PROBLEM SET IX
“FUN WITH FRACTIONS”
DUE FRIDAY, 2 DECEMBER
We’ve talked a lot in this course about how we specify real numbers using rational approximations. Here, we
study a way to do this using a beautiful, recursively-defined sequence of rationals.
Definition. Suppose (an )n≥0 a sequence of integers. Assume that for every integer n ≥ 1, the integer an is positive;
R>0 by the
the integer a0 is free to be negative or zero. For every integer n ≥ 1, define a function fn : R≥0
formula
1
fn (u) :=
.
an + u
Define functions Kn : R≥0
R for integers n ≥ 0 recursively in the following manner. Set K0 (u) := a0 + u, and for
any n ≥ 1, set Kn (u) := Kn−1 ( fn (u)). Slightly more informally, we may write that for any u ∈ R, one has
1
Kn (u) = a0 +
.
1
a1 +
a2 + .
.
1
.
+
1
an + u
Armed with this, define a sequence (kn )n≥0 of rational numbers via the formula
kn := Kn (0).
We will sometimes write [a0 ; a1 , a2 , . . . , an ] for the rational number kn , whence we have
1
[a0 ; a1 , a2 , . . . , an ] = a0 +
.
[a1 ; a2 , a3 , . . . , an ]
Exercise 64. Set p−1 := 1, q−1 := 0, p0 := a0 , and q0 := 1. Now define, for any integer n ≥ 1,
pn := an pn−1 + pn−2
and
qn := an qn−1 + qn−2 .
Show that for any integer n ≥ 0, one has qn > 0 and
[a0 ; a1 , a2 , . . . , an ] =
pn
qn
.
Exercise 65. Show that for any positive integer n,
pn qn−1 − pn−1 qn = (−1)n−1 ,
whence
[a0 ; a1 , a2 , . . . , an ] − [a0 ; a1 , a2 , . . . , an−1 ] =
(−1)n−1
.
qn qn−1
Conclude also that for any positive integer n, the natural numbers | pn | and qn are relatively prime.
Exercise 66. Show that the subsequence (k2m ) m≥0 is an increasing sequence, and show that (k2m+1 ) m≥0 is a decreasing sequence. Show also that for any nonnegative integers r and s, one has k2r < k2s +1 . Deduce that each of these
subsequences converges, and that
lim inf kn = lim k2m
n→∞
m→∞
and
lim sup kn = lim k2m+1 .
n→∞
1
m→∞
2
DUE FRIDAY, 2 DECEMBER
Exercise 67. Show that for any integer n ≥ 2, one has
qk ≥ 2(k−1)/2 .
Deduce that the sequence (kn )n≥0 converges.
We shall simply write [a0 ; a1 , a2 , . . . ] for the limit limn→∞ [a0 ; a1 , a2 , . . . , an ] = limn→∞ kn . We can think of this as
giving meaning to the regular continued fraction
1
[a0 ; a1 , a2 , . . . ] = a0 +
a1 +
a2 +
Exercise 68. Show that one has
[a0 ; a1 , a2 , . . . ] = a0 +
.
1
1
..
.
1
[a1 ; a2 , a3 , . . . ]
;
conclude that for any positive integer a, one has
[a; a, a, . . . ] =
p
1
a + a2 + 4 .
2
We now come to the most important result in this problem set: it turns out that any irrational number can be
written as a regular continued fraction in a unique way.
Exercise 69. Show that for any irrational number x, there exists a unique sequence (an )n≥0 of integers such that for
any n ≥ 1, one has an ≥ 1 and
x = [a0 ; a1 , a2 , . . . ].
This is the regular continued fraction representation of x.
Note that this establishes a bijection between the set of irrational numbers and the set of sequences (an )n≥0 of
integers such that for any n ≥ 1, one has an ≥ 1.
Not only can any irrational number be exhibited as a limit [a0 ; a1 , a2 , . . . ] in a unique manner, but, moreover, the
rational approximations to the real number [a0 ; a1 , a2 , . . . ] provided by the rationals [a0 ; a1 , a2 , . . . , an ] are in a sense
ideal, as the next exercise demonstrates.
Exercise? 70. Show that for any irrational number x and any integer n ≥ 2, one has
1
qn (qn + qn+1 )
< |[a0 ; a1 , a2 , . . . ] − [a0 ; a1 , a2 , . . . , an ]| <
1
qn qn+1
Use this to show that the irrationality exponent∗ of any irrational number can be computed by means of the formula
µ([a0 ; a1 , a2 , . . . ]) = 1 + lim sup
n→∞
log qn+1
log qn
.
Remark. This isn’t an exercise, but I wanted to share with you some regular continued fraction representations of
interesting irrational numbers, just to give you a feel for them.
(a) The so-called Golden Ratio φ has an exceptionally pleasing/boring continued fraction representation. [What
would you expect, right?] Here it is:
p
1+ 5
φ=
= [1; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, . . . ].
2
This follows from Exercise 68.
∗
Remember him? from the first problem set? That’s right; he’s back!
PROBLEM SET IX
“FUN WITH FRACTIONS”
3
p
(b) Here’s the continued fraction representation of 2. Set a0 := 1, and for n ≥ 1, set an := 2. Then
p
2 = [a0 ; a1 , a2 , . . . ] = [1; 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, . . . ].
You can prove this!
(c) It turns out that an irrational number is quadratic (i.e., a root of a quadratic polynomial with integer coefficients)
if and only if its regular continued fraction representation is periodic. For instance, if a0 = 80, and if for any
n ≥ 1, one has

