一门课的两个目标

18.100B 要教两件事：怎样写出严密的数学证明，以及怎样证明定理。这两件事听起来抽象，所以第一节课用一个具体例子把它们落地。

设 $f$ 是区间 $[a,b]$ 上的连续函数，$f(a)<0<f(b)$。介值定理断言：存在 $c\in(a,b)$，使得 $f(c)=0$。把图画出来，曲线从横轴下方出发，最后跑到上方，看起来非穿过横轴不可。

课程用了一个形象的比喻：连续就像用粉笔在桌上画曲线，笔不能离开桌面，也不能跳跃。既然笔从负的一侧连续地走到正的一侧，途中一定要经过 $0$。

直觉到了这里就停住了。真正的问题是：把这个直觉变成证明，需要用实数的哪一条性质？

为什么不能停在直觉

看起来显然是假的东西也会出错，只要基础不够严密，很快就会陷入「好像成立，但不知道是不是真的成立」的境地。逐步、严密地建立每一步，不仅保证结论正确，还让人更容易预测什么应该成立。

于是问题一层层往下追：

1. 证明介值定理，需要实数的什么性质？
2. 什么是实数？特别地，$\sqrt 2$ 是什么？它真的是一个数吗？
3. 这些性质能不能公理化地写清楚？

第三问的答案就是本讲的主线。实数 $\mathbb R$ 是一个完备的有序域，其中包含有理数 $\mathbb Q$。「完备」留到后面的课；这一讲把「域」和「有序」讲透。

时间轴

数系：记号约定

• 自然数 $\mathbb N=\{1,2,3,\dots\}$。这门课的自然数从 $1$ 开始，不含 $0$。
• 整数 $\mathbb Z=\{0,\pm1,\pm2,\pm3,\dots\}$。
• 有理数 $\mathbb Q=\{\,m/n : m\in\mathbb Z,\ n\in\mathbb N\,\}$。

有理数的定义里分母取自然数，因此分母恒为正，后面比较大小的时候不必讨论符号。

有理数上的三种结构

表示与相等

有理数写成商 $m/n$，但写法不唯一：$1/2=2/4=3/6$。两个商代表同一个有理数，当且仅当交叉相乘相等：

$$\frac{m_1}{n_1}=\frac{m_2}{n_2} \iff m_1n_2=m_2n_1 .$$

这是全部讨论的起点：有理数不是「一对整数」，而是「一对整数所在的等价类」。

加法与乘法

$$\frac{m_1}{n_1}+\frac{m_2}{n_2}=\frac{m_1n_2+m_2n_1}{n_1n_2}, \qquad \frac{m_1}{n_1}\cdot\frac{m_2}{n_2}=\frac{m_1m_2}{n_1n_2}.$$

分子是整数（可能是负的），分母是自然数，所以结果仍是有理数。看起来理所当然，但这里藏着一个必须验证的问题：如果换一个分数写法，结果会不会变？这就是下一节的「良定义」。

