184 IV Continued Fractions and Their Uses
Sinceη>1, its continued fraction expansion has the form [an,an+ 1 ,...], where
an≥1. It follows thatξhas the continued fraction expansion
[a 0 ,a 1 ,...,an− 1 ,an,...],resp. [a 0 ,a 1 ,...,an− 1 − 1 , 1 ,an,...].In either casep′/q′andp/qare consecutive convergents ofξandηis the correspond-
ing complete quotient.
A complex numberζis said to beequivalentto a complex numberωif there exist
integersa,b,c,dwithad−bc=±1suchthat
ζ=(aω+b)/(cω+d),andproperly equivalentif actuallyad−bc=1. Thenωis also equivalent, resp.
properly equivalent, toζ,since
ω=(dζ−b)/(−cζ+a).By takinga =d =1andb = c = 0, we see that any complex numberζ is
properly equivalent to itself. It is not difficult to verify also that ifζis equivalent toω
andωequivalent toχ,thenζis equivalent toχ, and the same holds with ‘equivalence’
replaced by ‘proper equivalence’. Thus equivalence and proper equivalence are indeed
‘equivalence relations’.
For any coprime integersb,d, there exist integersa,csuch thatad−bc=1. Since
b/d=(a· 0 +b)/(c· 0 +d),it follows that any rational number is properly equivalent to 0, and hence any two
rational numbers are properly equivalent. The situation is more interesting for irra-
tional numbers:
Proposition 2Two irrational numbersξ,ηare equivalent if and only if their con-
tinued fraction expansions[a 0 ,a 1 ,a 2 ,...],[b 0 ,b 1 ,b 2 ,...]have the same ‘tails’, i.e.
there exist integers m≥ 0 and n≥ 0 such that
am+k=bn+k for all k≥ 0.Proof If the continued fraction expansions ofξ andηhave the same tails, then
some complete quotientξm ofξ coincides with some complete quotientηnofη.
Butξ is equivalent toξm,sinceξ = (pm− 1 ξm+pm− 2 )/(qm− 1 ξm+qm− 2 )and
pm− 1 qm− 2 −pm− 2 qm− 1 =(− 1 )m, and similarlyηis equivalent toηn. Henceξandη
are equivalent.
Suppose on the other hand thatξandηare equivalent. Then
η=(aξ+b)/(cξ+d)for some integersa,b,c,dsuch thatad−bc=±1. By changing the signs of all four
we may suppose thatcξ+d>0. From the relation
ξ=(pn− 1 ξn+pn− 2 )/(qn− 1 ξn+qn− 2 )