PROBLEM SET X
“FRACTIONAL”
DUE TUESDAY,  NOVEMBER 
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; the integer a0 is free to be negative or zero. For every integer
n ≥ 1, define a function fn : R≥0 . R>0 by the formula
fn (u) :=
1
.
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 write [a0 ; a1 , a2 , . . . , an ] for the rational number kn , whence we have
[a0 ; a1 , a2 , . . . , an ] = a0 +
1
.
[a1 ; a2 , a3 , . . . , an ]
Exercise . 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 . Show that for any positive integer n,
pn qn−1 − pn−1 qn = (−1)n−1 ,


DUE TUESDAY,  NOVEMBER 
whence
(−1)n−1
.
qn qn−1
Conclude also that for any positive integer n, the natural numbers |pn | and qn are relatively
prime.
[a0 ; a1 , a2 , . . . , an ] − [a0 ; a1 , a2 , . . . , an−1 ] =
Exercise . 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 .
m→∞
n→∞
Exercise . Show that for any integer n ≥ 2, one has
qk ≥ 2(k−1)/2 .
Deduce that the sequence (kn )n≥0 converges.
Notation. We shall simply write [a0 ; a1 , a2 , . . . ] for the limit
lim [a0 ; a1 , a2 , . . . , an ] = lim kn .
n→∞
n→∞
We can think of this as giving meaning to the regular continued action
1
[a0 ; a1 , a2 , . . . ] = a0 +
.
1
a1 +
a2 +
1
..
.
Exercise . Show that one has
[a0 ; a1 , a2 , . . . ] = a0 +
1
;
[a1 ; a2 , a3 , . . . ]
conclude that for any positive integer a, one has
)
√
1(
[a; a, a, . . . ] =
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 . 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 , . . . ].
is is the regular continued action 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.
PROBLEM SET X
“FRACTIONAL”

Exercise⋆ . Show that for any irrational number x and any integer n ≥ 2, one has
1
1
< |[a0 ; a1 , a2 , . . . ] − [a0 ; a1 , a2 , . . . , an ]| <
qn (qn + qn+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
Example. is 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) e so-called Golden Ratio φ has an exceptionally pleasing/boring continued fraction
representation. Here it is:
√
1+ 5
φ=
= [1; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, . . . ].
2
Why?
√
(b) Here’s the continued fraction representation of 2. Set a0 := 1, and for n ≥ 1, set
an := 2. en
√
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;


1
if n = 11m ± 3 or if n = 11m ± 5;



3
if n = 11m ± 4.
then
√
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.
en
e = [a0 ; a1 , a2 , . . . ] = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, . . . ].
Remember him? from the first problem set? at’s right; he’s back!

DUE TUESDAY,  NOVEMBER 
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/ℓ) − exp(−1/ℓ)
= [0; ℓ, 3ℓ, 5ℓ, 7ℓ, 9ℓ, . . . ].
exp(1/ℓ) + exp(−1/ℓ)
(e) e 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, . . . ].
∑
(f ) We have seen that both log n and nj=1 1/j increase without bound as n → ∞.
eir difference, however, converges as n → ∞. e Euler–Mascheroni constant γ is
the limit:


n
∑
1
.
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 aj ’s:
αn (x) := (a0 a1 · · · an )1/n .
Let’s call a positive irrational x a K-number if the sequence (αn (x))n≥0 converges to the
real number
)log2 (j)
∞ (
∏
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, . . . ]
PROBLEM SET X
“FRACTIONAL”

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 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 Knumbers.)
As for Khinchin’s constant K — to date, it’s not even known whether it’s rational. Eerie.