1. ## A strange inequality

I saw this one in an analysis text. The author didn't say where this inequality comes from or where it is used. I'd like to know more about it if anyone has information on it. Looks like some kind of number theoretic inequality since $\displaystyle \sqrt {p}$ with $\displaystyle p$ prime is always irrational.

Let $\displaystyle a \in \mathbb{N}$ be such that $\displaystyle \sqrt{a}$ is irrational. We want to prove there is some $\displaystyle c > 0$ such that for all integers $\displaystyle p, q$ with $\displaystyle q > 0$, we have $\displaystyle |q \sqrt{a}-p| > \frac{c}{q}$.

I have no clue on how to prove this one. The hint was to rationalize. So $\displaystyle (q \sqrt{a}-p)(-q\sqrt{a}-p)=-q^2a+p^2.$ Then $\displaystyle |p^2-q^2a| \geq 0$ but supposing $\displaystyle |p^2-q^2a| = 0$, we have $\displaystyle |p-q\sqrt{a}||p+q\sqrt{a}| = 0$ and $\displaystyle |p-q\sqrt{a}| = 0$ implies $\displaystyle q\sqrt{a}$ is an integer which is a contradiction. Therefore $\displaystyle |p^2-q^2a| > 0$ and $\displaystyle |p^2-q^2a|$ is an integer. So $\displaystyle |p^2-q^2a| \geq 1$.

2. Originally Posted by Jes
Let $\displaystyle a \in \mathbb{N}$ be such that $\displaystyle \sqrt{a}$ is irrational. We want to prove there is some $\displaystyle c > 0$ such that for all integers $\displaystyle p, q$ with $\displaystyle q > 0$, we have $\displaystyle |q \sqrt{a}-p| < \frac{c}{q}$.
the inequality in this form that you wrote is false because, for example, it will fail if you put q = 1 and let p go to infinity. the correct form of the inequality is: $\displaystyle |q \sqrt{a}-p| > \frac{c}{q}. \ \ \ \ \color{red}(*)$

$\displaystyle \color{red}(*)$ is a special case of Liouville's approximation theorem : if $\displaystyle \alpha \notin \mathbb{Q}$ is a root of a polynomial of degree $\displaystyle n$ in $\displaystyle \mathbb{Z}[x],$ then there exists $\displaystyle c > 0$ such that for all integers $\displaystyle p, q$ with $\displaystyle q > 0,$ we'll

have: $\displaystyle |q \alpha -p| > \frac{c}{q^{n-1}}$.

Proof of $\displaystyle \color{red}(*)$: let $\displaystyle \sqrt{a}=\alpha$ and choose $\displaystyle 0 < c < \frac{1}{2\alpha + 1}.$ let p and q be integers with q > 0. we consider two cases:

case 1. $\displaystyle \frac{p}{q} \notin (\alpha - 1, \alpha + 1):$ in this case we'll have $\displaystyle |q\alpha - p| \geq q \geq \frac{1}{q} > \frac{c}{q}.$

case 2. $\displaystyle \frac{p}{q} \in (\alpha - 1, \alpha +1):$ then we'll get $\displaystyle |p| < (\alpha + 1)q.$ now since $\displaystyle \alpha \notin \mathbb{Q},$ we have $\displaystyle q^2\alpha^2 - p^2 \neq 0.$ thus: $\displaystyle 1 \leq \ |q^2\alpha^2 - p^2|=|q\alpha + p||q\alpha - p|.$ threrefore:

$\displaystyle |q\alpha - p| \geq \frac{1}{|q\alpha + p|} \geq \frac{1}{q \alpha + |p|} > \frac{1}{q(2\alpha + 1)} > \frac{c}{q}. \ \ \ \Box$

3. Thanks a lot. I need to be more careful when writing stuff down. Luckily, you were familiar enough to spot the obvious error.

It seems a more general version of the inequality can be used to construct a transcendental number and thereby give a constructive proof of the existence of transcendental numbers. Great stuff. Thanks for the link.