回顾：还缺哪一块

上一讲为介值定理准备了「有序域」，但明确留下了「完备」没有定义。本讲补上这一块，并回答一个具体问题：$\sqrt 2$ 到底是什么数？

先把上一讲的结构压缩成一张表：

本讲新增的传递性很关键：若 $x<y$ 且 $y<z$，则 $x<z$。它让「小于」可以串起来用，后面几乎所有不等式论证都靠它。

时间轴

$\sqrt 2$ 不是有理数

$\sqrt 2$ 是满足 $x>0$ 且 $x^2=2$ 的数。要说明它不属于 $\mathbb Q$，就是证明：不存在 $m\in\mathbb Z$、$n\in\mathbb N$，使 $(m/n)^2=2$。

命题。 不存在有理数 $x$ 使 $x^2=2$。

证明（反证）。 假设存在 $x=m/n$，其中 $m\in\mathbb Z$、$n\in\mathbb N$，且 $x^2=2$。

首先可以假设 $m$ 与 $n$ 没有公因子。理由：有理数的表示不唯一，若分子分母有公因子 $d>1$，把两者同除以 $d$，得到同一个有理数的另一组表示；反复约分，最终可以取到互素的 $m,n$。这一步不是可有可无的，它是后面产生矛盾的支点。

由 $m^2/n^2=2$ 得

$$m^2=2n^2 .$$

右边是偶数，所以 $m^2$ 是偶数。偶数的平方根仍是偶数：若 $m$ 为奇数，设 $m=2k+1$，则 $m^2=4k^2+4k+1$ 为奇数，矛盾。因此 $m$ 是偶数，写 $m=2m_1$，其中 $m_1\in\mathbb Z$。

代入：

$$(2m_1)^2=2n^2 \implies 4m_1^2=2n^2 \implies 2m_1^2=n^2 .$$

于是 $n^2$ 是偶数，同理 $n$ 也是偶数。

现在 $m$ 与 $n$ 都被 $2$ 整除，与「$m,n$ 互素」矛盾。所以假设不成立，$\sqrt 2\notin\mathbb Q$。$\square$

证明只用了两件事：有理数的约分、以及「若整数的平方是偶数则整数是偶数」。后者本身也值得写清楚，因为它是证明的一部分。

$\sqrt 2$ 在哪里：作为序列的极限

既然 $\mathbb Q$ 里没有 $\sqrt 2$，就要问：实数怎样容纳它？

先看十进制展开：

$$\sqrt 2=1.41421356\ldots$$

逐位截断得到一串数

$$1,\quad 1.4,\quad 1.41,\quad 1.414,\quad 1.4142,\ \dots$$

每一个截断都是有理数，而且随着位数增加越来越接近 $\sqrt 2$。所以一种想法是：把 $\sqrt 2$ 定义为这串数的极限。

这引出两个后面要花很多时间的概念：数列与极限。「越来越接近」只是直觉，真正要写出来，需要 $\varepsilon$ 语言，那是第 4 讲的内容。这里只借用它说明完备性要解决的问题：有理数没有 $\sqrt 2$，实数用「极限」把这个洞补上。

上界与最小上界

上界

设 $S$ 是有序集，$A\subseteq S$。称 $M\in S$ 是 $A$ 的上界，如果

$$a\le M \quad\text{对一切 } a\in A .$$

注意 $M$ 不必属于 $A$。例子：

• $S=\mathbb N$，$A=\{1,2,3\}$。则 $4$ 是上界，$3$ 也是上界；$2$ 不是，因为 $3>2$。
• $S=\mathbb Q$，$A=\mathbb N$。则 $A$ 没有上界——不管取多大的有理数，总有更大的自然数。

最小上界

设 $A$ 有上界。称 $M$ 是 $A$ 的最小上界（或上确界，记 $\sup A$），如果

1. $M$ 是 $A$ 的上界；
2. 对 $A$ 的任意上界 $M_1$，都有 $M\le M_1$。

换成一句更常用的话：$M$ 是最小的上界。它同时承担两个角色——够大（压得住 $A$ 的所有元素），又够小（没有任何更小的数还能压住）。

这里有一个容易被忽略的细节：最小上界不一定属于 $A$。例如 $A=\{x\in\mathbb Q: x<1\}$ 的 $\sup A=1$，而 $1\notin A$。

