. Assume $7^{k} - 1 = 6m$ for some $m \in \mathbb{Z}$.

归纳假设。设某 $m \in \mathbb{Z}$ 使 $7^{k} - 1 = 6m$。

Inductive step. Manipulate $7^{k+1} - 1$ to expose a factor of $6$:

归纳步骤。变形 $7^{k+1} - 1$，凑出 $6$ 的因子：

$$ 7^{k+1} - 1 \;=\; 7 \cdot 7^{k} - 1 \;=\; 7\,(7^{k} - 1) + 7 - 1 \;=\; 7 \cdot 6m + 6 \;=\; 6\,(7m + 1). $$

Since $7m + 1 \in \mathbb{Z}$, $7^{k+1} - 1$ is divisible by $6$.

$7m + 1 \in \mathbb{Z}$，故 $7^{k+1} - 1$ 被 $6$ 整除。

Conclusion. By induction, $6 \mid (7^{n} - 1)$ for all $n \ge 1$. $\square$

结论。由归纳法，对一切 $n \ge 1$，$6 \mid (7^{n} - 1)$。 $\square$

Pitfall — the IH must be used陷阱 —— IH 必须显式使用 The single most common deduction in IB induction proofs: the student writes "by IH, $\sum_{k=1}^{n} = \ldots$" without actually substituting the hypothesis into the inductive step. The mark scheme requires you to explicitly point out where the IH is being used (e.g. by underbracing $1 + 2 + \cdots + k$ and writing "by IH $= k(k+1)/2$").IB 归纳证明里最常见的扣分项：学生写了"由 IH，$\sum_{k=1}^{n} = \ldots$"却没真正把假设代入归纳步骤。评分细则要求明确指出 IH 用在了哪一步（如在 $1 + 2 + \cdots + k$ 上加下花括号并标注"由 IH = $k(k+1)/2$"）。
▸ Going deeper — why induction works▸ 深入 —— 为什么归纳法成立

Mathematical induction is a theorem of $\mathbb{Z}^{+}$: if $S \subseteq \mathbb{Z}^{+}$ contains $1$ and is closed under "successor" (i.e. $k \in S \Rightarrow k+1 \in S$), then $S = \mathbb{Z}^{+}$. The base case puts $n_{0}$ into the set $S = \{n : P(n) \text{ holds}\}$, and the inductive step makes $S$ closed under successor — so $S$ contains every positive integer from $n_{0}$ onward. This is the "domino effect": tip the first domino (base case), and arrange each domino to knock the next over (inductive step), then they all fall.

数学归纳法本质上是 $\mathbb{Z}^{+}$ 的一条定理：若 $S \subseteq \mathbb{Z}^{+}$ 含 $1$，且对"后继"封闭（$k \in S \Rightarrow k+1 \in S$），则 $S = \mathbb{Z}^{+}$。基础步骤把 $n_{0}$ 放入 $S = \{n : P(n) \text{ 成立}\}$，归纳步骤让 $S$ 对后继封闭 —— 所以 $S$ 包含从 $n_{0}$ 起的所有正整数。这就是"多米诺效应"：先推倒第一张骨牌（基础步骤），再确保每张能推倒下一张（归纳步骤），然后整列必倒。

In the induction proof of $1 + 3 + 5 + \cdots + (2n-1) = n^{2}$, the inductive step shows:在证明 $1 + 3 + 5 + \cdots + (2n-1) = n^{2}$ 时，归纳步骤要证明：
A5.3 · Q1
$k^{2} + (2k+1) = k^{2} + 2k$
$k^{2} + (2k+1) = (k+1)^{2}$
$k^{2} + (2k-1) = (k+1)^{2}$
$n^{2} = (n+1)^{2}$ for some $n$
By IH, sum to $k$ terms $= k^{2}$. Adding the next odd $(2(k+1) - 1) = 2k + 1$: $k^{2} + (2k+1) = k^{2} + 2k + 1 = (k+1)^{2}$.由 IH，前 $k$ 项之和 $= k^{2}$。再加下一项 $(2(k+1) - 1) = 2k + 1$：$k^{2} + (2k+1) = k^{2} + 2k + 1 = (k+1)^{2}$。
The $(k+1)$-th odd number is $2(k+1) - 1 = 2k + 1$. Add it to the IH sum $k^{2}$: $k^{2} + 2k + 1 = (k+1)^{2}$. ✓第 $(k+1)$ 个奇数是 $2(k+1) - 1 = 2k + 1$。加入 IH 之和：$k^{2} + 2k + 1 = (k+1)^{2}$。 ✓
Topic A5.4 · Partial FractionsTopic A5.4 · 部分分式

