Bounds for rational approximations to square roots

A consequence of Dirichlet’s approximation theorem is that for any real number $x\in \mathbb{R}$, there exist infinitely many rational approximations $\frac{p}{q}\in \mathbb{Q}$ that satisfy

In particular, the convergents of the continued fraction for $x$ satisfy this inequality. Hurwitz’s theorem gives a slightly tighter bound and states that there exist infinitely many rational approximations $\frac{p}{q}\in \mathbb{Q}$ that satisfy

Surprisingly, this bound is tight, since for the golden ratio $\varphi =\frac{1+\sqrt{5}}{2}$ since for any $c<\frac{1}{\sqrt{5}}$ there exist only finitely many solutions $\frac{p}{q}\in \mathbb{Q}$ to $|\varphi -\frac{p}{q}|<c\cdot \frac{1}{q^2}$.

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 $\sqrt{2}$: any rational approximation $\frac{p}{q}\in \mathbb{Q}$ to $\sqrt{2}$ satisfies

He proves this by supposing that $|\sqrt{2}-\frac{p}{q}|\le \frac{1}{4}\cdot \frac{1}{q^2}$ and deriving a contradiction. Multiplying both sides with $\sqrt{2}+\frac{p}{q}$ gives

where the last inequality follows from

Finally, multiplying by $q^2$ gives

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

This bound can obviously be improved, since there are places where inequalities are used that are not tight. For example, we use $2\sqrt{2}+\frac{1}{4}<4$, while $2\sqrt{2}+\frac{1}{4}\approx 3.078$ is much closer to $3$ than to $4$.

We can derive a tighter bound by supposing $|\sqrt{2}-\frac{p}{q}|<ϵ\cdot \frac{1}{q^2}$ for some $ϵ>0$ and following the same steps until. We end up with

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

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

Theorem: Let $n\in \mathbb{N}$ be a natural number that is not a square. Then any $\frac{p}{q}\in \mathbb{Q}$ satisfies

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

Proof: Set $ϵ=\sqrt{n+1}-\sqrt{n}$ and suppose there exists a $\frac{p}{q}\in \mathbb{Q}$ with

Multiply both sides by $\sqrt{n}+\frac{p}{q}=2\sqrt{n}-(\sqrt{n}-\frac{p}{q})<2\sqrt{n}+ϵ$ to get

If we substitute $ϵ=\sqrt{n+1}-\sqrt{n}$ in ${ϵ}^2+2\sqrt{n}ϵ$ and simplify we see ${ϵ}^2+2\sqrt{n}ϵ=1$. So the right side of this equation equals $\frac{1}{q^2}$. Multiplying both sides by $q^2$ gives us

Like before, this is a contradiction: it implies that $|n{q}^2-p^2|=0$, so $|n-\frac{p^2}{q^2}|=0$, so $\frac{p^2}{q^2}=n$, so $\frac{p}{q}=\sqrt{n}$. However, by assumption $n$ is not a square, so $\sqrt{n}$ is irrational. So this is a contradiction.
$\square$

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

From a high-level perspective we have used the expression

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

• For any $x\in \mathbb{R}$ there are infinitely many $\frac{p}{q}\in \mathbb{Q}$ with $e_x(\frac{p}{q})<\frac{1}{\sqrt{5}}$.
• For any natural number $n\in \mathbb{N}$ that is not a square root we have $e_{\sqrt{n}}(\frac{p}{q})>\sqrt{n+1}-\sqrt{n}$