Math 104B, Number Theory, Winter 2003.
Lecture 20. Quadratic Irrationals.
Aquadratic irrational xis an irrational number solving a quadratic equation with
integer coefficients. It can be written in the form x=a+b√nwhere nis an integer
which is not a perfect square, aand bare rational, and b6= 0. Conversely every number
which can be written in this form is a quadratic irrational. The representation a+b√n
is not unique, for example 1 + 2√3 = 1 + √12. However, if x=a+b√nis a quadratic
irrational then aand b√nare uniquely determined by xand hence the conjugate
a−b√nis well defined.
(To see this, suppose a+b√n=c+d√mwhere mis not a perfect square. Then
(b√n)2= ((c−a) + d√m)2is rational. But then (c−a)d√mis rational, so a=c.)
By a suitable choice of n, we can write any quadratic irrational in the form
(*) A+√n
B,where A, B ∈Z, B|A2−n.
For example
a
b+c
d√m=abd2+p(b2cd)2m
(bd)2.
Theorem 11.4.1. If xhas the form in (*) then complete quotients xkof the continued
fraction expansion of xhave the form
Ak+√n
Bk
,where Ak, Bk∈Z, Bk|A2
k−n.
We follow the proof in the book.
Theorem (Lagrange). The continued fraction expansion of a quadratic irrational is
eventually periodic.
Before proving this, we remark that the converse of this is easy to prove. We already
showed that a purely periodic continued fraction [a0, . . . , an] is a quadratic irrational.
But if x=a+b√nthen 1/x = ¯x/(a2−b2n) is also a quadratic irrational, so a continued
fraction of the form [b0, . . . , bm, a0, . . . , an] is a quadratic irrational.
Proof of the Theorem. Let xbe a continued fraction which we write as (A+√n)/B
where Aand Bare integers with B|A2−n. Then write xk= (Ak+√n)/Bkfor the
kth complete quotient of x. We aim to show that for ksufficiently large, the integers
Akand Bksatisfy
(1) 0 < Ak<√n, 0< Bk<2√n.
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