Bounds for rational approximations to square roots

A consequence of Dirichlet’s approximation theorem is that for any real number x∈Rx \in \mathbb{R}, there exist infinitely many rational approximations pq∈Q\frac{p}{q} \in \mathbb{Q} that satisfy
∣x−pq∣<1q2 |x - \frac{p}{q}| < \frac{1}{q^2}

In particular, the convergents of the continued fraction for xx satisfy this inequality. Hurwitz’s theorem gives a slightly tighter bound and states that there exist infinitely many rational approximations pq∈Q\frac{p}{q} \in \mathbb{Q} that satisfy
∣x−pq∣<15⋅1q2 |x - \frac{p}{q}| < \frac{1}{\sqrt{5}} \cdot \frac{1}{q^2}

Surprisingly, this bound is tight, since for the golden ratio ϕ=1+52\phi = \frac{1 + \sqrt{5}}{2} since for any c<15c < \frac{1}{\sqrt{5}} there exist only finitely many solutions pq∈Q\frac{p}{q} \in \mathbb{Q} to ∣ϕ−pq∣<c⋅1q2|\phi - \frac{p}{q}| < c \cdot \frac{1}{q^2}.

It turns out that in general, there are bounds on how well rational numbers can approximate algebraic numbers. This is the basis of Liouville’s theorem, which was used to construct the first known transcedental numbers.

In Frits Beukers’ “Getaltheorie voor Beginners”, he briefly mentions an explicit bound for 2\sqrt{2}: any rational approximation pq∈Q\frac{p}{q} \in \mathbb{Q} to 2\sqrt{2} satisfies
∣2−pq∣>14⋅1q2 |\sqrt{2} - \frac{p}{q}| > \frac{1}{4} \cdot \frac{1}{q^2}

He proves this by supposing that ∣2−pq∣≤14⋅1q2|\sqrt{2} - \frac{p}{q}| \leq \frac{1}{4} \cdot \frac{1}{q^2} and deriving a contradiction. Multiplying both sides with 2+pq\sqrt{2} + \frac{p}{q} gives
∣2−p2q2∣≤2+pq4⋅1q2<1q2 |2 - \frac{p^2}{q^2}| \leq \frac{\sqrt{2} + \frac{p}{q}}{4} \cdot \frac{1}{q^2} < \frac{1}{q^2}

where the last inequality follows from
2+pq=22−(2−pq)<22+14q2≤22+14<4 \sqrt{2} + \frac{p}{q} = 2 \sqrt{2} - (\sqrt{2} - \frac{p}{q}) < 2 \sqrt{2} + \frac{1}{4q^2} \leq 2 \sqrt{2} + \frac{1}{4} < 4

Finally, multiplying by q2q^2 gives
∣2q2−p2∣<1 | 2q^2 - p^2 | < 1

Now pp and qq are integers, so the left side is integer. The only integer with absolute value less than one is zero. However, this implies that q2p2=2\frac{q^2}{p^2} = 2 so pq=2\frac{p}{q} = \sqrt{2}, which is a contradiction since 2\sqrt{2} is irrational.

This bound can obviously be improved, since there are places where inequalities are used that are not tight. For example, we use 22+14<42 \sqrt{2} + \frac{1}{4} < 4, while 22+14≈3.0782 \sqrt{2} + \frac{1}{4} \approx 3.078 is much closer to 33 than to 44.

We can derive a tighter bound by supposing ∣2−pq∣<ϵ⋅1q2| \sqrt{2} - \frac{p}{q} | < \epsilon \cdot \frac{1}{q^2} for some ϵ>0\epsilon > 0 and following the same steps until. We end up with
∣2q2−p2∣<ϵ(22+ϵ) | 2q^2 - p^2 | < \epsilon(2 \sqrt{2} + \epsilon)

Now, we want to pick ϵ\epsilon so that the right hand side becomes one. So we solve ϵ(22+ϵ)=1\epsilon (2 \sqrt{2} + \epsilon) = 1 and find the positive solution ϵ=3−2\epsilon = \sqrt{3} - \sqrt{2}. By this, we have found the tighter bound ∣2−pq∣>(3−2)⋅1q2=12+3⋅1q2|\sqrt{2} - \frac{p}{q}| > (\sqrt{3} - \sqrt{2}) \cdot \frac{1}{q^2} = \frac{1}{\sqrt{2} + \sqrt{3}} \cdot \frac{1}{q^2}.

In fact, we can generalize this for any irrational n\sqrt{n}:

Theorem: Let n∈Nn \in \mathbb{N} be a natural number that is not a square. Then any pq∈Q\frac{p}{q} \in \mathbb{Q} satisfies
∣n−pq∣>cn⋅1q2 | \sqrt{n} - \frac{p}{q} | > c_n \cdot \frac{1}{q^2}

with cn=n+1−n=1n+n+1c_n = \sqrt{n + 1} - \sqrt{n} = \frac{1}{\sqrt{n} + \sqrt{n + 1}}.

Proof: Set ϵ=n+1−n\epsilon = \sqrt{n + 1} - \sqrt{n} and suppose there exists a pq∈Q\frac{p}{q} \in \mathbb{Q} with
∣n−pq∣<ϵ⋅1q2 | \sqrt{n} - \frac{p}{q} | < \epsilon \cdot \frac{1}{q^2}

Multiply both sides by n+pq=2n−(n−pq)<2n+ϵ\sqrt{n} + \frac{p}{q} = 2 \sqrt{n} - (\sqrt{n} - \frac{p}{q}) < 2 \sqrt{n} + \epsilon to get
∣n−p2q2∣<ϵ2+2nϵq2 | n - \frac{p^2}{q^2} | < \frac{\epsilon^2 + 2\sqrt{n} \epsilon}{q^2}

If we substitute ϵ=n+1−n\epsilon = \sqrt{n + 1} - \sqrt{n} in ϵ2+2nϵ\epsilon^2 + 2\sqrt{n} \epsilon and simplify we see ϵ2+2nϵ=1\epsilon^2 + 2\sqrt{n} \epsilon = 1. So the right side of this equation equals 1q2\frac{1}{q^2}. Multiplying both sides by q2q^2 gives us
∣nq2−p2∣<1 |nq^2 - p^2| < 1

Like before, this is a contradiction: it implies that ∣nq2−p2∣=0|nq^2 - p^2| = 0, so ∣n−p2q2∣=0|n - \frac{p^2}{q^2}| = 0, so p2q2=n\frac{p^2}{q^2} = n, so pq=n\frac{p}{q} = \sqrt{n}. However, by assumption nn is not a square, so n\sqrt{n} is irrational. So this is a contradiction.
□\square

I wonder if this can be generalized to nnth roots and numbers of the form a+bnc\frac{a + b \sqrt{n}}{c}.

From a high-level perspective we have used the expression
ex(pq)=q2∣x−pq∣ e_x(\frac{p}{q}) = q^2 |x - \frac{p}{q}|

to measure how well a rational number pq∈Q\frac{p}{q} \in \mathbb{Q} approximates a real number. We can now rephrase some results in terms of exe_x: