Number Theory: An Introduction to Mathematics

188 IV Continued Fractions and Their Uses

As an application of Proposition 4 we prove

Proposition 5Let d be a positive integer which is not a square and m an integer such
that 0 <m^2 <d. If x,y are positive integers such that

x^2 −dy^2 =m,

then x/y is a convergent of the irrational number

√

d.

Proof Suppose first thatm>0. Thenx/y>

√

dand

0 <x/y−

√

d=m/(xy+y^2

√

d)<

√

d/ 2 y^2

√

d= 1 / 2 y^2.

Hencex/yis a convergent of

√

d, by Proposition 4.

Suppose next thatm<0. Theny/x> 1 /

√

dand

0 <y/x− 1 /

√

d=−m/d(xy+x^2 /

√

d)< 1 /

√

d(xy+x^2 /

√

d)< 1 / 2 x^2.

Hencey/xis a convergent of 1/

√

d.But,since1/

√

d= 0 + 1 /

√

d, the convergents
of 1/

√

dare 0/1 and the reciprocals of the convergents of

√

d. 


In the next section we will show that the continued fraction expansion of

√

dhas
a particularly simple form.
It was shown by Vahlen (1895) that at least one of any two consecutive convergents
ofξsatisfies the inequality of Proposition 4. Indeed, since consecutive convergents lie
on opposite sides ofξ,

|pn/qn−ξ|+|pn− 1 /qn− 1 −ξ|=|pn/qn−pn− 1 /qn− 1 |

= 1 /qnqn− 1 ≤ 1 / 2 qn^2 + 1 / 2 q^2 n− 1 ,

with equality only ifqn=qn− 1. This proves the assertion, except whenn=1and
q 1 =q 0 =1. But in this casea 1 = 1 , 1 ≤ξ 1 <2 and hence

|ξ−p 1 /q 1 |=|ξ−a 0 − 1 |= 1 −ξ 1 −^1 < 1 / 2.
It was shown by Borel (1903) that at least one of any three consecutive convergents
ofξsatisfies the sharper inequality

|ξ−p/q|< 1 /

√

5 q^2.

In fact this is obtained by takingr=1 in the following more general result, due to
Forder (1963) and Wright (1964).

Proposition 6Letξbe an irrational number with the continued fraction expansion
[a 0 ,a 1 ,...]and the convergents pn/qn. If, for some positive integer r ,

|ξ−pn/qn|≥ 1 /(r^2 + 4 )^1 /^2 qn^2 for n=m− 1 ,m,m+ 1 ,

then am+ 1 <r.

Proof If we puts=(r^2 + 4 )^1 /^2 /2, thensis irrational. For otherwise 2swould be an
integer and from( 2 s+r)( 2 s−r)=4 we would obtain 2s+r=4, 2s−r=1and
hencer= 3 /2, which is a contradiction.
By the hypotheses of the proposition,
1 /qm− 1 qm=|pm− 1 /qm− 1 −pm/qm|=|ξ−pm− 1 /qm− 1 |+|ξ−pm/qm|
≥(qm−−^21 +qm−^2 )/ 2 s