Equivalent definition for irrational numbers I just read an answer in a recently hot question here, which claims that a number $x$ is irrational if and only if 
For all natural numbers $q'$ there exists a rational number $p/q$ such that $q > q'$ and $0<|x - p/q| < 1/q^2$. 
I proved that if a number is rational then it satisfies the negation of the above statement.
Now I have to prove that if a number is irrational then it satisfies the statement. I though about rational number sequences approximating $x$ such that the denominator can be made arbitrarily large (example: dyadic rationals) but I can't find a way to get $|x - p/q| < 1/q^2$. So I'm stuck.
Thanks in advance
 A: We need the following lemma:
Lemma: Let $x>0$ be any real number. Then given some positive integer $Q$, we may find positive integers $p$ and $q$ with $1\leq q\leq Q$ such that $|x-\frac{p}{q}|\leq\frac{1}{Qq}\leq\frac{1}{q^2}$.
Given this, we prove the following:
Theorem: If $x$ is irrational then there are infinitely many rational numbers $\frac{p}{q}$ such that $|x-\frac{p}{q}|\leq\frac{1}{q^2}$.
Pf.: Given some $p_0,q_0$ satisfying the inequality in question, we know that for irrational $x$, $|x-\frac{p_0}{q_0}|\neq0$ so that we may choose $\frac{1}{Q}<|x-\frac{p_0}{q_0}|$ and then by the above we have that there exist $q_1,p_1$ such that $|x-\frac{p_1}{q_1}|\leq\frac{1}{q_1Q}\leq\frac{1}{q_1^2}$ but where $\frac{1}{q_1Q}\leq\frac{1}{Q}<|x-\frac{p_0}{q_0}|$ so that $p_1$ and $q_1$ cannot both be equal to $p_0$ and $q_0$. 
Proving the lemma is the difficult part and serves as a good exercise. A good hint is to use the pigeonhole principle.
A: The inequality $$\left|\alpha-\frac pq\right|\lt\frac{1}{q^2}$$ admits infinitely many solutions where $\alpha$ is irrational (a Dirichlet's theorem easily proved using the pigeonhole principle). Hence take $q\gt q'$ and you finish.