完备有序域

定义很短：

完备有序域是一个有序域，其中每个有界子集都有最小上界。

「有界」指有上界。课程说明，只定义上界而不定义下界并不损失一般性：把集合换成它的「反向」，上界就变成下界；两种说法等价，多写一遍是冗余。当然，也可以对 $-A=\{-a:a\in A\}$ 使用上确界来得到下确界。

接下来是一个不证明的定理：

定理。 存在一个最小的完备有序域，它包含有理数 $\mathbb Q$。

这个域就是实数域 $\mathbb R$。证明它需要从 $\mathbb Q$ 出发构造 $\mathbb R$（例如用戴德金分割或柯西序列），超出了本课范围；课程把它当作出发点。

有了这条定理，上一讲的介值定理才终于有了所需的性质。但「完备」这个定义是否真的能推出我们想要的东西？课程选择用 $\sqrt 2$ 来检验。

在 $\mathbb R$ 中造出 $\sqrt 2$

思路：不直接解方程，而是把 $\sqrt 2$ 定义为某个集合的最小上界，再验证这个上界确实平方等于 $2$。

集合 $A$

$$A=\{\,x\in\mathbb R : x>0,\ x^2<2\,\}.$$

先检查定义用得上：

• 非空：$1\in A$，因为 $1>0$ 且 $1^2=1<2$。
• 有上界：若 $x\in A$，则 $x>0$ 且 $x^2<2$。由于 $2^2=4>2$，必有 $x<2$。为什么不可能是 $x\ge2$？因为 $x\ge2>0$ 时 $x^2\ge4>2$，与 $x^2<2$ 矛盾。所以 $2$ 是上界。

由完备性，$\sup A$ 存在。令

$$x=\sup A .$$

课程特意不把它记作 $\sqrt 2$：如果在证明途中就叫它 $\sqrt 2$，大脑会自动使用「$\sqrt 2$ 的平方等于 $2$」这条还没证明的性质。

由 $1\in A$ 且 $x$ 是上界，得 $1\le x$，所以 $x>0$。

剩下要证 $x^2=2$。分成两个不等式。

第一步：$x^2\le 2$

反设 $x^2>2$。于是 $x^2-2>0$，可以取一个很小的正数 $h$，满足

$$0<h<\min\left\{x,\ \frac{x^2-2}{2x}\right\}.$$

（$x>0$，所以右端为正，这样的 $h$ 存在。）计算

$$(x-h)^2=x^2-2xh+h^2\ge x^2-2xh .$$

因为 $h<\dfrac{x^2-2}{2x}$，两边乘以 $2x>0$ 得 $2xh<x^2-2$，即 $x^2-2xh>2$。于是

$$(x-h)^2>2 .$$

现在任取 $a\in A$。由 $a>0$、$x-h>0$（因为 $h<x$）以及 $a^2<2<(x-h)^2$，开方保序给出 $a<x-h$。所以 $x-h$ 是 $A$ 的上界，而 $x-h<x$，与 $x$ 是最小上界矛盾。

因此 $x^2\le 2$。

第二步：$x^2\ge 2$

反设 $x^2<2$。于是 $2-x^2>0$。取 $h$ 满足

$$0<h<\min\left\{1,\ \frac{2-x^2}{1+2x}\right\}.$$

计算

$$(x+h)^2=x^2+2xh+h^2 .$$

因为 $0<h<1$，有 $h^2<h$。所以

$$(x+h)^2<x^2+2xh+h=x^2+h(1+2x).$$

由 $h(1+2x)<2-x^2$ 得

$$(x+h)^2<x^2+(2-x^2)=2 .$$

同时 $x+h>0$，所以 $x+h\in A$。但 $x+h>x$，与 $x$ 是 $A$ 的上界矛盾。

因此 $x^2\ge 2$。

两步合起来，$x^2=2$。于是 $\sup A$ 就是 $\sqrt 2$，实数中确实存在这个数。

这两步的构造值得记住：要证明「上确界满足某个等式」，就用两次反证——若偏大，往左挪一点仍压得住集合；若偏小，往右挪一点就落进集合。挪动的幅度由 $x^2$ 与 $2$ 的差决定。

