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。
Direct Proof & Counter-Example直接证明与反例 SL 1.6
- 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.
- 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。
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。 $\square$
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。
Prove: for all real $x$, $x^2 + 1 \ge 2x$.证明：对所有实数 $x$，$x^2 + 1 \ge 2x$。
Rearrange to isolate zero on one side.
移项，使一侧为零。
$$ x^2 + 1 - 2x \ge 0 \;\Longleftrightarrow\; x^2 - 2x + 1 \ge 0 $$Recognise a perfect square.
识别完全平方。
$$ x^2 - 2x + 1 = (x-1)^2 $$Conclude. $(x-1)^2 \ge 0$ for every real $x$ (a square is never negative), so $x^2+1-2x \ge 0$, i.e. $x^2+1 \ge 2x$. $\square$
得证。对每个实数 $x$，$(x-1)^2 \ge 0$（平方永不为负），故 $x^2+1-2x \ge 0$，即 $x^2+1 \ge 2x$。 $\square$
Proof by Contradiction反证法 SL 1.6
- 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).
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）。
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。
Prove that there are infinitely many prime numbers.证明质数有无穷多个。
Suppose, for contradiction, that there are only finitely many primes: $p_1, p_2, \ldots, p_n$.
反设质数只有有限多个：$p_1, p_2, \ldots, p_n$。
Construct a new number from all of them.
用它们全部构造一个新数。
$$ N = p_1 p_2 \cdots p_n + 1 $$Examine N's remainder under each $p_i$. Dividing $N$ by any $p_i$ leaves remainder $1$ (since $p_1\cdots p_n$ is exactly divisible by $p_i$), so no $p_i$ in our supposedly complete list divides $N$.
检查 $N$ 除以每个 $p_i$ 的余数。$N$ 除以任一 $p_i$ 余数为 $1$（因 $p_1\cdots p_n$ 能被 $p_i$ 整除），故列表中没有一个 $p_i$ 能整除 $N$。
Contradiction. Every integer greater than 1 has a prime factor, so $N$ has some prime factor $q$. But $q$ cannot be any $p_i$ (Step 3), so $q$ is a prime missing from our "complete" list — contradicting the assumption. Therefore there are infinitely many primes. $\square$
矛盾。每个大于 1 的整数都有质因数，故 $N$ 必有质因数 $q$。但 $q$ 不可能是任何 $p_i$（第 3 步），所以 $q$ 是遗漏于"完整"列表之外的质数——与假设矛盾。故质数有无穷多个。 $\square$
Mathematical Induction数学归纳法 HL AHL 1.6
- Base case. Verify $P(n_{0})$ directly (usually $n_{0} = 1$).
- Inductive hypothesis (IH). Assume $P(k)$ is true for some $k \ge n_{0}$.
- Inductive step. Show $P(k) \Rightarrow P(k+1)$, using the IH explicitly.
- 基础步骤（
base case）。直接验证 $P(n_{0})$（通常 $n_{0} = 1$）。 - 归纳假设（
inductive hypothesis，IH）。设某个 $k \ge n_{0}$ 时 $P(k)$ 成立。 - 归纳步骤（
inductive step）。显式使用 IH，证明 $P(k) \Rightarrow P(k+1)$。
principle of mathematical induction），$P(n)$ 对所有 $n \ge n_{0}$ 成立。" $\square$
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。 $\square$
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 $n \ge 1$，$6 \mid (7^{n} - 1)$。 $\square$
▸ 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。
Prove that $2^{n} > n$ for all $n \in \mathbb{Z}^{+}$.证明对所有正整数 $n$，$2^{n} > n$。
Base case ($n=1$). $2^1 = 2 > 1$. ✓
基础步骤（$n=1$）。$2^1 = 2 > 1$。 ✓
IH. Assume $2^{k} > k$ for some $k \ge 1$.
归纳假设。设某 $k \ge 1$ 时 $2^{k} > k$。
Inductive step. Multiply the IH by 2 (a valid step since both sides are positive), then bound the result below by $k+1$.
