Statement

For a fixed α\alpha, there is a constant cα>0c_\alpha>0 such that

m+nαcα1+m2+n2for all (m,n)Z2{(0,0)}.|m+n\alpha|\ge\frac{c_\alpha}{1+\sqrt{m^2+n^2}} \quad\text{for all }(m,n)\in\mathbb Z^2\setminus\{(0,0)\}.
Proof

With a,b,ca,b,c and α\alpha' as in the conjugate identity, the product a(m+nα)(m+nα)a(m+n\alpha)(m+n\alpha') is a nonzero integer and has absolute value at least one. Also m+nαCαm2+n2|m+n\alpha'|\le C_\alpha\sqrt{m^2+n^2}. Dividing gives the claimed bound.

Rational approximation

There is consequently c ⁣>0c'\!>0 such that αp/qc/q2|\alpha-p/q|\ge c'/q^2 for all integers pp and q1q\ge1. When p/qα1|p/q-\alpha|\le1, apply the linear-form estimate and use p(α+1)q|p|\le(|\alpha|+1)q; when the difference exceeds one, reduce the constant. This is the badly approximable property of quadratic irrationals.

References