次序

因为分母都是正数，交叉相乘不会改变不等号方向：

$$\frac{m_1}{n_1}<\frac{m_2}{n_2} \iff m_1n_2<m_2n_1 .$$

如果分母允许为负，这一步就要额外小心符号。课程特意把分母定义成自然数，正是为了避免这种麻烦。

良定义：一个必须验证的细节

定义一个运算，不只是写出一条公式，还要证明结果不依赖代表元的选取。

命题。 若

$$\frac{m_1}{n_1}=\frac{m_2}{n_2}, \qquad \frac{p_1}{q_1}=\frac{p_2}{q_2},$$

则

$$\frac{m_1}{n_1}\cdot\frac{p_1}{q_1} = \frac{m_2}{n_2}\cdot\frac{p_2}{q_2}.$$

证明。 由假设的相等关系，交叉相乘得到

$$m_1n_2=m_2n_1, \qquad p_1q_2=p_2q_1 .$$

把这两条等式相乘：

$$(m_1n_2)(p_1q_2)=(m_2n_1)(p_2q_1).$$

重排乘法次序（整数乘法满足交换律与结合律）：

$$(m_1p_1)(n_2q_2)=(m_2p_2)(n_1q_1).$$

按有理数相等的判据，这正好说明两个乘积代表同一个有理数。$\square$

加法的验证完全同理，只是写起来更繁琐；课程选择乘法，只是因为它在黑板上更短。

这个证明同时示范了本课要求的证明粒度：每一步都要能指出它用的是哪条已知事实。这里用到的只有三件事：有理数相等的判据、等式两边可以相乘、整数乘法的交换律与结合律。

域的定义：五条、五条、再加一条

现在把有理数上观察到的结构抽象出来。一个域是一个集合 $F$，配上两个运算 $+$ 与 $\cdot$，满足下面的公理。

加法的五条

1. 封闭性：$x,y\in F\Rightarrow x+y\in F$。
2. 交换律：$x+y=y+x$。
3. 结合律：$(x+y)+z=x+(y+z)$。
4. 零元：存在 $0\in F$，使 $0+x=x$ 对一切 $x$ 成立。
5. 加法逆元：对每个 $x$，存在 $-x$，使 $x+(-x)=0$。

乘法的五条

1. 封闭性：$x,y\in F\Rightarrow xy\in F$。
2. 交换律：$xy=yx$。
3. 结合律：$(xy)z=x(yz)$。
4. 单位元：存在 $1\in F$，使 $1\cdot x=x$ 对一切 $x$ 成立。
5. 乘法逆元：对每个 $x\ne0$，存在 $x^{-1}$，使 $x\cdot x^{-1}=1$。

第五条有一个不能省略的补丁：必须排除 $0$。$0$ 不可能有乘法逆元——若 $0\cdot u=1$，用分配律会推出 $0=1$，而域通常还要求 $1\ne0$。所以公理写成「除 $0$ 以外每个元素都有乘法逆元」。

第十一条：分配律

前十条公理各自只管一个运算，彼此之间几乎没有联系。唯一的例外是「乘法逆元要提到零元」，那也只是顺带提到。真正把两种运算缝在一起的是分配律：

$$x(y+z)=xy+xz .$$

这条公理看起来平淡，却是唯一允许「展开」与「提取公因式」的规则。后面证明不等式乘正数时，它会是关键的一步。

第一个定理：零元唯一

公理只保证零元存在，没有保证只有一个。定理补上这一点。

定理。 任意域 $F$ 中，加法单位元唯一。

证明。 设 $0_1$ 与 $0_2$ 都是加法单位元，即对任意 $x$，

$$0_1+x=x, \qquad 0_2+x=x .$$

考虑 $0_1+0_2$ 这一个表达式，用两种方式计算：

• 把 $0_2$ 当作「任意元素 $x$」，用 $0_1$ 是零元：$0_1+0_2=0_2$。
• 先交换次序，把 $0_1$ 当作 $x$，用 $0_2$ 是零元：$0_2+0_1=0_1$。

由交换律，$0_1+0_2=0_2+0_1$，于是

$$0_2=0_1+0_2=0_2+0_1=0_1 .$$

所以 $0_1=0_2$。$\square$

这个证明只有三行，却展示了证明的典型套路：找一个同时涉及两个对象、又能用两种方式计算的表达式。课程说，学会这种写法需要时间，不要因为一开始写不出来就气馁。

同样地，加法逆元、乘法单位元、乘法逆元也都可以证明唯一。

哪些集合是域

• $\mathbb Q$ 是域：加法、乘法、次序都满足上面全部公理。
• $\mathbb N$ 不是域：连加法零元 $0$ 都不在里面，而且多数元素没有加法逆元。
• $\mathbb Z$ 不是域：它差一点，但在乘法逆元上失败。取 $2\in\mathbb Z$，若有逆元 $u$，则 $2u=1$，即 $u=1/2$，而 $1/2\notin\mathbb Z$。

