MT 430 Intro to Number Theory
PROBLEM SET 6 SOLUTIONS
Problem 1. Evaluate the infinite continued fractions:
(1) h2, 3, 1, 1, 1, 1, · · · i
(2) h2, 1, 2, 1, 2, 1, · · · i
(3) h1, 3, 1, 2, 1, 2, 1, 2, · · · i
Solution:
√ (1) h2, 3, 1, 1, 1, 1, · · · i. Recall from the lecture that ϕ =
h1i = ( 5 + 1)/2 is the golden ratio. It follows that
√
√
1
ϕ
7 5 + 11
25 − 5
h2, 3, 1, 1, 1, 1, · · · i = 2 +
=2+
= √
=
3ϕ + 1
10
3 + ϕ1
3 5+5
(2) Set x = h2, 1, 2, 1, 2, 1, · · · i. Then x = h2, 1, xi and so
x=2+
1
.
1 + x1
x=2+
x
x+1
Simplifying, we get
or
x2 − 2x − 2 = 0.
√
Solving this equation, we get x = 1 + 3.
(3)
√
h1, 3, 1, 2, 1, 2, 1, 2, · · · i = h1, 3, 1, h2, 1ii = h1, 3, 1, 3 + 1i.
We compute
√
h1, 3, 1, 3 + 1i = 1 +
1
3+
1
=3−
√
3
1+ √ 1
3+1
(after doing some tedious simplifications).
Problem 2. Find the irrational number having continued fraction expansion h9, 9, 18i.
Solution: First, we calculate x = h9, 18i. To do this, note that
x=9+
1
18 +
1
x
=9+
1
x
163x + 9
=
.
18x + 1
18x + 1
2
PROBLEM SET 6 SOLUTIONS
Solving the resulting quadratic equation 18x2 − 162x − 9 = 0 or 2x2 −
18x − 1 = 0. The only positive root of this equation is
√
9 + 83
x=
.
2
We now have
√
1
2
√ = 83.
h9, 9, 18i = 9 + = 9 +
x
9 + 83
√
Problem 3. Expand 15 into an infinite simple continued fraction.
√
√
√
Solution 1: We have b 15c = 3. Since 1/( 15 − 3) = 15+3
, we have
6
√
1
15 = 3 + √
.
15+3
6
√
√
= 1. Since 1/( 15+3
− 1) = 15 + 3, we have
6
√
15 + 3
1
=1+ √
.
6
15 + 3
√
√
√
Clearly, b 15 + 3c =√6. Hence 15 + 3 = 6 + ( 15 − 3). We have
already encountered 15 − 3 before, so we conclude that
√
15 = h3, 1, 6i.
√
We also have b
15+3
c
6
Solution 2: √More generally, we’ll compute the continued fraction expansion of d where d = (m √
+ 1)2 −√1 = m2 + 2m.
√ First, consider the
reduced quadratic √
irrational d + b dc = m + d. Let m0 = m and
q0 = 1, a0 = bm + dc = 2m. Then using the recursive formulae from
lectures, we compute
m1 = a0 q0 − m0 = 2m − m = m
d − m21
= (m2 + 2m) − m2 = 2m
q0
√
m+ d
a1 = b
c=1
2m
m2 = a1 q1 − m1 = 2m − m = m
q1 =
d − m22
(m2 + 2m) − m2
=
=1
q1
2m
√
m+ d
a2 = b
c = 2m
1
q2 =
PROBLEM SET 6 SOLUTIONS
3
√
From now on, the
patter
repeats,
so
that
m+
d = h2m, 1, 2m, 1, . . . i =
√
h2m, 1i. Hence d = hm, 1, 2mi. Summarizing
√
m2 + 2m = hm, 1, 2mi
for any positive integer m.