The Gaps in the Rationals

The irrationality of $\sqrt2$ is the prototype for an entire branch of number theory studying which numbers are rational, irrational, or transcendental — questions central to Diophantine approximation and to modern results like the proofs that $\pi$ and $e$ are irrational (and transcendental). The Pell equation $p^2 - 2q^2 = \pm1$ that appeared in the second worked example reappears in algorithms for computing square roots efficiently, in continued-fraction based cryptographic attacks (e.g. on RSA via continued fraction factorization methods), and in the theory of quadratic number fields $\mathbb{Q}(\sqrt2)$ used throughout algebraic number theory. More broadly, the structural lesson — that an ordered field can fail to have the least-upper-bound property — is the entire reason real analysis exists as a distinct subject from algebra. Every theorem that depends on completeness (the Intermediate Value Theorem, the Extreme Value Theorem, the convergence of Cauchy sequences, the existence of limits of bounded monotonic sequences) ultimately traces back to filling exactly the kind of gap exhibited by $A = \{q \in \mathbb{Q}: q^2 <2\}$. In computer science and numerical analysis, floating-point arithmetic is itself an approximation scheme reminiscent of the rational approximations $b_n$ in Example 1 — finite-precision surrogates chasing a real target they can never exactly reach, a discretized echo of the ancient gap in $\mathbb{Q}$.

Ask the tutor: How does the least-upper-bound property, once assumed as an axiom for $\mathbb{R}$, actually guarantee that $\sqrt2$ exists as a real number?

Picture the rational numbers as points marked on a number line — not just the integers, but every fraction $1/2, 22/7, -5/3$, and so on. There are infinitely many rationals between any two others: the rationals are dense. It is tempting to think that a line made of infinitely many densely packed points has no room left for gaps. But density is not the same as completeness. Consider the set $A$ of positive rationals $q$ with $q^2 < 2$: this includes $1, 1.4, 1.41, 1.414, 1.4142, \dots$ These numbers creep steadily toward "the square root of $2$," each one a better rational approximation than the last. Geometrically, $\sqrt{2}$ is a perfectly definite point — it is the length of the diagonal of a unit square, constructible with a compass and straightedge. Yet as a number, when we ask "which rational is it?", the answer is: none of them. The sequence of approximations converges toward a location on the line where no rational number sits. The rational line has a pinhole exactly there, invisible under magnification of any finite precision, but real.

The failure is not about a lack of precision — it is a structural fact about $\mathbb{Q}$ itself. This is the seed of everything that follows: to fill such gaps, mathematicians eventually build the real numbers precisely by declaring that every nonempty, bounded-above set of numbers must have a least upper bound. Understanding why $\mathbb{Q}$ fails this property, concretely and rigorously, is the goal of this article.

Example 1: Locating $\sqrt2$ between rationals via the correction map. Let's use the map $b \mapsto b' = \dfrac{b^2+2}{2b}$ from the Lemma's proof to squeeze in on $\sqrt 2$ starting from a deliberately poor, asymmetric guess $b_0 = 3$ (not the "nice" guess $1.5$).

• $b_0 = 3$: check $b_0^2 = 9 > 2$. Compute

• $b_1 = \frac{11}{6}$: check $b_1^2 = \frac{121}{36} \approx 3.361 > 2$. Compute

• $b_2 = \frac{193}{132}$: check $b_2^2 = \frac{37249}{17424} \approx 2.1381 > 2$. Compute

Since $17424 = 264\cdot 66$, this simplifies to $\dfrac{72097}{264\cdot 193} = \dfrac{72097}{50952} \approx 1.41508$. Each $b_n$ is rational (an explicit ratio of integers, confirmed above), each satisfies $b_n^2 > 2$ by Step 2 of the Lemma, and the sequence is strictly decreasing by Step 1. Numerically $b_3 \approx 1.41508$ is already within $0.0009$ of $\sqrt2 \approx 1.41421$. This is a concrete demonstration: we can get arbitrarily rationally close to $\sqrt2$, but by the Theorem, we never land on it — the target $x^2=2$ is never hit exactly by any $b_n$, consistent with $\sqrt2 \notin \mathbb{Q}$.

Example 2: A rational candidate that "almost" works, checked exactly. A common near-miss is $x = \frac{7}{5}$, since $\left(\frac{7}{5}\right)^2 = \frac{49}{25} = 1.96$, tantalizingly close to $2$. Let's verify precisely why this — and every rational — must fail, using the exact machinery of the Theorem rather than decimal intuition. Suppose someone proposes $x = \frac{99}{70}$ (a classical convergent of $\sqrt2$'s continued fraction) as an exact square root of $2$. Compute:

Is $\frac{9801}{4900} = 2$? Cross-multiply: this would require $9801 = 2 \cdot 4900 = 9800$. But $9801 \neq 9800$ — off by exactly $1$. So $x=\frac{99}{70}$ fails, missing by the smallest possible margin at this scale, since $|9801 - 9800|=1$. This is not a coincidence: the proof of the Theorem shows no fraction $p/q$ in lowest terms can satisfy $p^2 = 2q^2$ exactly, because such an equation forces both $p$ and $q$ to be even, violating lowest terms. The near-misses like $\frac{99}{70}$ (with $p^2 - 2q^2 = \pm1$) are solutions to the Pell equation $p^2 - 2q^2 = \pm 1$, which has infinitely many integer solutions — explaining why arbitrarily good rational approximations to $\sqrt2$ exist even though an exact one never does.