158 if n = 11m;




13
if n = 11m ± 1;


an = 5
if n = 11m ± 2;




if n = 11m ± 3 or if n = 11m ± 5;

1
3
if n = 11m ± 4.
then
p
1 + 13 37
=
[a0 ; a1 , a2 , . . . ]
=
[158, 13, 5, 1, 3, 1, 1, 3, 1, 5, 13, 158, 13, 5, 1, 3, 1, 1, 3, 1, 5, 13, 158, 13, 5, 1, 3, 1, 1, 3, 1, 5, 13, . . . ].
(d) Here’s the regular continued fraction representation of e := exp(1). Set a0 := 2, and for n ≥ 1, set
(
2m if n = 3m − 1;
an :=
1
otherwise.
Then
e = [a0 ; a1 , a2 , . . . ] = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, . . . ].
You can obtain can obtain pretty good estimates of e from this, e.g.:
517656
49171
= k14 < e < k13 =
.
190435
18089
Unsurprisingly, the regular continued fraction representation of e was discovered by Euler, as was the formula
exp(1/`) − exp(−1/`)
exp(1/`) + exp(−1/`)
= [0; `, 3`, 5`, 7`, 9`, . . . ].
(e) The beautiful expression for the continued fraction of e may lead you to believe that π should admit a similarly
pleasant continued fraction representation. Sadly, it seems that this is not the case; here are the first few:
π = [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, 1, 84, 2, 1, 1, 15, 3, 13, 1, 4, 2, 6, 6, 99, 1, 2, 2, 6, 3, 5, 1, 1, . . . ].
Yeah. I don’t see a pattern there either. It’s not any better when you look at 2π:
2π = [6; 3, 1, 1, 7, 2, 146, 3, 6, 1, 1, 2, 7, 5, 5, 1, 4, 1, 2, 42, 5, 31, 1, 1, 1, 6, 2, 2, 4, 3, 12, 49, 1, 5, 1, 12, 1, 1, 1, 2, 3, 1, . . . ]
P
(f) We have seen that both log n and nj=1 1/ j increase without bound as n → ∞. Their difference, however, converges as n → ∞. The Euler–Mascheroni constant γ is the limit:


n 1
X


lim − log n +
.
n→∞
j
j =1
It too has an utterly mystifying regular continued fraction representation:
γ = [0; 1, 1, 2, 1, 2, 1, 4, 3, 13, 5, 1, 1, 8, 1, 2, 4, 1, 1, 40, 1, 11, 3, 7, 1, 7, 1, 1, 5, 1, 49, 4, 1, 65, 1, 4, 7, 11, 1, 399, 2, 1, . . . ].
Remark. Let me end with one very unsettling fact. Suppose x a positive irrational number. Write
x = [a0 ; a1 , a2 , . . . ].
For each approximation [a0 ; a1 , a2 , . . . , an ] to x given this way, let’s form the geometric mean of the a j ’s:
1/n
αn (x) := a0 a1 · · · an
.
4
DUE FRIDAY, 2 DECEMBER
Let’s call a positive irrational x a K-number if the sequence (αn (x))n≥0 converges to the real number
‚
Œlog2 ( j )
∞
Y
1
K =
1+
j ( j + 2)
j =1
=
=
2.6854520010 6530644530 9714835481 7956938203 8229399446 2953051152 · · ·
[2; 1, 2, 5, 1, 1, 2, 1, 1, 3, 10, 2, 1, 3, 2, 24, 1, 3, 2, 3, 1, 1, 1, 90, 2, 1, 12, 1, 1, 1, 1, 5, 2, 6, 1, 6, 3, 1, 1, 2, 5, 2, 1, 2, 1, 1, . . . ]
called Khinchin’s constant. Note that K doesn’t dependent upon x at all, and I haven’t offered any explanation for this
suspicious-looking formula, so this seems like madness; it would be reasonable for you to expect that there are very
few K-numbers. But you’d be wrong: remarkably, Khinchin showed that almost all positive irrational numbers are
K-numbers. As of today, however, no particular example of a K-number is known, though computational evidence
suggests that numbers whose regular continued fraction representation do not have nice
p patterns should all be
K-numbers. (For instance, we expect that π, 2π, and γ are all K-numbers, but e and 2 are known not to be
K-numbers.) As for Khinchin’s constant K — to date, it’s not even known whether it’s rational. Eerie.