Number Theory: An Introduction to Mathematics

204 IV Continued Fractions and Their Uses

Proof For anyz∈Cwe writez=x+iy,wherex,y∈R. We show first that no
two points ofFare equivalent. Assume on the contrary that there exist distinct points
z,z′∈Fwithy′≥ysuch that

z′=(az+b)/(cz+d)

for somea,b,c,d∈Zwithad−bc=1. Ifc=0, thena=d=±1,b=0and
z′=z+b/d, which is impossible forz,z′∈F. Hencec=0. Since

y′=y/|cz+d|^2 ,

we have|cz+d|≤1. Thus|z+d/c|≤ 1 /|c|, which is impossible not only if|c|≥ 2
but also ifc=±1.
We now show that anyz 0 ∈His equivalent to a point of theclosureF ̄=F∪∂F.
We can choosem 0 ∈Zso thatz 1 =z 0 +m 0 satisfies|x 1 |≤ 1 /2. If|z 1 |≥1, there is
nothing more to do. Thus we now suppose|z 1 |<1. Putz 2 =− 1 /z 1 .Then

y 2 =y 1 /|z 1 |^2 >y 1

and actuallyy 2 ≥ 2 y 1 ify 1 ≤ 1 /2, since then|z 1 |^2 ≤ 1 / 4 + 1 / 4 = 1 /2. We now
repeat the process, withz 2 in place ofz 0 , and choosem 2 ∈Zso thatz 3 =z 2 +m 2
satisfies|x 3 |≤ 1 /2. Fromz 3 =(m 2 z 1 − 1 )/z 1 we obtain

|z 3 |^2 ={(m 2 x 1 − 1 )^2 +(m 2 y 1 )^2 }/(x 12 +y 12 ).

Assume|z 3 |<1. Thenm 2 =0andalsom 2 =±1, since| 1 ±x 1 |≥ 1 / 2 ≥|x 1 |.
If|m 2 |≥2, then|z 3 |^2 ≥ 4 |y 1 |^2 and hencey 1 < 1 /2. Thus in passing fromz 1 toz 3
we obtain eitherz 3 ∈F ̄ory 3 =y 2 ≥ 2 y 1. Hence, after repeating the process finitely
many times we must obtain a pointz 2 k+ 1 ∈F ̄.

Proposition 13 implies that the sets{g(F ̄):g∈Γ}form atilingofH,since

H=∪

g∈Γ

g(F ̄), g(F)∩g′(F)=∅ifg,g′∈Γandg=g′.

–1 –1/2 (^0) 1/ 2 1
F
Fig. 1.Fundamental domain forΓ.