Partial Fractions部分分式分解 HL AHL 1.11

Decompose $P(x)/Q(x)$ (deg $P$ < deg $Q$) into a sum of simpler fractions. One term per factor of $Q$:
  • Linear factor $(ax + b)$ → $\dfrac{A}{ax+b}$
  • Repeated linear $(ax+b)^{2}$ → $\dfrac{A_{1}}{ax+b} + \dfrac{A_{2}}{(ax+b)^{2}}$
  • Irreducible quadratic $(ax^{2}+bx+c)$ → $\dfrac{Ax+B}{ax^{2}+bx+c}$
Multiply through by $Q(x)$, then either substitute convenient $x$-values (e.g. roots of factors) or equate coefficients of like powers of $x$.
把 $P(x)/Q(x)$（deg $P$ < deg $Q$）拆成更简单分式之和。$Q$ 的每个因式贡献一项：
  • 线性因式 $(ax + b)$ → $\dfrac{A}{ax+b}$
  • 重复线性 $(ax+b)^{2}$ → $\dfrac{A_{1}}{ax+b} + \dfrac{A_{2}}{(ax+b)^{2}}$
  • 不可约二次（irreducible quadratic）$(ax^{2}+bx+c)$ → $\dfrac{Ax+B}{ax^{2}+bx+c}$
两边同乘 $Q(x)$，然后或代入合适的 $x$ 值（如各因式的根），或比较 $x$ 同次幂系数。
Worked Example A5.4a — Two distinct linear factorsA5.4a 例题 —— 两个不同线性因式

Decompose $\dfrac{5x - 4}{x^{2} - x - 2}$ into partial fractions.将 $\dfrac{5x - 4}{x^{2} - x - 2}$ 分解为部分分式。

Factor the denominator.

分母因式分解。

$$ x^{2} - x - 2 \;=\; (x - 2)(x + 1). $$

Set up the decomposition.

列分解式。

$$ \frac{5x - 4}{(x-2)(x+1)} \;=\; \frac{A}{x - 2} + \frac{B}{x + 1}. $$

Multiply through by $(x-2)(x+1)$.

两边乘以 $(x-2)(x+1)$。

$$ 5x - 4 \;=\; A(x + 1) + B(x - 2). $$

Substitute convenient $x$-values. $x = 2$: $5(2) - 4 = 6 = 3A$, so $A = 2$. $x = -1$: $5(-1) - 4 = -9 = -3B$, so $B = 3$.

代入合适的 $x$ 值。$x = 2$：$5(2) - 4 = 6 = 3A$，故 $A = 2$。$x = -1$：$5(-1) - 4 = -9 = -3B$，故 $B = 3$。

Result.

结果。

$$ \frac{5x - 4}{(x-2)(x+1)} \;=\; \frac{2}{x - 2} + \frac{3}{x + 1}. $$
Worked Example A5.4b — Repeated linear factorA5.4b 例题 —— 重复线性因式

Decompose $\dfrac{2x + 5}{(x + 1)^{2}}$.分解 $\dfrac{2x + 5}{(x + 1)^{2}}$。

Set up — one term per power.

列式 —— 每个幂次一项。

$$ \frac{2x + 5}{(x + 1)^{2}} \;=\; \frac{A_{1}}{x + 1} + \frac{A_{2}}{(x + 1)^{2}}. $$

Multiply through.

两边相乘。

$$ 2x + 5 \;=\; A_{1}(x + 1) + A_{2}. $$

$x = -1$: $2(-1) + 5 = 3 = A_{2}$. Expand: $A_{1}x + (A_{1} + A_{2}) = 2x + 5$. Coefficient of $x$: $A_{1} = 2$.

$x = -1$：$2(-1) + 5 = 3 = A_{2}$。展开：$A_{1}x + (A_{1} + A_{2}) = 2x + 5$。$x$ 系数：$A_{1} = 2$。

Result.