这组例子说明：从 $\mathbb N$ 到 $\mathbb Z$ 补上了加法逆元，从 $\mathbb Z$ 到 $\mathbb Q$ 补上了乘法逆元。每一步都在填补一类「缺失的数」。

有序域：次序如何与运算相容

有序集

集合 $S$ 上的一个序，是使任意两个元素都能比较的关系 $<$：对任意 $x,y\in S$，下面三种情况恰有一种成立：

$$x=y, \qquad x<y, \qquad y<x .$$

这条性质通常称为三歧性（trichotomy）。

有序域

一个域如果同时是有序集，并且序与两个运算相容，就称为有序域：

• 与加法相容：$x<y\Rightarrow x+z<y+z$ 对任意 $z$ 成立。
• 与乘法相容：$x>0$ 且 $y>0\Rightarrow xy>0$。

$\mathbb Q$ 是有序域：有理数的次序可以比较任意两个数，并且上面两条都成立。

这两条公理能推出很多熟悉的不等式规则。比如 $x>0$ 时 $-x<0$：把 $x>0$ 写成 $x>0+0$……更直接地，由 $0<x$ 与加法相容性，两边加 $-x$ 得 $-x<0$。又如 $x<y$ 且 $z<0$ 时不等号反向（见理解检验第 4 题）。

定理：不等式两边乘正数

定理。 设 $F$ 是有序域，$x<y$，$z>0$，则 $xz<yz$。

证明。 目标是把结论化归到两条已知公理上。

第一步，把 $xz<yz$ 改写。两边同加 $-(xz)$，由加法相容性，

$$xz<yz \iff xz+(-xz)<yz+(-xz) \iff 0<yz-xz .$$

第二步，用分配律提出公因子 $z$：

$$yz-xz=(y-x)z .$$

第三步，把 $y-x$ 的符号定下来。由 $x<y$，两边同加 $-x$：

$$0<y-x .$$

第四步，$y-x>0$ 且 $z>0$，由乘法相容性立即得到

$$(y-x)z>0 .$$

于是 $0<yz-xz$，即 $xz<yz$。$\square$

整个证明没有用到任何「显然」：每一步都对应一条公理——加法相容、分配律、加法相容、乘法相容。这就是公理化写法的价值：结论不再依赖画图，而是从少数几条规则机械地推出。

本节结论

• 证明介值定理需要实数的完备性；本讲先处理「有序域」。
• 有理数有三种结构：加法、乘法、次序；其中加法与乘法必须先证明良定义。
• 域 = 加法五条 + 乘法五条 + 分配律；乘法逆元必须排除 $0$。
• 零元唯一，证明方法是同一个表达式算两次。
• $\mathbb Q$ 是有序域；$\mathbb N$、$\mathbb Z$ 都不是域。
• 有序域中，$x<y$ 且 $z>0$ 推出 $xz<yz$；乘负数时不等号方向要反。

理解检验

1. 为什么验证乘法的良定义时，必须同时更换两个因子的表示？只换其中一个为什么不够？

参考答案：乘法把两个等价类映到第三个等价类，而公式是写在代表元上的：$[m_1/n_1]\cdot[p_1/q_1]=[m_1p_1/(n_1q_1)]$。良定义要求：无论从两个等价类中各取哪个代表元，得到的结果类都相同。只换一个因子，只验证了「结果与第一个因子的代表元无关」，没有验证「与第二个因子的代表元无关」，也没有验证「两个因子同时换代表元」时的一致性。只有两个位置上的替换都任意时，公式才真正定义了等价类之间的运算。

2. 仿照零元唯一的证明，证明加法逆元唯一。

参考答案：设 $x$ 有两个加法逆元 $u$、$v$，即 $x+u=0$ 且 $x+v=0$。考虑 $u$：

$$u=u+0=u+(x+v)=(u+x)+v=0+v=v .$$

这里依次用了：$0$ 是零元、$x+v=0$、结合律、$u+x=x+u=0$、$0$ 是零元。于是 $u=v$。证明结构与零元唯一完全相同：构造一个能用两种方式化简的表达式。