推论：$\mathbb Q$ 不完备

把上面的论证整理一下就能得到：

推论。 有理数域 $\mathbb Q$ 不是完备的。

证明。 设 $F$ 是任一完备的有序域，且包含 $\mathbb Q$。上面的构造只用到了域运算、次序与完备性，所以 $\sqrt 2=\sup\{x\in F:x>0,x^2<2\}$ 必在 $F$ 中。但第一段已证明 $\sqrt 2\notin\mathbb Q$。若 $\mathbb Q$ 完备，取 $F=\mathbb Q$ 就矛盾。$\square$

再看集合 $A_{\mathbb Q}=\{x\in\mathbb Q:x>0,x^2<2\}$。它在 $\mathbb Q$ 中有上界（例如 $2$），却没有最小上界。验证如下：设 $q\in\mathbb Q$ 是它的任一上界。因为 $1\in A_{\mathbb Q}$，有 $q\ge1>0$。若 $q^2<2$，则 $q\in A_{\mathbb Q}$，与 $q$ 是上界矛盾；若 $q^2=2$，则 $q$ 是有理数的平方根，已证不可能。所以 $q^2>2$。

令

$$q'=\frac{2/q+q}{2}=\frac{q^2+2}{2q}.$$

$q'$ 是有理数且为正。比较大小：

$$q'<q \iff q^2+2<2q^2 \iff 2<q^2,$$

成立。再看 $q'$ 是不是上界：任取 $x\in A_{\mathbb Q}$，则 $x^2<2$，而

$$(q')^2=\frac{(q^2+2)^2}{4q^2}>2 \iff (q^2-2)^2>0,$$

由 $q^2>2$ 成立。于是 $x^2<2<(q')^2$，开方得 $x<q'$。所以 $q'$ 是更小的有理上界。

每个有理上界都对应一个更小的有理上界，因此不存在最小上界。这正是「$\mathbb Q$ 有空隙」的精确含义：空隙就在 $\sqrt 2$ 的位置。

阿基米德性质

定理（阿基米德性质）。 对任意实数 $x$，存在自然数 $n$ 使 $n>x$。

直觉上显然：自然数没有上限。但「显然」恰恰是需要用完备性证明的东西。

证明（反证）。 假设存在实数 $x$，使得对一切 $n\in\mathbb N$ 都有 $n\le x$。于是 $x$ 是 $\mathbb N$ 的上界，$\mathbb N$ 作为 $\mathbb R$ 的子集有上界。

由完备性，令

$$y=\sup\mathbb N .$$

因为 $y$ 是上界，对每个 $n\in\mathbb N$ 都有 $n\le y$。另一方面 $n+1$ 也是自然数，所以

$$n+1\le y \implies n\le y-1 \quad\text{对一切 } n .$$

这说明 $y-1$ 也是 $\mathbb N$ 的上界。但 $y-1<y$，与 $y$ 是最小上界矛盾。$\square$

这个证明很短，却用尽了三条性质：$\mathbb N$ 对 $+1$ 封闭、序的传递性、以及最小上界的「最小」。课程强调，这类证明正是分析课的基本功。

推论：有理数在实数中稠密

推论。 若 $x<y$ 是实数，则存在有理数 $m/n$ 使

$$x<\frac mn<y .$$

证明。 令 $\beta=y-x>0$。由阿基米德性质，存在 $n\in\mathbb N$ 使 $n>1/\beta$，等价地

$$\frac1n<\beta .$$

现在取最小的整数 $m$ 使 $m/n>x$。这样的 $m$ 存在：由阿基米德性质可取整数 $k>nx$，则 $k/n>x$，所以满足条件的整数集合非空；整数集有下界，故有最小元。

由 $m$ 的最小性，$m-1$ 不满足条件，即

$$\frac{m-1}{n}\le x .$$

于是

$$\frac mn=\frac{m-1}{n}+\frac1n\le x+\frac1n<x+\beta=y .$$

结合 $m/n>x$，得到 $x<m/n<y$。$\square$

几何上就是：取一根长度 $1/n$ 的「尺子」，它比 $x$ 到 $y$ 的距离还短；从 $x$ 出发向右数格子，第一个越过 $x$ 的格子点一定还没越过 $y$。