结果。

$$ \frac{2x + 5}{(x + 1)^{2}} \;=\; \frac{2}{x + 1} + \frac{3}{(x + 1)^{2}}. $$
▸ Going deeper — why partial fractions are worth the effort▸ 深入 —— 部分分式为何值得学

Partial fractions matter because they decouple a hard rational integrand into pieces that have known antiderivatives. The decomposition $\dfrac{1}{x(x+1)} = \dfrac{1}{x} - \dfrac{1}{x+1}$ instantly delivers $\int \dfrac{dx}{x(x+1)} = \ln|x| - \ln|x+1| + C$. This is the bridge from Topic A5 into integral calculus (Topic E3) — and the reason partial fractions are an HL prerequisite.

部分分式的价值在于把一个复杂的有理被积函数解耦为若干已知原函数的部件。如 $\dfrac{1}{x(x+1)} = \dfrac{1}{x} - \dfrac{1}{x+1}$ 直接给出 $\int \dfrac{dx}{x(x+1)} = \ln|x| - \ln|x+1| + C$。这是 A5 通向积分（Topic E3）的桥梁 —— 也正是 HL 把部分分式列入前置技能的原因。

Decompose $\dfrac{1}{x(x+2)}$ into partial fractions.将 $\dfrac{1}{x(x+2)}$ 分解为部分分式。
A5.4 · Q1
$\dfrac{1/2}{x} - \dfrac{1/2}{x+2}$
$\dfrac{1}{x} - \dfrac{1}{x+2}$
$\dfrac{1}{x} + \dfrac{1}{x+2}$
$\dfrac{2}{x} - \dfrac{2}{x+2}$
Set $\tfrac{1}{x(x+2)} = \tfrac{A}{x} + \tfrac{B}{x+2}$. Multiply: $1 = A(x+2) + Bx$. $x = 0$: $A = 1/2$. $x = -2$: $B = -1/2$.设 $\tfrac{1}{x(x+2)} = \tfrac{A}{x} + \tfrac{B}{x+2}$。相乘：$1 = A(x+2) + Bx$。$x = 0$：$A = 1/2$。$x = -2$：$B = -1/2$。
Multiply both sides by $x(x+2)$: $1 = A(x+2) + Bx$. Plugging $x = 0$ gives $A = 1/2$; $x = -2$ gives $B = -1/2$.两边乘以 $x(x+2)$：$1 = A(x+2) + Bx$。$x = 0$ 给出 $A = 1/2$；$x = -2$ 给出 $B = -1/2$。
Topic A5.5 · Linear SystemsTopic A5.5 · 线性方程组

$3 \times 3$ Linear Systems三元线性方程组 HL AHL 1.15

Gaussian elimination on the augmented matrix. For $$ \begin{cases} a_{1}x + b_{1}y + c_{1}z = d_{1} \\ a_{2}x + b_{2}y + c_{2}z = d_{2} \\ a_{3}x + b_{3}y + c_{3}z = d_{3} \end{cases} $$ form $[A \mid \mathbf{b}]$ and apply row operations: $R_{i} \to R_{i} + \lambda R_{j}$, $R_{i} \leftrightarrow R_{j}$, $R_{i} \to \lambda R_{i}$ ($\lambda \ne 0$) until upper-triangular (row echelon). Then back-substitute. Three outcomes:
  • Unique solution — three pivots.
  • Infinite solutions — fewer pivots than unknowns, parametrise the free variable(s).
  • No solution — a row of the form $[\,0\;\;0\;\;0 \mid c\,]$ with $c \ne 0$ (inconsistent).
在增广矩阵（augmented matrix）上做 Gauss 消元。对 $$ \begin{cases} a_{1}x + b_{1}y + c_{1}z = d_{1} \\ a_{2}x + b_{2}y + c_{2}z = d_{2} \\ a_{3}x + b_{3}y + c_{3}z = d_{3} \end{cases} $$ 写出 $[A \mid \mathbf{b}]$，使用初等行变换：$R_{i} \to R_{i} + \lambda R_{j}$、$R_{i} \leftrightarrow R_{j}$、$R_{i} \to \lambda R_{i}$（$\lambda \ne 0$），直至上三角（行阶梯形，row echelon）。然后回代。三种结果：
  • 唯一解 —— 三个主元。
  • 无穷多解 —— 主元少于未知数，把自由变量参数化。
  • 无解 —— 出现 $[\,0\;\;0\;\;0 \mid c\,]$（$c \ne 0$），方程组不相容（inconsistent）。
Worked Example A5.5 — Gaussian eliminationA5.5 例题 —— Gauss 消元

Solve the system $\begin{cases} x + 2y - z = 4 \\ 2x + y + z = 3 \\ x - y + 3z = -1 \end{cases}$.求解方程组 $\begin{cases} x + 2y - z = 4 \\ 2x + y + z = 3 \\ x - y + 3z = -1 \end{cases}$。

Augmented matrix.