3. 整数为什么不是域？请指出它具体违反了哪一条公理，并给出反例。

参考答案：违反乘法的「逆元」公理。域要求每个非零元素都有乘法逆元；在 $\mathbb Z$ 中取 $2\ne0$，若有 $u\in\mathbb Z$ 使 $2u=1$，则 $u=1/2$，但 $1/2\notin\mathbb Z$。所以 $2$ 在 $\mathbb Z$ 中没有乘法逆元。（另外 $\mathbb N$ 连加法零元都没有，失败得更彻底。）

4. 设 $x<y$ 且 $z<0$，证明 $xz>yz$——不等号方向反过来了。

参考答案：先定符号。由 $z<0$ 与加法相容性，两边加 $-z$ 得 $0<-z$，即 $-z>0$。由 $x<y$ 得 $y-x>0$。两个正数相乘为正：

$$(y-x)(-z)>0 .$$

用分配律展开：$(y-x)(-z)=-(y-x)z=xz-yz$。于是 $xz-yz>0$，即 $xz>yz$。关键点是「乘负数」被翻译成了「乘一个正数 $-z$，再取相反数」，不等式方向因此翻转。

5. 有理数已经满足全部有序域公理，为什么仍不足以证明介值定理？

参考答案：介值定理需要的是完备性——$\mathbb R$ 的每个非空有上界子集都有最小上界。$\mathbb Q$ 是有序域，但有空隙：不存在有理数 $c$ 使 $c^2=2$。考虑函数 $f(x)=x^2-2$，定义在 $\mathbb Q$ 上。它在 $x=1$ 处取 $-1<0$，在 $x=2$ 处取 $2>0$，且作为多项式函数在 $\mathbb Q$ 上连续；但方程 $x^2=2$ 在 $\mathbb Q$ 中无解，所以找不到 $c\in\mathbb Q$ 使 $f(c)=0$。这说明「连续 + 端点异号」还不足以保证过零，必须再假定数系没有空隙。补充：这个反例是标准的，课上没有展开，它正是下一讲要定义的完备性所排除的情形。

与前后课的关系

本讲只建立了「有序域」，把「完备」留空。下一讲补上完备性（最小上界性质），说明它如何排除 $\mathbb Q$ 式的空隙，从而让介值定理成为可证明的定理。第三讲转向证明的写法本身，系统讲解量词、反证与 $\varepsilon$ 论证，为后面数列极限做准备。

整门课的主线就此确定：每一条分析结论，都要能追溯到实数的公理。

Two goals for one course

18.100B teaches two things: how to write a rigorous mathematical proof, and how to prove theorems. Both sound abstract, so the first lecture makes them concrete with a single example.

Let $f$ be continuous on $[a,b]$ with $f(a)<0<f(b)$. The intermediate value theorem asserts that some $c\in(a,b)$ satisfies $f(c)=0$. On the picture, the curve starts below the axis and ends above it, so it seems bound to cross.

The course offers a physical image: continuity is like drawing the graph with a piece of chalk that may never leave the table and may never jump. If the chalk travels continuously from the negative side to the positive side, it must pass through $0$.

Intuition stops here. The real question is: which property of the real numbers does this argument use?

Why intuition is not enough

Things that look obvious can still fail when the foundations are not airtight; with loose foundations one quickly reaches a point where something "seems true" without anyone being able to tell whether it is true. Building every step rigorously not only guarantees correctness, it also makes it far easier to predict what should be true.

So the questions unfold:

1. Which property of the reals makes the IVT provable?
2. What is a real number? In particular, what is $\sqrt 2$ — is it a number at all?
3. Can these properties be written down as axioms?

The answer to the third question is the spine of the lecture. The reals $\mathbb R$ form a complete ordered field containing the rationals $\mathbb Q$. Completeness is postponed; this lecture treats "field" and "ordered" in full.