本节结论

• $\sqrt 2\notin\mathbb Q$；反证法的支点是「先约分到互素」。
• 完备性 = 每个有界子集都有最小上界（上确界）。
• $\mathbb R$ 是最小的完备有序域，它包含 $\mathbb Q$。
• 在 $\mathbb R$ 中，$\sup\{x>0:x^2<2\}$ 存在且平方为 $2$，这就是 $\sqrt 2$ 的构造。
• $\mathbb Q$ 不是完备的：同一集合在 $\mathbb Q$ 中没有最小上界。
• 阿基米德性质：自然数无上界；它由完备性推出，而不是假设。
• 有理数在实数中稠密：任意开区间 $(x,y)$ 内都有有理数。

理解检验

1. 证明 $\sqrt 2$ 不是有理数时，为什么可以先假设 $m$ 与 $n$ 互素？这一步在证明中起什么作用？

参考答案：有理数的表示不唯一，$m/n$ 与 $(dm)/(dn)$ 表示同一个数。若 $m,n$ 有公因子 $d>1$，就同除以 $d$，反复进行直到互素；这不改变所表示的有理数，所以「存在有理数 $x$ 使 $x^2=2$」等价于「存在互素的 $m,n$ 使 $(m/n)^2=2$」。它的作用是制造矛盾：论证会推出 $2$ 同时整除 $m$ 和 $n$，这与互素直接冲突。若不做约分，$m,n$ 同乘 $2$ 后仍然满足 $m^2=2n^2$，就得不出矛盾。

2. 在证明 $x^2\le2$ 时，若 $x^2>2$，为什么选 $h<\dfrac{x^2-2}{2x}$ 就能让 $x-h$ 仍是上界？请把逻辑补全。

参考答案：目标是找 $h>0$ 使 $(x-h)^2>2$。展开得 $(x-h)^2=x^2-2xh+h^2\ge x^2-2xh$，所以只要 $x^2-2xh>2$ 即可，即 $h<\dfrac{x^2-2}{2x}$（这里 $x>0$，不等号方向不变）。此时对任意 $a\in A$，有 $a^2<2<(x-h)^2$；又 $a>0$、$x-h>0$，开方保序得 $a<x-h$。所以 $x-h$ 是 $A$ 的上界。但 $x-h<x$，与 $x$ 是最小上界矛盾。还需保证 $x-h>0$，故同时要求 $h<x$，这由取 $\min\{x,(x^2-2)/(2x)\}$ 保证。

3. 在证明 $x^2\ge2$ 时，为什么要额外要求 $h<1$？如果没有这个条件，论证会在哪一步断掉？

参考答案：这一步要证 $x+h\in A$，即 $(x+h)^2<2$。展开为 $x^2+2xh+h^2$，其中 $h^2$ 无法直接控制。条件 $0<h<1$ 给出 $h^2<h$，于是 $(x+h)^2<x^2+2xh+h=x^2+h(1+2x)$，把二次项压成了一次项。再取 $h<\dfrac{2-x^2}{1+2x}$ 就得到 $(x+h)^2<2$。若没有 $h<1$，$h^2<h$ 不成立，就无法把 $h^2$ 吸收进 $h(1+2x)$，估计失效。

4. 阿基米德性质的证明中，为什么「$y-1$ 也是上界」会构成矛盾？

参考答案：$y$ 被定义为 $\mathbb N$ 的最小上界，即任何上界都不小于 $y$。而 $y-1<y$，若它也是上界，就出现了一个比最小上界更小的上界，直接违反最小性。矛盾说明「$\mathbb N$ 有上界」的假设不成立。证明中「$n+1$ 也是自然数」这一步保证了 $n\le y-1$ 对所有 $n$ 成立，从而 $y-1$ 确实是上界。

5. 用稠密性证明：任意两个实数之间有无穷多个有理数。

参考答案：设 $x<y$。稠密性给出有理数 $q_1$ 使 $x<q_1<y$。对区间 $(q_1,y)$ 再用一次稠密性，得有理数 $q_2$ 使 $q_1<q_2<y$。如此反复：已知 $x<q_k<y$，对 $(q_k,y)$ 用稠密性得到 $q_{k+1}$，满足 $q_k<q_{k+1}<y$。每一步都严格递增，因此得到无穷多个互不相同的有理数，全部落在 $(x,y)$ 内。这里只需要稠密性可以反复使用，而不需要选择公理之类的额外工具，因为每次都是在具体的非空开区间上做一次构造。