列增广矩阵。

12-14
2113
1-13-1

Eliminate column 1 below the pivot. $R_{2} \to R_{2} - 2R_{1}$; $R_{3} \to R_{3} - R_{1}$.

消去第 1 列主元以下。$R_{2} \to R_{2} - 2R_{1}$；$R_{3} \to R_{3} - R_{1}$。

12-14
0-33-5
0-34-5

Eliminate column 2 below the pivot. $R_{3} \to R_{3} - R_{2}$.

消去第 2 列主元以下。$R_{3} \to R_{3} - R_{2}$。

12-14
0-33-5
0010

Back-substitute. Row 3: $z = 0$. Row 2: $-3y + 3z = -5 \Rightarrow y = 5/3$. Row 1: $x + 2(5/3) - 0 = 4 \Rightarrow x = 4 - 10/3 = 2/3$.

回代求解。第 3 行：$z = 0$。第 2 行：$-3y + 3z = -5 \Rightarrow y = 5/3$。第 1 行：$x + 2(5/3) - 0 = 4 \Rightarrow x = 4 - 10/3 = 2/3$。

Solution. $x = \tfrac{2}{3}$, $y = \tfrac{5}{3}$, $z = 0$.

解。$x = \tfrac{2}{3}$、$y = \tfrac{5}{3}$、$z = 0$。

Verify. $\tfrac{2}{3} + 2 \cdot \tfrac{5}{3} - 0 = \tfrac{2 + 10}{3} = 4$ ✓; $2 \cdot \tfrac{2}{3} + \tfrac{5}{3} + 0 = \tfrac{4 + 5}{3} = 3$ ✓; $\tfrac{2}{3} - \tfrac{5}{3} + 0 = -1$ ✓.

验证。$\tfrac{2}{3} + 2 \cdot \tfrac{5}{3} - 0 = \tfrac{2 + 10}{3} = 4$ ✓；$2 \cdot \tfrac{2}{3} + \tfrac{5}{3} + 0 = \tfrac{4 + 5}{3} = 3$ ✓；$\tfrac{2}{3} - \tfrac{5}{3} + 0 = -1$ ✓。

Three outcomes — a worked taxonomy三种结果 —— 分类速查
  • Unique solution. Row reduction yields an upper-triangular matrix with non-zero diagonals (three pivots). Example: above.
  • 唯一解。消元后得对角非零的上三角矩阵（三个主元）。例如上题。
  • Infinite solutions. A row becomes $[\,0\;\;0\;\;0 \mid 0\,]$. The system has a free variable; parametrise it (e.g. set $z = t$) and back-substitute to get $x, y$ in terms of $t$.
  • 无穷多解。出现 $[\,0\;\;0\;\;0 \mid 0\,]$。方程组有自由变量；将其参数化（如设 $z = t$），回代得 $x, y$ 关于 $t$ 的表达式。
  • No solution. A row becomes $[\,0\;\;0\;\;0 \mid c\,]$ with $c \ne 0$. The system is inconsistent — write "no solution."
  • 无解。出现 $[\,0\;\;0\;\;0 \mid c\,]$（$c \ne 0$）。方程组不相容，直接写"无解（no solution）"。
After row-reducing a $3\times 3$ augmented matrix, you find the last row is $[\,0\;\;0\;\;0 \mid 5\,]$. What does this tell you?某 $3\times 3$ 增广矩阵化简后，最后一行为 $[\,0\;\;0\;\;0 \mid 5\,]$。这说明什么？
A5.5 · Q1
Unique solution $z = 5$唯一解 $z = 5$
Infinite solutions parametrised by $z$关于 $z$ 参数化的无穷多解
No solution (the system is inconsistent)无解（方程组不相容）
Need more rows to decide还需要更多行才能判断
$[\,0\;\;0\;\;0 \mid 5\,]$ represents the equation $0 = 5$, which is false. The system is inconsistent — no solution exists.$[\,0\;\;0\;\;0 \mid 5\,]$ 对应方程 $0 = 5$，恒假。方程组不相容，无解。
The row reads $0x + 0y + 0z = 5$, i.e. $0 = 5$ — a contradiction. No solution.该行表示 $0x + 0y + 0z = 5$，即 $0 = 5$，矛盾。无解。
Topic A5.6 · RREFTopic A5.6 · 行简化阶梯形