归纳步骤。把 IH 两边乘以 2（两边均为正，此步合法），再把结果向下界定为 $k+1$。
$$ 2^{k+1} = 2 \cdot 2^k > 2k \ge k + 1 \quad (\text{since } k \ge 1) $$So $2^{k+1} > k+1$, proving $P(k) \Rightarrow P(k+1)$. By induction, $2^n > n$ for all $n \ge 1$. $\square$ This example shows induction working on an inequality, not just an equation — the inductive step needs a small extra inequality ($2k \ge k+1$, true since $k\ge1$) rather than pure algebraic simplification.
故 $2^{k+1} > k+1$，证得 $P(k) \Rightarrow P(k+1)$。由归纳法，对所有 $n \ge 1$，$2^n > n$。 $\square$ 这个例子展示归纳法用于不等式而非单纯等式——归纳步骤需要额外一个小不等式（$2k \ge k+1$，因 $k\ge1$ 而成立），而不只是代数化简。
Partial Fractions部分分式分解 HL AHL 1.11
- 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 $(ax^{2}+bx+c)$ → $\dfrac{Ax+B}{ax^{2}+bx+c}$
- 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）$(ax^{2}+bx+c)$ → $\dfrac{Ax+B}{ax^{2}+bx+c}$
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。
$$ \frac{5x - 4}{(x-2)(x+1)} \;=\; \frac{2}{x - 2} + \frac{3}{x + 1}. $$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 $\dfrac{3x+1}{(x-1)(x^2+1)}$.分解 $\dfrac{3x+1}{(x-1)(x^2+1)}$。
Set up — a quadratic factor gets a linear numerator.
列式 —— 二次因式对应线性分子。
$$ \frac{3x+1}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1} $$Multiply through and substitute $x=1$ first (kills the quadratic term).
两边相乘，先代入 $x=1$（消去二次项）。
$$ 3x+1 = A(x^2+1) + (Bx+C)(x-1) $$ $$ x=1: \quad 4 = 2A \Rightarrow A = 2 $$Equate coefficients of $x^2$ and the constant term to find B, C.
比较 $x^2$ 项与常数项系数，求 B、C。
$$ x^2: \; 0 = A + B \Rightarrow B = -2 \qquad \text{const:} \; 1 = A - C \Rightarrow C = 1 $$Result.
结果。
$$ \frac{3x+1}{(x-1)(x^2+1)} = \frac{2}{x-1} + \frac{-2x+1}{x^2+1} $$$3 \times 3$ Linear Systems三元线性方程组 HL AHL 1.15
- 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).
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）。
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。
| 1 | 2 | -1 | 4 |
| 2 | 1 | 1 | 3 |
| 1 | -1 | 3 | -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。$R_{2} \to R_{2} - 2R_{1}$；$R_{3} \to R_{3} - R_{1}$。
| 1 | 2 | -1 | 4 |
| 0 | -3 | 3 | -5 |
| 0 | -3 | 4 | -5 |
Eliminate column 2 below the pivot. $R_{3} \to R_{3} - R_{2}$.
消去第 2 列主元以下。$R_{3} \to R_{3} - R_{2}$。
| 1 | 2 | -1 | 4 |
| 0 | -3 | 3 | -5 |
| 0 | 0 | 1 | 0 |
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。$\tfrac{2}{3} + 2 \cdot \tfrac{5}{3} - 0 = \tfrac{2 + 10}{3} = 4$ ✓；$2 \cdot \tfrac{2}{3} + \tfrac{5}{3} + 0 = \tfrac{4 + 5}{3} = 3$ ✓；$\tfrac{2}{3} - \tfrac{5}{3} + 0 = -1$ ✓。
- 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）"。
Solve $\begin{cases} x + y + z = 3 \\ 2x - y + z = 0 \\ 3x + 2z = 3 \end{cases}$.求解 $\begin{cases} x + y + z = 3 \\ 2x - y + z = 0 \\ 3x + 2z = 3 \end{cases}$。
Eliminate column 1: $R_2 \to R_2 - 2R_1$, $R_3 \to R_3 - 3R_1$.
消去第 1 列：$R_2 \to R_2 - 2R_1$，$R_3 \to R_3 - 3R_1$。
$$ \begin{cases} x+y+z=3 \\ -3y-z=-6 \\ -3y-z=-6 \end{cases} $$Row 3 is now identical to row 2 — it carries no new information. This signals infinitely many solutions with one free variable.
