Why ℝ Has No Holes

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 $|x_5-\sqrt2| = \frac{e_5}{x_5+\sqrt2} < \frac{2.7\times 10^{-24}}{2.8} < 10^{-24}$.

$b_n = \left(\frac{n+1}{n}\right)^n$ 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 $e$.

Identification of the two $e$'s first.

Write $S = \sum_{k=0}^{\infty} \frac{1}{k!}$ and expand $b_n$ by the binomial theorem as in the eseries scene (compound problem): $b_n = \sum_{k=0}^{n} a_{n,k}$ with $a_{n,k} = \frac{1}{k!}\prod_{j=0}^{k-1}(1 - \frac{j}{n}) \le \frac{1}{k!}$.

Fix $K \le n$: then $b_n \ge \sum_{k=0}^{K} a_{n,k}$, and $a_{n,k} \to \frac{1}{k!}$ as $n \to \infty$ for each fixed $k$, so $\liminf b_n \ge \sum_{k=0}^{K} \frac{1}{k!}$.

Also $b_n \le \sum_{k=0}^{n} \frac{1}{k!} \le S$, so $\limsup b_n \le S$.

Letting $K \to \infty$ forces $\lim b_n = S$: the $e$ defined here as $\lim b_n$ is the series number $S$.

With the two definitions identified, 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 $\sqrt2$.

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.

Let $\varepsilon > 0$.

Since $x_n \to \sqrt2$, choose $N$ with $|x_n - \sqrt2| < \varepsilon/2$ for all $n \ge N$.

Then for all $m, n \ge N$, $|x_m - x_n| \leq |x_m - \sqrt2| + |x_n - \sqrt2| < \varepsilon/2 + \varepsilon/2 = \varepsilon$.

Hence $(x_n)$ 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.