Reduced Row-Echelon Form行简化阶梯形（RREF） HL AHL 1.15

RREF takes Gaussian elimination one step further. Two extra conditions on top of row-echelon form:
  1. Every pivot equals $1$ (rescale rows as needed).
  2. Every pivot is the only non-zero entry in its column (eliminate above each pivot, not just below).
Once a matrix is in RREF, the solution reads off directly — no back-substitution. On the GDC, the rref() function on the matrix menu computes this in one step.
RREF（reduced row-echelon form）在行阶梯形之上再加两条要求：
  1. 每个主元都等于 $1$（必要时把整行乘以非零常数）。
  2. 每个主元在其所在列是唯一的非零元素（不仅消下方，也消上方）。
化成 RREF 后，解可直接读出，不必回代。GDC 上"矩阵"菜单中的 rref() 一步出结果。
Worked Example A5.6 — Reach RREF from row-echelonA5.6 例题 —— 由行阶梯形化为 RREF

Continue from the row-echelon end of A5.5 and convert to RREF, reading off the solution directly.从 A5.5 的行阶梯形继续化为 RREF，直接读出解。

Start (row echelon).

起点（行阶梯形）。

12-14
0-33-5
0010

Step 1 — make pivot of row 2 equal to $1$. $R_{2} \to -\tfrac{1}{3} R_{2}$.

第 1 步 —— 把第 2 行主元化为 $1$。$R_{2} \to -\tfrac{1}{3} R_{2}$。

12-14
01-15/3
0010

Step 2 — eliminate above pivot of column 3. $R_{1} \to R_{1} + R_{3}$; $R_{2} \to R_{2} + R_{3}$.

第 2 步 —— 消去第 3 列主元上方。$R_{1} \to R_{1} + R_{3}$；$R_{2} \to R_{2} + R_{3}$。

1204
0105/3
0010

Step 3 — eliminate above pivot of column 2. $R_{1} \to R_{1} - 2 R_{2}$.

第 3 步 —— 消去第 2 列主元上方。$R_{1} \to R_{1} - 2 R_{2}$。

1002/3
0105/3
0010

Read off. $x = \tfrac{2}{3}$, $y = \tfrac{5}{3}$, $z = 0$ — matches the back-substitution result of A5.5.

直接读出。$x = \tfrac{2}{3}$、$y = \tfrac{5}{3}$、$z = 0$ —— 与 A5.5 回代结果一致。

▸ Going deeper — why RREF is unique▸ 深入 —— RREF 的唯一性

A given matrix has many row-echelon forms (the pivots can be scaled differently, the rows ordered differently above the pivot) but exactly one RREF. This uniqueness is what makes RREF a canonical form — two different sequences of valid row operations always converge to the same final matrix. It's also why the GDC's rref() button is well-defined: there's no ambiguity in what to return.

同一个矩阵可以有许多种行阶梯形（主元可按不同常数缩放，主元上方各行的次序也可不同），但只有唯一一个 RREF。这一唯一性使 RREF 成为标准形（canonical form）—— 两条不同的合法行变换序列总收敛到同一个最终矩阵。这也是 GDC 上 rref() 按钮定义良好的根源：返回结果无歧义。

