Bound on difference between irrational square root and rational number

Solution 1:

I suppose that $d\in \mathbb{N}$, $d\geq 2$. Let $p/q$ a positive rational, $p,q\in \mathbb{N}$.

Suppose first that $\displaystyle |\sqrt{d}-p/q|\leq 1$. Then we get $\displaystyle p/q\leq \sqrt{d}+1$, and we have $\displaystyle \sqrt{d}+p/q\leq 2\sqrt{d}+1$.

Now:

$$|\sqrt{d}-p/q|=\frac{|q^2d-p^2|}{q^2(\sqrt{d}+p/q)}\geq \frac{|q^2d-p^2|}{q^2(2\sqrt{d}+1)} $$

Now $q^2 d-p^2 $ is not $0$, in $\mathbb{Z}$, hence $\displaystyle |q^2 d-p^2|\geq 1$, and $\displaystyle |\sqrt{d}-p/q|\geq \frac{1}{q^2(2\sqrt{d}+1)} $

If $\displaystyle |\sqrt{d}-p/q|\geq 1$, then obviously $\displaystyle |\sqrt{d}-p/q|\geq \frac{1}{q^2(2\sqrt{d}+1)}$, so this inequality is true for all $p,q$.

Solution 2:

$\sqrt2$ Your problem is part of a glorious history in mathematics. This begins with Liouville, who showed in 1844 that for every irrational algebraic $\alpha$ of degree $\ge 2$, there is a constant $c=c(\alpha)\gt 0$ such that $|\alpha-\frac {p}{q}|>$ $\frac{c}{q^n}$ for all rational $\frac{p}{q}\space (q\gt 0)$ (your post is the particular case $n=2$ and you can find the proof for n in many books of diophantine approximation).

His theorem allows Liouville to find the first known transcendental number and states, in other words, that the measure of irrationality of the irrational algebraic $\alpha$ cannot be greater than its degree.

This measure of irrationality of $\alpha$ it's pretty smaller but it had to wait until 1955 to be known, grace to Klaus Friedrich Roth whose work it is a pinnacle of mathematics.

The exponent $n$ of Liouville was improved on several important occasions which we summarize below.

Joseph Liouville (1809-1882): exponent n (1844).

Axel Thue (1863-1922): exponent $\frac {n}{2} +1$ (1909).

Carl Ludwig Siegel (1896-1981): exponent $2\sqrt n$ (1929).

Freeman John Dyson (1923- ): exponent $\sqrt{2n}$ (1947).

Klaus Friedrich Roth (1925- ): exponent 2 (1955). The best exponent. A deep result which earned its author the Fields Medal.

"The achievement is one that speaks for itself: it closes a chapter, and a new chapter is now opened. Roth’s theorem settles a question which is both of a fundamental nature and of extreme difficulty. It will stand as a landmark in mathematics for as long as mathematics is cultivated" (Harold Davenport’speech on the presentation to the Fields Medal at the International Congress of Edinburgh, 1958).

Solution 3:

Let $ d $ be a positive integer which isn't a square, and consider a rational $ \frac{p}{q} $ ($ p, q $ integers, $ q > 0 $).

Looking at $ \left| \sqrt{d} - \frac{p}{q} \right| $, it is $ \dfrac{1}{q} |q\sqrt{d} - p | $ $ = \dfrac{1}{q} \dfrac{|q^2 d - p^2|}{|q\sqrt{d}+p|} $. As $ \sqrt{d} $ is irrational, numerator $ |q^2 d - p^2 | $ is non-zero (and an integer), making $ | q^2 d - p^2 | \geq 1 $.

So $ \left| \sqrt{d} - \dfrac{p}{q} \right| \geq \dfrac{1}{q^2} \dfrac{1}{\left| \sqrt{d} + \frac{p}{q} \right|}$. Notice denominator $$\left| \sqrt{d} + \frac{p}{q} \right|= \left| 2\sqrt{d} -\sqrt{d} + \frac{p}{q} \right|\leq 2\sqrt{d} + \left| -\sqrt{d} + \frac{p}{q} \right|,$$ giving us

$$ \left| \sqrt{d} - \dfrac{p}{q} \right| \geq \dfrac{1}{q^2} \left(\dfrac{1}{2\sqrt{d} + \left| \sqrt{d} - \frac{p}{q} \right|} \right).$$

Writing $ t := \left| \sqrt{d} - \dfrac{p}{q} \right| $, this is $ t \geq \dfrac{1}{q^2 (2\sqrt{d} + t)} $. So $ q^2 t^2 + q^2 2\sqrt{d} t - 1 \geq 0 $.

Quadratic on LHS has roots $ \dfrac{-q \sqrt{d} \pm \sqrt{ q^2 d + 1 }}{q} $, so the inequality becomes $$ \left(t-\frac{-q\sqrt{d} - \sqrt{q^2 d+1}}{q} \right) \left( t - \frac{-q\sqrt{d}+\sqrt{q^2 d+1}}{q} \right) \geq 0.$$

The first factor is anyways $ > 0 $, giving $ t \geq \dfrac{-q\sqrt{d} + \sqrt{q^2 d + 1}}{q} $. Now RHS $$ \dfrac{-q\sqrt{d} + \sqrt{q^2 d + 1}}{q} = \dfrac{1}{q(q\sqrt{ d} + \sqrt{q^2 d + 1})}=\dfrac{1}{q^2\left(\sqrt{d} + \sqrt{d + \dfrac{1}{q^2}}\right)}\geq \dfrac{1}{q^2(\sqrt{d} + \sqrt{d+1})}= \dfrac{\sqrt{d+1}-\sqrt{d}}{q^2}$$

So finally, we have

$$ \left| \sqrt{d} - \dfrac{p}{q} \right| \geq \dfrac{\sqrt{d+1}-\sqrt{d}}{q^2} $$

for any integers $ p, q $ with $ q > 0 $.