Explain briefly how a positive irrational number x gives rise to a continued fraction
a0+a1+a2+a3+…111
with the aj non-negative integers and aj⩾1 for j⩾1.
Show that, if we write
(pnqnpn−1qn−1)=(a0110)(a1110)⋯(an−1110)(an110)
then
qnpn=a0+a1+a2+an−1+an1111
for n⩾0.
Use the observation [which need not be proved] that x lies between pn/qn and pn+1/qn+1 to show that
∣pn/qn−x∣⩽1/qnqn+1
Show that qn⩾Fn where Fn is the nth Fibonacci number (thus F0=F1=1, Fn+2=Fn+1+Fn), and conclude that
∣∣∣∣∣qnpn−x∣∣∣∣∣⩽FnFn+11