A matrix in RREF must satisfy all of: pivots are $1$; each pivot is the only non-zero entry in its column; pivots step right as we go down; all-zero rows are at the bottom. Which condition is extra to RREF that is not required of plain row echelon?RREF 必须满足：主元为 $1$、每个主元是其所在列唯一的非零元素、主元逐行向右移动、全零行在底部。下列哪条是 RREF新增而行阶梯形不要求的？
A5.6 · Q1
Each pivot is the only non-zero entry in its column每个主元是其所在列唯一的非零元素
Pivots step right as we go down主元逐行向右移动
All-zero rows are at the bottom全零行在底部
No two rows are identical没有两行完全相同
Row echelon already requires the staircase pattern and zero rows at the bottom. RREF adds two requirements: pivots $= 1$, and each pivot is the only non-zero entry in its column (eliminate above each pivot too).行阶梯形已要求阶梯形态和全零行底部。RREF 额外要求：主元 $= 1$，且每个主元是其所在列唯一的非零元素（主元上方也要消零）。
The "pivot is unique in its column" requirement is what distinguishes RREF from plain row echelon. Plain echelon allows non-zero entries above pivots."主元在列中唯一"是 RREF 区别于普通行阶梯形的关键。普通行阶梯形允许主元上方有非零元。
Exam Tactics考试技巧

Exam Strategy & Common Pitfalls考试策略与常见陷阱

Induction (Paper 1B)归纳法（Paper 1B）
  • Always write all three lines. Base + IH + IS — even if one of them is two lines long. Mark schemes deduct for missing structure.
  • 三段话一律全写。基础步骤 + 归纳假设 + 归纳步骤 —— 哪怕其中一段只有两行。结构缺失会扣分。
  • Underbrace the IH when you use it. Show where in the inductive step the assumption is substituted. Examiners want a visible IH use.
  • 用 IH 时打"下花括号"。在归纳步骤里标出哪一步代入了假设。考官要看见 IH 被用了。
  • Conclude with the principle. "By the principle of mathematical induction, $P(n)$ holds for all $n \ge n_{0}$." Skipping this costs the final $R1$ mark.
  • 结论里要引"原理"。"由数学归纳法原理，$P(n)$ 对所有 $n \ge n_{0}$ 成立。"不写这句要丢最后 1 分（$R1$）。
Linear systems (Paper 1B / Paper 3)线性方程组（Paper 1B / Paper 3）
  • Paper 1B is no-calc. Stick to Gaussian elimination by hand; keep fractions. Watch sign errors on the negative-pivot scaling steps.
  • Paper 1B 不可用计算器。手工 Gauss 消元，分数照写。负主元缩放那一步特别容易出符号错。
  • Paper 3 lets the GDC do RREF. Type the augmented matrix, press rref(), read off the solution. Method marks still require showing the matrix set-up.
  • Paper 3 允许用 GDC 算 RREF。输入增广矩阵，按 rref()，读出答案。方法分仍要看到增广矩阵列式。
  • Three-outcome classification. Always end with "unique / infinite / no solution" — examiners want this taxonomy explicitly.
  • 三类结果分类。结尾必明确写"唯一解 / 无穷多解 / 无解"。考官要看到分类。
Active Recall主动回忆

Flashcards闪卡

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Direct proof structure?直接证明结构？
Assume $P$. Then ... Therefore $Q$. $\square$设 $P$ 成立。则 ... 故 $Q$ 成立。 $\square$
To disprove "for all $x$, $P(x)$"?反驳"对所有 $x$, $P(x)$"？
One counter-example $x_{0}$ with $P(x_{0})$ false.一个反例 $x_{0}$，使 $P(x_{0})$ 不成立。
Proof by contradiction setup?反证法的开头？
"Suppose, for contradiction, that $\neg P$.""反设 $\neg P$ 成立。"
Induction — three lines?归纳法三段？
(1) Base $P(n_{0})$. (2) IH: assume $P(k)$. (3) IS: show $P(k) \Rightarrow P(k+1)$.(1) 基础 $P(n_{0})$。(2) IH：设 $P(k)$。(3) IS：证 $P(k) \Rightarrow P(k+1)$。
$1 + 2 + \cdots + n = ?$$1 + 2 + \cdots + n = ?$
$$\tfrac{n(n+1)}{2}$$
$\sqrt{2}$ irrational — contradiction with?$\sqrt{2}$ 无理 —— 与何矛盾？
$\gcd(a, b) = 1$, but both turn out even.$\gcd(a, b) = 1$，但 $a, b$ 同为偶数。
Partial fraction setup for $\dfrac{1}{(x-a)(x-b)}$?$\dfrac{1}{(x-a)(x-b)}$ 部分分式？
$$\dfrac{A}{x-a} + \dfrac{B}{x-b}$$
Setup for $\dfrac{1}{(x-a)^{2}}$ form?$\dfrac{1}{(x-a)^{2}}$ 形式的设法？
$$\dfrac{A_{1}}{x-a} + \dfrac{A_{2}}{(x-a)^{2}}$$
Three allowed row operations?三种合法行变换？
Swap; scale ($\lambda \ne 0$); add multiple of another row.行互换；按 $\lambda \ne 0$ 缩放；加另一行的倍数。
$[\,0\;\;0\;\;0 \mid 5\,]$ means?$[\,0\;\;0\;\;0 \mid 5\,]$ 含义？
$0 = 5$ — inconsistent — no solution.$0 = 5$ —— 不相容 —— 无解。
RREF vs row echelon — extras?RREF 比行阶梯形多了？
Pivots $= 1$; each pivot is alone in its column.主元 $= 1$；每个主元在其所在列单独存在。
GDC button for RREF?GDC 上 RREF 的功能键？
rref(matrix)
Mixed Practice综合练习