与前后课的关系

本讲补上了「完备」，$\mathbb R$ 的定义终于完整：完备的有序域。但到目前为止，「极限」还只是直观说法。第三讲先把证明的写法系统化（量词、反证、$\varepsilon$ 论证），第四讲才正式定义数列收敛，并用本讲的完备性去证明单调收敛定理与柯西收敛定理。

可以这样看这条线索：第 1 讲给出公理，第 2 讲补全公理并演示如何从公理造数，第 3–6 讲开始用这些公理去证明关于极限的定理。

What was still missing

The previous lecture prepared an ordered field for the intermediate value theorem but left "complete" undefined. This lecture supplies it and answers a concrete question: what kind of number is $\sqrt 2$?

First, compress the previous lecture into a table:

Transitivity is new here: if $x<y$ and $y<z$ then $x<z$. It is what allows inequalities to be chained, and nearly every later argument uses it.

Timeline

$\sqrt 2$ is not rational

$\sqrt 2$ is the positive number satisfying $x^2=2$. To show it is not in $\mathbb Q$ means showing that no $m\in\mathbb Z$, $n\in\mathbb N$ satisfy $(m/n)^2=2$.

Claim. No rational $x$ satisfies $x^2=2$.

Proof (by contradiction). Suppose $x=m/n$ with $m\in\mathbb Z$, $n\in\mathbb N$ and $x^2=2$.

We may assume $m$ and $n$ share no common factor. Why: a rational has many representations, and if $d>1$ divides both, dividing both by $d$ gives another representation of the same rational. Repeating, we arrive at coprime $m,n$. This reduction is not cosmetic; it is the hinge of the contradiction.

From $m^2/n^2=2$,

$$m^2=2n^2 .$$

The right side is even, so $m^2$ is even. An even square has an even root: if $m=2k+1$ were odd then $m^2=4k^2+4k+1$ would be odd. Hence $m=2m_1$ with $m_1\in\mathbb Z$.

Substituting,

$$(2m_1)^2=2n^2 \implies 4m_1^2=2n^2 \implies 2m_1^2=n^2 .$$

So $n^2$ is even, and by the same argument $n$ is even.

Now $2$ divides both $m$ and $n$, contradicting coprimality. The assumption fails, so $\sqrt 2\notin\mathbb Q$. $\square$

Only two facts are used: cancellation of common factors in a rational, and "an integer whose square is even is even". The latter deserves to be spelled out, since it is part of the proof.

Where $\sqrt 2$ lives: a limit of a sequence

If $\mathbb Q$ does not contain $\sqrt 2$, how do the reals accommodate it?

Look at the decimal expansion:

$$\sqrt 2=1.41421356\ldots$$

Truncating digit by digit gives

$$1,\quad 1.4,\quad 1.41,\quad 1.414,\quad 1.4142,\ \dots$$

Each truncation is rational, and the later ones sit closer and closer to $\sqrt 2$. So one idea is: define $\sqrt 2$ as the limit of this sequence.

That introduces two notions that will occupy much of the course: sequences and limits. "Getting closer" is only intuition; writing it down needs the $\varepsilon$ language of Lecture 4. Here it only illustrates what completeness must accomplish: the rationals have a hole at $\sqrt 2$, and the reals fill it with a limit.

Upper bounds and least upper bounds

Upper bounds

Let $S$ be an ordered set and $A\subseteq S$. An element $M\in S$ is an upper bound for $A$ if

$$a\le M \quad\text{for all } a\in A .$$

$M$ need not belong to $A$. Examples:

• $S=\mathbb N$, $A=\{1,2,3\}$. Then $4$ is an upper bound, and so is $3$; $2$ is not, because $3>2$.
• $S=\mathbb Q$, $A=\mathbb N$. Then $A$ has no upper bound: however large a rational is chosen, some natural number is larger.

Least upper bounds

Suppose $A$ has an upper bound. Then $M$ is the least upper bound (or supremum, written $\sup A$) if

1. $M$ is an upper bound for $A$;
2. for every upper bound $M_1$ of $A$, $M\le M_1$.

