Why is irrationality mesure of irrational numbers at least 2

Question

For any $x \in \mathbb{R}$, define its irrationality measure $\mu(x)$ to be the smallest number $\mu$ such that the inequality $|x-\dfrac{p}{q}| > \dfrac{1}{q^{\mu+\epsilon}}$ holds for any $\epsilon > 0$ and any rational number $\dfrac{p}{q}$ with sufficiently large $q$.

Literature says that $\mu(x)=1$ when $x$ is rational and $\mu(x) \geq 2$ otherwise. But it's not clear to me why $\mu(x) < 2$ can't be true for irrational numbers x. For example, why is something like $\mu(x) = \sqrt{2}$ not possible?

Answer 1

I don't know all the proof details, but here are a few cool, famous facts about approximating irrational $x$:

• Write the continued fraction $x=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\ddots}}}$ with $a_0\in\Bbb Z,\,a_1,\,\dots\in\Bbb N$. Then $a_0,\,a_0+\frac{1}{a_1},\,a_0+\frac{1}{a_1+\frac{1}{a_2}}$ etc. alternately under- and overestimate $x$ (OK, this one's easy to prove).
• These fractions $\frac{a_0}{1},\,\frac{a_0a_1+1}{a_1},\,a_0+\frac{a_0a_1a_2+a_0+a_2}{a_1a_2+1}$ etc. each approximate $x$ more closely than other fractions of smaller positive denominator.
• What's more, some $x$-dependent $c>0$ satisfies $|x-p/q|<c/q^2$, with $p/q$ any of the above fractions in its lowest terms.
• We can set $c\le\tfrac{1}{\sqrt{5}}$; this inequality is saturated iff $x=\frac{a+b\varphi}{c+d\varphi}$ for integers $a,\,b,\,c,\,d$ (with $c,\,d$ not both $0$ and $ad\ne bc$), where $\varphi$ is the golden ratio. Such $x$ are therefore often called the "most irrational" numbers, which is an amusing concept because they're of algebraic degree $2$, yet "more irrational" than all transcendental numbers.