第 3 行现与第 2 行完全相同——不带来新信息。这说明存在一个自由变量，方程组有无穷多解。
Parametrise the free variable. Let $z = t$. From row 2: $-3y - t = -6 \Rightarrow y = \tfrac{6-t}{3} = 2 - \tfrac{t}{3}$.
把自由变量参数化。设 $z = t$。由第 2 行：$-3y - t = -6 \Rightarrow y = \tfrac{6-t}{3} = 2 - \tfrac{t}{3}$。
Back-substitute into row 1.
代回第 1 行。
$$ x = 3 - y - z = 3 - \left(2-\tfrac{t}{3}\right) - t = 1 - \tfrac{2t}{3} $$Solution set. $(x,y,z) = \left(1-\tfrac{2t}{3},\ 2-\tfrac{t}{3},\ t\right)$ for any $t \in \mathbb{R}$ — a line of solutions, not a single point.
解集。$(x,y,z) = \left(1-\tfrac{2t}{3},\ 2-\tfrac{t}{3},\ t\right)$，$t$ 取任意实数——是一条解的直线，而非单一的点。
Reduced Row-Echelon Form行简化阶梯形（RREF） HL AHL 1.15
- Every pivot equals $1$ (rescale rows as needed).
- Every pivot is the only non-zero entry in its column (eliminate above each pivot, not just below).
rref() function on the matrix menu computes this in one step.
reduced row-echelon form）在行阶梯形之上再加两条要求：
- 每个主元都等于 $1$（必要时把整行乘以非零常数）。
- 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）。
rref() 一步出结果。
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）。
| 1 | 2 | -1 | 4 |
| 0 | -3 | 3 | -5 |
| 0 | 0 | 1 | 0 |
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 $1$。$R_{2} \to -\tfrac{1}{3} R_{2}$。
| 1 | 2 | -1 | 4 |
| 0 | 1 | -1 | 5/3 |
| 0 | 0 | 1 | 0 |
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。$R_{1} \to R_{1} + R_{3}$；$R_{2} \to R_{2} + R_{3}$。
| 1 | 2 | 0 | 4 |
| 0 | 1 | 0 | 5/3 |
| 0 | 0 | 1 | 0 |
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。$R_{1} \to R_{1} - 2 R_{2}$。
| 1 | 0 | 0 | 2/3 |
| 0 | 1 | 0 | 5/3 |
| 0 | 0 | 1 | 0 |
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。
Exam Strategy & Common Pitfalls考试策略与常见陷阱
- 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（$R1$）。
- 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。
Flashcards闪卡
rref(matrix)Unit A5 — Practice Quiz单元 A5——练习测验
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。
- Lay out a direct proof using the "Assume ... Then ... Therefore" template用"设…则…故"模板写出一个直接证明
- Construct a counter-example to disprove a universal statement构造反例反驳全称命题
- Reproduce the proof that $\sqrt{2}$ is irrational (the lowest-terms contradiction)复现 $\sqrt{2}$ 无理的反证（最简分数矛盾）
- HL Write a complete induction proof: base, IH stated, IS using IH, conclusion citing the principle写出完整归纳证明：基础步骤、IH 明确陈述、IS 显式用 IH、结论引用归纳法原理
- HL Apply induction to sum formulas, divisibility, and inequalities将归纳法用于求和公式、整除性、不等式
- HL Decompose $P(x)/Q(x)$ into partial fractions for distinct linear factors对不同线性因式的 $P(x)/Q(x)$ 做部分分式分解
- HL Handle repeated linear factors with the $A_{1}/(\cdot) + A_{2}/(\cdot)^{2}$ pattern用 $A_{1}/(\cdot) + A_{2}/(\cdot)^{2}$ 模板处理重复线性因式
- HL Solve $3 \times 3$ linear systems by Gaussian elimination + back-substitution用 Gauss 消元 + 回代法解 $3 \times 3$ 线性方程组
- HL Reduce a matrix all the way to RREF and read the solution off directly将矩阵化为 RREF 并直接读出解
- HL Identify the three outcomes: unique, infinite, no solution from the reduced matrix shape由化简后矩阵形状辨别"唯一/无穷/无解"三种结果
- HL Parametrise free variables in the infinite-solution case在无穷多解情形下，将自由变量参数化
- HL Use the GDC
rref()function and verify against hand calculation使用 GDC 的rref()并与手算结果核对
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）。
How to use this guide本指南使用说明
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：
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。
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。