In one phrase: $M$ is the smallest upper bound. It plays two roles at once — large enough to dominate every element of $A$, small enough that nothing smaller can.

One detail is easy to miss: the least upper bound need not belong to $A$. For $A=\{x\in\mathbb Q:x<1\}$, $\sup A=1$, but $1\notin A$.

Complete ordered fields

The definition is short:

A complete ordered field is an ordered field in which every bounded subset has a least upper bound.

"Bounded" means bounded above. The lecture notes that defining only upper bounds loses nothing: replace a set by its "mirror image" and upper bounds become lower bounds, so the two notions are equivalent and stating both would be redundant. Concretely, lower bounds and infima come from applying suprema to $-A=\{-a:a\in A\}$.

Next comes a theorem that is not proved here:

Theorem. There exists a smallest complete ordered field containing the rationals $\mathbb Q$.

That field is the real numbers $\mathbb R$. Proving existence requires building $\mathbb R$ from $\mathbb Q$ (Dedekind cuts, or Cauchy sequences), which lies outside the course; the lecture takes it as a starting point.

With this theorem the intermediate value theorem finally has the property it needs. But does the definition of completeness really deliver what we want? The lecture tests it on $\sqrt 2$.

Constructing $\sqrt 2$ in $\mathbb R$

The plan: do not solve an equation. Define $\sqrt 2$ as a least upper bound, then verify that this bound really squares to $2$.

The set $A$

$$A=\{\,x\in\mathbb R : x>0,\ x^2<2\,\}.$$

Check that the definition applies:

• Nonempty: $1\in A$, since $1>0$ and $1^2=1<2$.
• Bounded above: if $x\in A$ then $x>0$ and $x^2<2$. Since $2^2=4>2$, necessarily $x<2$. Indeed $x\ge2>0$ would force $x^2\ge4>2$. So $2$ is an upper bound.

By completeness, $\sup A$ exists. Set

$$x=\sup A .$$

The lecture deliberately avoids writing $\sqrt 2$ during the proof: using that name would tempt one to invoke "$x^2=2$" before it has been proved.

Since $1\in A$ and $x$ is an upper bound, $1\le x$, so $x>0$.

It remains to prove $x^2=2$, in two inequalities.

Step one: $x^2\le 2$

Suppose instead that $x^2>2$. Then $x^2-2>0$, so we may choose a small positive $h$ with

$$0<h<\min\left\{x,\ \frac{x^2-2}{2x}\right\}.$$

(The right side is positive because $x>0$, so such an $h$ exists.) Compute

$$(x-h)^2=x^2-2xh+h^2\ge x^2-2xh .$$

Since $h<\dfrac{x^2-2}{2x}$, multiplying by $2x>0$ gives $2xh<x^2-2$, i.e. $x^2-2xh>2$. Hence

$$(x-h)^2>2 .$$

Now take any $a\in A$. Because $a>0$, $x-h>0$ (since $h<x$), and $a^2<2<(x-h)^2$, taking square roots preserves the order and yields $a<x-h$. So $x-h$ is an upper bound for $A$, and $x-h<x$, contradicting minimality of $x$.

Therefore $x^2\le2$.

Step two: $x^2\ge2$

Suppose instead that $x^2<2$. Then $2-x^2>0$. Choose $h$ with

$$0<h<\min\left\{1,\ \frac{2-x^2}{1+2x}\right\}.$$

Compute

$$(x+h)^2=x^2+2xh+h^2 .$$

Since $0<h<1$, we have $h^2<h$, so

$$(x+h)^2<x^2+2xh+h=x^2+h(1+2x).$$

From $h(1+2x)<2-x^2$,

$$(x+h)^2<x^2+(2-x^2)=2 .$$

Also $x+h>0$, hence $x+h\in A$. But $x+h>x$, contradicting that $x$ is an upper bound.

Therefore $x^2\ge2$.

Together, $x^2=2$. So $\sup A$ is $\sqrt 2$: the number genuinely exists in $\mathbb R$.

The device is worth remembering: to prove that a supremum satisfies an equation, argue twice by contradiction — if it is too large, shift left a little and it still dominates the set; if too small, shift right a little and it lands inside the set. The size of the shift is governed by the gap between $x^2$ and $2$.