Timeline

Number systems: conventions

• Natural numbers $\mathbb N=\{1,2,3,\dots\}$. In this course they start at $1$; zero is not included.
• Integers $\mathbb Z=\{0,\pm1,\pm2,\pm3,\dots\}$.
• Rationals $\mathbb Q=\{\,m/n : m\in\mathbb Z,\ n\in\mathbb N\,\}$.

The denominator is taken from $\mathbb N$, so it is always positive; comparisons never need a sign discussion.

Three structures on the rationals

Representation and equality

A rational is written as a quotient $m/n$, but the presentation is not unique: $1/2=2/4=3/6$. Two quotients denote the same rational exactly when cross-multiplication agrees:

$$\frac{m_1}{n_1}=\frac{m_2}{n_2} \iff m_1n_2=m_2n_1 .$$

This is where everything starts: a rational is not a pair of integers but the equivalence class of such a pair.

Addition and multiplication

$$\frac{m_1}{n_1}+\frac{m_2}{n_2}=\frac{m_1n_2+m_2n_1}{n_1n_2}, \qquad \frac{m_1}{n_1}\cdot\frac{m_2}{n_2}=\frac{m_1m_2}{n_1n_2}.$$

The numerator is an integer (possibly negative) and the denominator is natural, so the result is again rational. This looks routine, but it hides a question that must be settled: does the answer change if a different representative is used? That is the notion of being well defined.

Order

Since denominators are positive, cross-multiplication preserves the inequality:

$$\frac{m_1}{n_1}<\frac{m_2}{n_2} \iff m_1n_2<m_2n_1 .$$

If negative denominators were allowed, the sign would need special care. Taking denominators from $\mathbb N$ removes that annoyance by construction.

Well defined: the detail that must be checked

Defining an operation is not just writing a formula; one must prove that the result does not depend on the chosen representatives.

Claim. If

$$\frac{m_1}{n_1}=\frac{m_2}{n_2}, \qquad \frac{p_1}{q_1}=\frac{p_2}{q_2},$$

then

$$\frac{m_1}{n_1}\cdot\frac{p_1}{q_1} = \frac{m_2}{n_2}\cdot\frac{p_2}{q_2}.$$

Proof. Cross-multiplying the hypotheses gives

$$m_1n_2=m_2n_1, \qquad p_1q_2=p_2q_1 .$$

Multiply the two equalities:

$$(m_1n_2)(p_1q_2)=(m_2n_1)(p_2q_1).$$

Rearranging factors (commutativity and associativity of integer multiplication):

$$(m_1p_1)(n_2q_2)=(m_2p_2)(n_1q_1).$$

By the criterion for equality of rationals, the two products denote the same rational. $\square$

The proof for addition is identical in structure, only messier to write; the lecture chooses multiplication purely because it is shorter on the board.

The proof also sets the expected level of detail: every step must name the fact it uses. Only three facts appear here: the equality criterion for rationals, the ability to multiply equalities, and commutativity and associativity of integer multiplication.

The axioms of a field: five, five, and one

Now abstract what was observed on $\mathbb Q$. A field is a set $F$ with two operations $+$ and $\cdot$.

Five axioms for addition

1. Closure: $x,y\in F\Rightarrow x+y\in F$.
2. Commutativity: $x+y=y+x$.
3. Associativity: $(x+y)+z=x+(y+z)$.
4. Zero element: there is $0\in F$ with $0+x=x$ for all $x$.
5. Additive inverse: for every $x$ there is $-x$ with $x+(-x)=0$.

Five axioms for multiplication

1. Closure: $x,y\in F\Rightarrow xy\in F$.
2. Commutativity: $xy=yx$.
3. Associativity: $(xy)z=x(yz)$.
4. Unit element: there is $1\in F$ with $1\cdot x=x$ for all $x$.
5. Multiplicative inverse: for every $x\ne0$ there is $x^{-1}$ with $x\cdot x^{-1}=1$.

The fifth axiom carries a patch that cannot be dropped: zero is excluded. Zero cannot have an inverse — from $0\cdot u=1$ the distributive law forces $0=1$, and a field requires $1\ne0$. Hence the axiom says "every element other than $0$ has a multiplicative inverse".

