Why ℝ Has No Holes

Every term of

x_{n+1}=\frac{x_n+2/x_n}{2}

is rational, yet the limit

\sqrt{2}

is not. The rationals

\mathbb{Q}

have a hole — shown at the idea level, without a full irrationality proof.

Examples & Exercises

With $x_0 = 1$ and $x_{n+1} = \frac{x_n + 2/x_n}{2}$ , write $x_5$ as an exact fraction and prove $|x_5 - \sqrt{2}| < 10^{-24}$ .

x_5 = \frac{886731088897}{627013566048}

|x_5 - \sqrt{2}| < 10^{-24}

n = 5

— five iterations get 24 decimal places right.

Show proof

Computing in fractions gives $x_1 = \frac{3}{2}$ , $x_2 = \frac{17}{12}$ , $x_3 = \frac{577}{408}$ , $x_4 = \frac{665857}{470832}$ , $x_5 = \frac{886731088897}{627013566048}$ , by applying $\frac{p}{q} \mapsto \frac{p^2+2q^2}{2pq}$ at each step.
Moreover $p_{n+1}^2 - 2q_{n+1}^2 = (p_n^2+2q_n^2)^2 - 8p_n^2q_n^2 = (p_n^2-2q_n^2)^2$ , and $p_1^2-2q_1^2 = 9-8 = 1$ , so $p_n^2-2q_n^2 = 1$ for all $n \geq 1$ . Hence $\gcd(p_n,q_n)=1$ — every fraction is in lowest terms.
Put $e_n = x_n^2 - 2$ . Expanding gives $e_{n+1} = \frac{e_n^2}{4x_n^2}$ . Since $x_1 = \frac32 > \sqrt2$ and $x_{n+1}-\sqrt2 = \frac{(x_n-\sqrt2)^2}{2x_n} \geq 0$ , we have $x_n \geq \sqrt2$ for all $n \geq 1$ , hence $e_{n+1} \leq \frac{e_n^2}{8}$ .
We get $e_1 = \frac14$ , $e_2 = \frac{1}{144}$ , $e_3 = \frac{1}{166464}$ , $e_4 = \frac{1}{221682772224} < 4.6\times 10^{-12}$ . Hence $e_5 \leq \frac{e_4^2}{8} < 2.7\times 10^{-24}$ , and $|x_5-\sqrt2| = \frac{e_5}{x_5+\sqrt2} < \frac{2.7\times 10^{-24}}{2.8} < 10^{-24}$ .

Show that every term of $b_n = \left(1+\frac1n\right)^n$ is rational, and explain why its irrational limit $e$ is another "hole" in $\mathbb{Q}$ .

b_n = \left(\frac{n+1}{n}\right)^n \in \mathbb{Q}, \qquad \lim_{n\to\infty} b_n = e \notin \mathbb{Q}

Babylonian iterates at

n=5

— the same phenomenon: rational terms homing in on an irrational limit.

Show proof

$b_n = \left(\frac{n+1}{n}\right)^n$ with $\frac{n+1}{n}$ rational. A rational number raised to a positive integer power is rational, so $b_n \in \mathbb{Q}$ for every $n$ .
The sequence $(b_n)$ is increasing and bounded above (taken as known — it is standard), hence convergent. Its limit is called $e$ .
That $e \notin \mathbb{Q}$ is an established theorem (proved, for instance, by estimating the remainder of $e = \sum_{k=0}^{\infty}\frac{1}{k!}$ ). So the rational sequence $(b_n)$ converges to no rational number — another hole in $\mathbb{Q}$ , distinct from the one at $\sqrt2$ .

Prove that the Babylonian sequence $(x_n)$ is Cauchy but has no limit in $\mathbb{Q}$ — this is what completeness is about.

|x_m - x_n| \to 0 \ \ (m,n \to \infty), \qquad \nexists\, r \in \mathbb{Q} : x_n \to r

n=8

— the terms get arbitrarily close to each other, yet no rational number is their destination.

Show proof

From the error analysis, $e_{n+1} \leq \frac{e_n^2}{8}$ with $e_1 = \frac14$ . Since $e_n \leq 2^{-n}$ by induction (true at $n=1$ , and $e_{n+1} \leq \frac{2^{-2n}}{8} = 2^{-(2n+3)} \leq 2^{-(n+1)}$ ), we get $e_n \to 0$ . With $|x_n-\sqrt2| = \frac{e_n}{x_n+\sqrt2} \leq \frac{e_n}{2}$ , we have $x_n \to \sqrt2$ as real numbers.
Hence $|x_m - x_n| \leq |x_m - \sqrt2| + |x_n - \sqrt2|$ becomes arbitrarily small for large $m, n$ : $(x_n)$ is a Cauchy sequence.
If $x_n \to r$ for some rational $r$ , uniqueness of limits forces $r = \sqrt2$ . But $\sqrt2 \notin \mathbb{Q}$ — a contradiction. So $(x_n)$ has no limit inside $\mathbb{Q}$ .
For reference, the classical proof that $\sqrt2 \notin \mathbb{Q}$ : if $\sqrt2 = \frac pq$ in lowest terms, then $p^2 = 2q^2$ , so $p$ is even; writing $p = 2k$ gives $q^2 = 2k^2$ , so $q$ is even too — contradicting lowest terms.
$\sqrt2 = \frac pq \ \Rightarrow\ p^2 = 2q^2 \ \Rightarrow\ p, q \text{ both even, contradiction}$