Unit A5 — Practice Quiz单元 A5——练习测验

Which of the following claims requires proof by contradiction rather than direct proof?下列哪个命题更适合用反证法而非直接证明？
Q1
"The sum of two even integers is even.""两个偶数之和为偶数。"
"$(n+1)^{2} = n^{2} + 2n + 1$ for all $n$.""对所有 $n$，$(n+1)^{2} = n^{2} + 2n + 1$。"
"There is no rational $r$ with $r^{2} = 2$.""不存在有理数 $r$ 使 $r^{2} = 2$。"
"$3 + 4 = 7$.""$3 + 4 = 7$。"
A non-existence claim ("there is no such $r$") is naturally handled by assuming such an $r$ exists and deriving a contradiction — the irrationality of $\sqrt{2}$."不存在某 $r$"这种否定性命题适合反证：假设其存在，推出矛盾 —— 即 $\sqrt{2}$ 的无理性证明。
Existence/non-existence statements ("there is no $r$ with...") are the canonical setup for contradiction.关于"存在/不存在"的命题是反证法的标准场景。
In an induction proof that $5^{n} + 7$ is divisible by $4$ for all $n \ge 1$, the inductive step uses the identity:证明 $5^{n} + 7$ 对一切 $n \ge 1$ 被 $4$ 整除时，归纳步骤用到的恒等式是：
Q2
$5^{k+1} + 7 = 5^{k} + 7 + 4 \cdot \text{stuff}$
$5^{k+1} + 7 = 5(5^{k} + 7) - 28$
$5^{k+1} = 5 \cdot 5^{k} + 7$
No identity needed — divisibility is preserved automatically不需要恒等式 —— 整除性自动成立
$5^{k+1} + 7 = 5 \cdot 5^{k} + 7 = 5(5^{k} + 7) - 5 \cdot 7 + 7 = 5(5^{k} + 7) - 28$. The first term is divisible by $4$ by IH, $28 = 4 \cdot 7$ is divisible by $4$, so the sum is divisible by $4$.$5^{k+1} + 7 = 5 \cdot 5^{k} + 7 = 5(5^{k} + 7) - 5 \cdot 7 + 7 = 5(5^{k} + 7) - 28$。第一项由 IH 被 $4$ 整除；$28 = 4 \cdot 7$ 也被 $4$ 整除；差仍被 $4$ 整除。
Rewrite $5^{k+1} + 7$ as $5(5^{k} + 7) - 28$ — both pieces are divisible by $4$ (IH gives the first; $28 = 4 \cdot 7$ gives the second).将 $5^{k+1} + 7$ 改写为 $5(5^{k} + 7) - 28$ —— 两部分均被 $4$ 整除（IH 给第一项；$28 = 4 \cdot 7$ 给第二项）。
Decompose $\dfrac{x + 1}{(x - 1)(x + 2)}$ into partial fractions.将 $\dfrac{x + 1}{(x - 1)(x + 2)}$ 分解为部分分式。
Q3
$\dfrac{2/3}{x - 1} + \dfrac{1/3}{x + 2}$
$\dfrac{1/3}{x - 1} + \dfrac{2/3}{x + 2}$
$\dfrac{1}{x - 1} + \dfrac{1}{x + 2}$
$\dfrac{2}{x - 1} - \dfrac{1}{x + 2}$
$x + 1 = A(x+2) + B(x-1)$. $x = 1$: $2 = 3A \Rightarrow A = 2/3$. $x = -2$: $-1 = -3B \Rightarrow B = 1/3$.$x + 1 = A(x+2) + B(x-1)$。$x = 1$：$2 = 3A \Rightarrow A = 2/3$。$x = -2$：$-1 = -3B \Rightarrow B = 1/3$。
Multiply through by $(x-1)(x+2)$: $x + 1 = A(x+2) + B(x-1)$. Plug $x = 1$ to find $A$, $x = -2$ to find $B$.两边乘 $(x-1)(x+2)$：$x + 1 = A(x+2) + B(x-1)$。代 $x = 1$ 得 $A$，$x = -2$ 得 $B$。