The eleventh axiom: distributivity

The first ten axioms each govern a single operation and barely interact. The only hint of contact is the mention of zero inside the inverse axiom. What actually ties the two operations together is distributivity:

$$x(y+z)=xy+xz .$$

It looks bland, but it is the only axiom that permits expanding a product or factoring a common term. It will be the decisive step when we multiply an inequality by a positive number.

First theorem: the zero element is unique

The axioms guarantee that a zero element exists, not that it is unique. The theorem supplies that.

Theorem. In any field $F$, the additive identity is unique.

Proof. Let $0_1$ and $0_2$ both be additive identities, so for every $x$,

$$0_1+x=x, \qquad 0_2+x=x .$$

Compute the single expression $0_1+0_2$ in two ways:

• Treat $0_2$ as the "arbitrary $x$" and use that $0_1$ is an identity: $0_1+0_2=0_2$.
• First commute, treat $0_1$ as $x$, and use that $0_2$ is an identity: $0_2+0_1=0_1$.

By commutativity $0_1+0_2=0_2+0_1$, hence

$$0_2=0_1+0_2=0_2+0_1=0_1 .$$

So $0_1=0_2$. $\square$

Three lines, but the pattern is the standard one: find an expression that involves both objects and can be evaluated in two ways. The lecturer notes that this style takes time to learn; struggling at first is normal.

The same argument, with the same skeleton, shows that additive inverses, multiplicative units and multiplicative inverses are unique.

Which sets are fields

• $\mathbb Q$ is a field: addition, multiplication and order satisfy all axioms.
• $\mathbb N$ is not: it does not even contain the additive identity $0$, and most elements have no additive inverse.
• $\mathbb Z$ is not: it comes close but fails on multiplicative inverses. Take $2\in\mathbb Z$; if $u$ satisfied $2u=1$ then $u=1/2$, and $1/2\notin\mathbb Z$.

The examples show a pattern: $\mathbb N\to\mathbb Z$ adds additive inverses, $\mathbb Z\to\mathbb Q$ adds multiplicative inverses. Each step fills in a missing kind of number.

Ordered fields: how order interacts with the operations

Ordered sets

An order on a set $S$ is a relation $<$ that compares any two elements: for all $x,y\in S$, exactly one of

$$x=y, \qquad x<y, \qquad y<x$$

holds. This is trichotomy.

Ordered fields

A field that is also an ordered set, with order compatible with both operations, is an ordered field:

• Compatibility with addition: $x<y\Rightarrow x+z<y+z$ for every $z$.
• Compatibility with multiplication: $x>0$ and $y>0\Rightarrow xy>0$.

$\mathbb Q$ is an ordered field: any two rationals are comparable, and both compatibility axioms hold.

These axioms generate the familiar rules of inequalities. For instance $x>0$ implies $-x<0$: from $0<x$, add $-x$ to both sides. And when $z<0$, multiplying an inequality reverses it (see question 4 below).

Theorem: multiplying an inequality by a positive number

Theorem. Let $F$ be an ordered field, $x<y$ and $z>0$. Then $xz<yz$.

Proof. Reduce the goal to the two compatibility axioms.

First, rewrite. Adding $-(xz)$ to both sides and using compatibility with addition,

$$xz<yz \iff xz+(-xz)<yz+(-xz) \iff 0<yz-xz .$$

Second, factor out $z$ with distributivity:

$$yz-xz=(y-x)z .$$

Third, determine the sign of $y-x$. From $x<y$, add $-x$ to both sides:

$$0<y-x .$$

Fourth, $y-x>0$ and $z>0$, so compatibility with multiplication gives

$$(y-x)z>0 .$$

Hence $0<yz-xz$, that is $xz<yz$. $\square$

Nothing in this proof is "obvious": each line invokes a named axiom — additive compatibility, distributivity, additive compatibility again, multiplicative compatibility. That is the point of the axiomatic style: the conclusion no longer rests on a picture but follows mechanically from a short list of rules.