Corollary: $\mathbb Q$ is not complete

Corollary. The rationals $\mathbb Q$ do not form a complete ordered field.

Proof. Let $F$ be any complete ordered field containing $\mathbb Q$. The construction above uses only field operations, order and completeness, so $\sqrt2=\sup\{x\in F:x>0,x^2<2\}$ must lie in $F$. But the first part of the lecture showed $\sqrt2\notin\mathbb Q$. If $\mathbb Q$ were complete, taking $F=\mathbb Q$ would be a contradiction. $\square$

Concretely, consider $A_{\mathbb Q}=\{x\in\mathbb Q:x>0,x^2<2\}$. It has upper bounds in $\mathbb Q$ (for instance $2$) but no least upper bound. Indeed, let $q\in\mathbb Q$ be any upper bound. Since $1\in A_{\mathbb Q}$, $q\ge1>0$. If $q^2<2$ then $q\in A_{\mathbb Q}$, contradicting that $q$ bounds the set; if $q^2=2$ then $q$ is a rational square root of $2$, which is impossible. Hence $q^2>2$.

Define

$$q'=\frac{2/q+q}{2}=\frac{q^2+2}{2q}.$$

Then $q'$ is rational and positive. Compare:

$$q'<q \iff q^2+2<2q^2 \iff 2<q^2,$$

which holds. Is $q'$ an upper bound? For any $x\in A_{\mathbb Q}$ we have $x^2<2$, and

$$(q')^2=\frac{(q^2+2)^2}{4q^2}>2 \iff (q^2-2)^2>0,$$

true because $q^2>2$. Hence $x^2<2<(q')^2$, so $x<q'$.

Every rational upper bound admits a smaller rational upper bound, so no least upper bound exists. This is the precise sense in which $\mathbb Q$ has a gap — and the gap sits exactly at $\sqrt2$.

The Archimedean property

Theorem (Archimedean property). For every real $x$ there is a natural number $n$ with $n>x$.

It looks obvious — the naturals are unbounded. But "obvious" is precisely what completeness has to prove.

Proof (by contradiction). Suppose some real $x$ satisfies $n\le x$ for all $n\in\mathbb N$. Then $x$ bounds $\mathbb N$ above, so $\mathbb N$, viewed as a subset of $\mathbb R$, is bounded above.

By completeness, let

$$y=\sup\mathbb N .$$

As an upper bound, $y\ge n$ for every $n\in\mathbb N$. But $n+1$ is also natural, so

$$n+1\le y \implies n\le y-1 \quad\text{for all } n .$$

Thus $y-1$ is also an upper bound for $\mathbb N$. Yet $y-1<y$, contradicting that $y$ is the least upper bound. $\square$

The proof is short but uses three things: closure of $\mathbb N$ under $+1$, transitivity of order, and minimality of the least upper bound. The lecture stresses that arguments of this kind are the basic craft of analysis.

Corollary: the rationals are dense in the reals

Corollary. If $x<y$ are real, there is a rational $m/n$ with

$$x<\frac mn<y .$$

Proof. Let $\beta=y-x>0$. By the Archimedean property there is $n\in\mathbb N$ with $n>1/\beta$, equivalently

$$\frac1n<\beta .$$

Now take the smallest integer $m$ with $m/n>x$. Such an $m$ exists: by the Archimedean property there is an integer $k>nx$, so $k/n>x$ and the set of admissible integers is nonempty; since it is bounded below, it has a least element.

By minimality of $m$, the integer $m-1$ does not satisfy the condition, so

$$\frac{m-1}{n}\le x .$$

Therefore

$$\frac mn=\frac{m-1}{n}+\frac1n\le x+\frac1n<x+\beta=y .$$

With $m/n>x$, this gives $x<m/n<y$. $\square$

Geometrically: choose a ruler of length $1/n$, shorter than the distance from $x$ to $y$. Walk rightward from $x$ in steps of that length; the first grid point past $x$ cannot yet have passed $y$.

