M3P14 LECTURE NOTES 12: DIOPHANTINE
APPROXIMATION
1. Liouville’s theorem
We’ve shown (in two different ways!) that for any irrational real number
α, there are infinitely many rational numbers pq with | pq − α| < q12 . What
happens if we ask for something stronger? For instance, can we find infinitely
many pq with | pq − α| < q1e for some e > 2?
It turns out that there for many irrational α, the answer is no. In particular, let’s make the following definition:
Definition 1.1. Let d be a positive integer. A complex number α is algebraic of degree d if there is a degree d (not necessarily monic) polynomial
P (x), with integer coefficients, such that P (α) = 0, and no such polynomial
of degree less than d.
We then have:
Theorem 1.2 (Liouville’s theorem). Let α be a real number that is algebraic
of degree d. Then for any real number e > d, there are at most finitely many
rational numbers pq such that | pq − α| < q1e .
Proof. Let P (x) be a polynomial of degree d, with integer coefficients, such
that P (α) = 0. Choose such that P (x) has no roots other than α on the
closed interval [α − , α + ].
Write P (x) = (x − α)Q(x); Q(x) is a monic polynomial with real coefficients, of degree d − 1. Since |Q(x)| is a continuous, real valued function,
there is a real number K > 0 such that |Q(x)| ≤ K on the compact set
[α − , α + ].
Now suppose we have pq with | pq − α| < q1e . There are only finitely many
q such that q13 ≥ , so we may assume q1e < .
Now on the one hand, we have:
p
p
p
1
|P ( )| = |( − α|Q( | < e K.
q
q
q
q
On the other hand, since P has degree d and integer coefficients, the denominator of P ( pq ) (when written in lowest terms) is a divisor of q d . But pq
is NOT a root of P (x), so P ( pq ) is nonzero and hence |P ( pq )| ≥ q1d .
Putting the inequalities together, we get:
1
p
1
≤ |P ( )| < e K.
q
q
qd
1
2
M3P14 LECTURE NOTES 12: DIOPHANTINE APPROXIMATION
Rewriting, we find q e−d < K; since e − d > 0 there are only finitely many q
for which this is possible.
2. Constructing transcendentals
Recall that a complex number α is transcendental if there is no polynomial
P (x), with integer coefficients, such that P (α) = 0, and algebraic otherwise.
The set of polynomials P (x) with integer coefficients is countable; since
each such polynomial has finitely many roots the set of algebraic numbers
is countable. Since the set of reals is uncountable this means that, in a very
strong sense, almost every real number is transcendental.
In spite of this it is very hard to give an example of a single real number
that is provably transcendental. (In fact, e and π are examples of transcendental numbers, but this is much harder than what we’ll do.)
Liouville’s theorem gives one approach to proving that a given number is
transcendental: show that it admits too many good rational approximations.
If we can show that for any e, there exist infinitely many pq such that | pq −α| <
1
q e , then Liouville’s theorem tells us that α can’t be algebraic of any degree,
and hence must be transcendental.
As an example, define a real number α by
α=
∞
X
1
.
10n!
n=1
This clearly converges.
We can find rational approximations to α simply by truncating the series;
for each k, let αk be the sum:
αk =
k
X
1
.
10n!
n=1
On one hand, αk is rational with denominator 10k! . On the other hand,
we have:
∞
X
1
2
|α − αk | =
< (k+1)! .
n!
10
10
n=k+1
d
2
< (101 k! . Thus
10(k+1)!
1 d
. Liouville’s theorem
(10k!
Fix a positive integer d. For any k > d, we have
there are infinitely many k such that |α − αk | <
thus tells us that α cannot be algebraic of degree d. Since d was arbitrary,
α must be transcendental.
3. Roth’s theorem
Liouville’s theorem tells us that algebraic numbers are difficult to approximate well by rationals. In fact, they are even more difficult to approximate
than Liouville’s theorem would suggest. For instance, we have:
M3P14 LECTURE NOTES 12: DIOPHANTINE APPROXIMATION
3
Theorem 3.1 (Roth’s Theorem). Suppose α is algebraic. Then for any >
1
0, there are only finitely many rational numbers pq such that | pq − α| < q2+
.
In other words, for algebraic numbers, you cannot do any better than
Dirichlet’s theorem.
Roth’s theorem is considerably harder to prove than Liouville’s, and we
will not give a proof in this course. Note that the stronger bound gives
proofs that additional real numbers are transcendental; for instance, one
can prove with Roth’s theorem that the number
∞
X
1
β=
103n
n=1
is transcendental, which is not possible with Liouville’s theorem.