A $3 \times 3$ system's augmented matrix row-reduces to [1,0,0|2; 0,1,0|-1; 0,0,0|0]. The solution set is:$3 \times 3$ 方程组的增广矩阵化简为 [1,0,0|2; 0,1,0|-1; 0,0,0|0]。解集为：
Q4
Unique: $(2, -1, 0)$唯一解：$(2, -1, 0)$
No solution无解
Infinite: $(2, -1, t)$ for $t \in \mathbb{R}$无穷多解：$(2, -1, t)$，$t \in \mathbb{R}$
Infinite: $(t, -t, 0)$ for $t \in \mathbb{R}$无穷多解：$(t, -t, 0)$，$t \in \mathbb{R}$
Row 3 is $0 = 0$ (no constraint). Rows 1 and 2 give $x = 2$ and $y = -1$. $z$ is free — parametrise as $z = t$.第 3 行 $0 = 0$（无约束）。前两行给出 $x = 2$、$y = -1$。$z$ 是自由变量 —— 参数化为 $z = t$。
The all-zero row means a free variable. $x = 2$, $y = -1$ are determined; $z$ is free.全零行表示有自由变量。$x = 2$、$y = -1$ 已确定；$z$ 自由。
In the induction step of "$n^{3} \ge 2n^{2}$ for $n \ge 2$," after assuming $k^{3} \ge 2k^{2}$, you need to show:证明 "$n^{3} \ge 2n^{2}$（$n \ge 2$）"时，已设 $k^{3} \ge 2k^{2}$，需证明：
Q5
$k^{3} + 3k^{2} \ge 2k^{2} + 2k$
$(k+1)^{2} \ge 2k$
$k^{3} \ge 2(k+1)^{2}$
$(k+1)^{3} \ge 2(k+1)^{2}$
The inductive step proves the same statement with $k$ replaced by $k+1$: $(k+1)^{3} \ge 2(k+1)^{2}$. Use the IH and the fact that $3k^{2} + 3k + 1 \ge 4k + 2$ for $k \ge 2$.归纳步骤证明把 $k$ 换成 $k+1$ 后的同一命题：$(k+1)^{3} \ge 2(k+1)^{2}$。用 IH，并证 $3k^{2} + 3k + 1 \ge 4k + 2$（$k \ge 2$ 时成立）。
The inductive step always restates the claim at $n = k+1$: here $(k+1)^{3} \ge 2(k+1)^{2}$.归纳步骤总是把命题写在 $n = k+1$ 处：本题即 $(k+1)^{3} \ge 2(k+1)^{2}$。
Self-check自查

Readiness Checklist备考清单

Tick each one when you can do it cold — without notes, without the formula box, on your first attempt.

每一条都要"裸做"做对（不看笔记、不看公式框、一次过）才打勾。

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下一步

IB Paper-Style PracticeIB 试卷风格练习

A5 Practice + Solutions are still on the roadmap (sibling of A1/A3/A4). When they ship they'll live under Practice Questions/Unit_A5_*.html. For now, the A1 and A3 practice sets cover adjacent Topic 1 patterns (induction-style sum identities appear in A1 Practice Q4, Q11).

A5 配套的 Practice + Solutions 仍在排期。上线后将位于 Practice Questions/Unit_A5_*.html。在此之前，A1 与 A3 的练习题覆盖了相邻 Topic 1 模式（归纳式求和恒等式见 A1 练习 Q4、Q11）。

A1 Practice →A1 练习题 → A3 Practice →A3 练习题 →
鼎睿学苑 · Dingrui Scholars

IB Math Analysis & Approaches HL — Unit A5: Proof & Algebraic Manipulation · First Assessment 2029IB 数学分析与方法 HL——单元 A5：证明与代数运算 · 首次考核 2029