What this lecture establishes

• $\sqrt2\notin\mathbb Q$; the pivot of the contradiction is reducing to coprime $m,n$.
• Completeness = every bounded subset has a least upper bound (supremum).
• $\mathbb R$ is the smallest complete ordered field, and it contains $\mathbb Q$.
• In $\mathbb R$, $\sup\{x>0:x^2<2\}$ exists and squares to $2$; that is the construction of $\sqrt2$.
• $\mathbb Q$ is not complete: the same set has no least upper bound among the rationals.
• The Archimedean property — the naturals are unbounded — follows from completeness rather than being assumed.
• The rationals are dense in the reals: every open interval $(x,y)$ contains a rational.

Comprehension check

1. When proving $\sqrt2$ irrational, why may we assume $m$ and $n$ are coprime? What role does it play?

Reference answer: A rational has many representations, since $m/n=(dm)/(dn)$. If $d>1$ divides both, divide both by $d$ and repeat until coprime; this does not change the rational, so "there is a rational $x$ with $x^2=2$" is equivalent to "there are coprime $m,n$ with $(m/n)^2=2$". The role is to create a contradiction: the argument forces $2$ to divide both $m$ and $n$, which directly contradicts coprimality. Without the reduction, multiplying a solution by $2$ in numerator and denominator still satisfies $m^2=2n^2$, so no contradiction would arise.

2. In the proof of $x^2\le2$, why does $h<(x^2-2)/(2x)$ make $x-h$ an upper bound? Complete the logic.

Reference answer: The goal is $h>0$ with $(x-h)^2>2$. Expanding, $(x-h)^2=x^2-2xh+h^2\ge x^2-2xh$, so it suffices that $x^2-2xh>2$, i.e. $h<(x^2-2)/(2x)$ (the direction is unchanged because $x>0$). Then for any $a\in A$, $a^2<2<(x-h)^2$; with $a>0$ and $x-h>0$, taking roots gives $a<x-h$. So $x-h$ bounds $A$ above. But $x-h<x$, contradicting minimality of $x$. One must also keep $x-h>0$, i.e. $h<x$, which the choice of $\min\{x,(x^2-2)/(2x)\}$ guarantees.

3. In the proof of $x^2\ge2$, why require $h<1$? Where would the argument break without it?

Reference answer: The goal is $x+h\in A$, i.e. $(x+h)^2<2$. Expanding gives $x^2+2xh+h^2$, and $h^2$ is not directly controlled. The condition $0<h<1$ gives $h^2<h$, hence $(x+h)^2<x^2+2xh+h=x^2+h(1+2x)$, replacing a quadratic term by a linear one. Choosing $h<(2-x^2)/(1+2x)$ then yields $(x+h)^2<2$. Without $h<1$ the inequality $h^2<h$ fails and $h^2$ cannot be absorbed into $h(1+2x)$, so the estimate collapses.

4. In the Archimedean proof, why does "$y-1$ is also an upper bound" contradict something?

Reference answer: $y$ is defined as the least upper bound of $\mathbb N$, so no upper bound is smaller than $y$. But $y-1<y$; if it were an upper bound there would be an upper bound smaller than the least one, violating minimality. The contradiction refutes the assumption that $\mathbb N$ is bounded above. The step "$n+1$ is natural" is what makes $n\le y-1$ hold for every $n$, so that $y-1$ really is an upper bound.

5. Use density to show that between any two reals there are infinitely many rationals.

Reference answer: Let $x<y$. Density gives a rational $q_1$ with $x<q_1<y$. Apply density again to $(q_1,y)$ to get a rational $q_2$ with $q_1<q_2<y$. Inductively, given $x<q_k<y$, density applied to $(q_k,y)$ produces $q_{k+1}$ with $q_k<q_{k+1}<y$. The sequence is strictly increasing, so the $q_k$ are distinct, and all lie in $(x,y)$. Only repeated use of density on concrete nonempty open intervals is needed — no extra set-theoretic tools.

Relation to neighbouring lectures

This lecture completes the definition of $\mathbb R$ as a complete ordered field. Limits, however, are still only intuitive. Lecture 3 systematises proof writing — quantifiers, contradiction, $\varepsilon$-arguments — and Lecture 4 finally defines convergence, after which completeness is used to prove the monotone convergence theorem and the Cauchy criterion.

The thread so far: Lecture 1 states the axioms, Lecture 2 completes them and shows how to build numbers from them, and Lectures 3–6 begin proving limit theorems with those axioms.