2 Diophantine Approximation 189and hence
qm^2 − 2 sqm− 1 qm+qm^2 − 1 ≤ 0.Furthermore, this inequality also holds whenqm− 1 ,qmare replaced byqm,qm+ 1. Con-
sequentlyqm− 1 /qmandqm+ 1 /qmboth satisfy the inequalityt^2 − 2 st+ 1 ≤0. Since
t^2 − 2 st+ 1 =(t−s+r/ 2 )(t−s−r/ 2 ),it follows that
s−r/ 2 <qm− 1 /qm<qm+ 1 /qm<s+r/ 2 ,the first and last inequalities being strict becausesis irrational. Hence
am+ 1 =qm+ 1 /qm−qm− 1 /qm<s+r/ 2 −(s−r/ 2 )=r. It follows from Proposition 6 withr=1 that, for any irrational numberξ,there
exist infinitely many rational numbersp/q=pn/qnsuch that
|ξ−p/q|< 1 /√
5 q^2.Here the constant
√
5 is best possible. For take anyc>√
- If there exists a rational
numberp/q, withq>0and(p,q)=1, such that
|ξ−p/q|< 1 /cq^2 ,thenp/qis a convergent ofξ, by Proposition 4. But for any convergentpn/qnwe have
|ξ−pn/qn|= 1 /qn(qnξn+ 1 +qn− 1 ).If we takeξ=τ:=( 1 +
√
5 )/2, then alsoξn+ 1 =τandpn=qn+ 1. Hence|τ−qn+ 1 /qn|= 1 /q^2 n(τ+qn− 1 /qn),whereτ+qn− 1 /qn→τ+τ−^1 =
√
5, sinceqn/qn− 1 →τ. Thus, for anyc>√
5,
there exist at most finitely many rational numbersp/qsuch that
|τ−p/q|< 1 /cq^2.It follows from Proposition 6 withr=2thatif|ξ−pn/qn|≥ 1 /√
8 qn^2 for all largen,