What this lecture establishes

• The IVT needs completeness; this lecture settles "ordered field" first.
• Rationals carry addition, multiplication and order, and the operations must be shown well defined.
• A field is five additive axioms, five multiplicative axioms and distributivity; the inverse axiom must exclude $0$.
• The zero element is unique, proved by evaluating one expression twice.
• $\mathbb Q$ is an ordered field; $\mathbb N$ and $\mathbb Z$ are not fields.
• In an ordered field, $x<y$ and $z>0$ imply $xz<yz$; multiplying by a negative reverses the inequality.

Comprehension check

1. Why must both factors change representatives when checking that multiplication is well defined? Why is changing only one not enough?

Reference answer: Multiplication maps a pair of equivalence classes to a third class, and the formula is written on representatives: $[m_1/n_1]\cdot[p_1/q_1]=[m_1p_1/(n_1q_1)]$. Well definedness requires that any choice of representatives from the two classes gives the same resulting class. Changing one factor only shows independence from the first factor's representative; it says nothing about the second, nor about the case where both change at once. Only when both slots may be replaced freely does the formula define an operation on classes rather than on presentations.

2. Imitate the proof that the zero element is unique to prove that additive inverses are unique.

Reference answer: Let $x$ have additive inverses $u$ and $v$, so $x+u=0$ and $x+v=0$. Compute $u$:

$$u=u+0=u+(x+v)=(u+x)+v=0+v=v .$$

The steps use, in order: $0$ is the identity, $x+v=0$, associativity, $u+x=x+u=0$, and $0$ is the identity. Hence $u=v$. The skeleton is identical to the zero-element proof: build one expression and simplify it two ways.

3. Why are the integers not a field? Name the axiom that fails and give a counterexample.

Reference answer: The multiplicative inverse axiom fails. A field requires an inverse for every nonzero element. Take $2\ne0$ in $\mathbb Z$; if $u\in\mathbb Z$ satisfied $2u=1$, then $u=1/2$, which is not an integer. So $2$ has no multiplicative inverse in $\mathbb Z$. ($\mathbb N$ fails even earlier — it lacks the additive identity.)

4. Suppose $x<y$ and $z<0$. Prove $xz>yz$ — the inequality reverses.

Reference answer: First determine signs. From $z<0$, add $-z$ to both sides to get $0<-z$, i.e. $-z>0$. From $x<y$ get $y-x>0$. The product of two positive elements is positive:

$$(y-x)(-z)>0 .$$

Expand: $(y-x)(-z)=-(y-x)z=xz-yz$. Hence $xz-yz>0$, i.e. $xz>yz$. The trick is to translate "multiply by a negative" into "multiply by the positive $-z$, then negate", which is exactly why the direction flips.

5. The rationals satisfy every ordered-field axiom. Why is that still not enough to prove the IVT?

Reference answer: The IVT needs completeness — every nonempty subset of $\mathbb R$ bounded above has a least upper bound. $\mathbb Q$ is an ordered field but has gaps: no rational $c$ satisfies $c^2=2$. Consider $f(x)=x^2-2$ on $\mathbb Q$. It takes $-1<0$ at $x=1$ and $2>0$ at $x=2$, and as a polynomial it is continuous on $\mathbb Q$, yet $x^2=2$ has no solution in $\mathbb Q$, so no $c\in\mathbb Q$ gives $f(c)=0$. Thus "continuous with opposite signs at the endpoints" does not suffice; one must also assume the number system has no gaps. Note: this standard counterexample is supplied for context, not developed in the lecture; it is precisely the situation that completeness rules out.

Relation to neighbouring lectures

This lecture builds "ordered field" and leaves "complete" open. The next lecture supplies completeness (the least upper bound property) and shows how it removes the gaps of $\mathbb Q$, making the IVT provable. Lecture 3 turns to the craft of proof itself — quantifiers, contradiction, and $\varepsilon$-arguments — preparing the ground for limits of sequences.

The theme of the whole course is now fixed: every statement in analysis must be traceable back to the axioms of the real numbers.