Problem Set 15: Liouville Numbers
Notation: N is the set on natural numbers, Z is the integers, and Z[x] is the ring of all real valued
polynomials whose coefficients are integers.
Definition: z is a Liouville Number iff  n  N  p, q  Z with q  1 and 0  z  (p/q)  1/q n .
Example: z  

k 1
2
 k!
is a Liouville number.
Proof: Given a natural number n let q  2 n ! and p  
z  (p/q) 

 k  n 1 2
 k!


 k  (n  1)! 2
k
2
1  (n  1)!
n
k 1
2
2
n ! k !
q
 n (n !)
 k 1 2
n
 1/q
n
 k!
.
, and
the first equality shows 0  z  (p/q) .
Lemma: Each Liouville number α is irrational.
Proof: To get a contradiction suppose α  c/d for some integers c and d with d > 0. Chose n so that
2
n 1
 d and then integers p and q as in the definition of Liouville number. Since
0  α  (p/q)  (cq  dp)/qd
1/q
n
the integer cq  dp is non-zero and cq  dp  1 . This gives
 α  (p/q)  (c q  d p)/q d  1/(q d) which implies 2
n 1
dq
n 1
, a contradicting q  2 .
Lemma: If f(x) is a real polynomial and a is any real, then there is a polynomial g(x) so that for all x,
f(x)  f(a)  (x  a) g(x) .
Proof: Write f(x) 
x
k
a
k
 k0 c k x
 (x  a) (x
f(x)  f(a) 
n
k 1
 ax
 k0 c k (x
n
k
k
k2
k
. Using Problem 1-2
2
a x
a )
k3
 ...  a
k2
xa
 k  0 c k ( x  a) g k (x)
n
k 1
)  (x  a) g k (x)
 ( x  a)
,
 k  0 c k g k (x) .
n
Theorem: If f(x)  Z[x] has degree n and α is an irrational root of f(x), then
 c  0  p, q  Z with q  0 and α  (p/q)  c/q
n
.
Proof: Write f(x)  f( α )  (x  α)g(x) as in the preceding lemma, set
M  max { g(x) : α  1  x  α  1 } ,
and let α 1 , α 2 , ... , α m be the roots of f(x) different from α . For c choose any number satisfying
0  c  min { 1, 1/M, α  α k : 1  k  m }
and to get a contradiction suppose
 p, q  Z with q  0 and
α  (p/q)  c/q
n
.
There are four steps.
α  (p/q)  c/q
(1) g(p/q)  1/c . Why?
g(p/q)
(2)
 M  1/c
f (p/q)  0 .
irrational,
n
 c  1 implies α  1  p/q  α  1
and thus
.
Why? If f(p/q) = 0 then p/q would be a rational root of f(x). Since
α  p/q
and thus
p/q  α i
α  (p/q)  c/q
for some i. But then
n
α
is
 c  α  α i , an
impossibility.
(3) f(p/q)  1/q n . Why? Write f(x) 
f(p/q)

 k  0 c k (p/q)
n
k

 k0 c k x
k
nk
n
n
 k0 c k p
n
Since f (p/q)  0 , the integer w 
k
q
n
 k0 c k p
k
q
/q
nk
is non-zero, w  1 and f(p/q)  w /q n  1 /q n .
(4) Finally 1/q n  f(p/q)  f( α )  f(p/q)  α  (p/q)
contradicting the assumption
α  (p/q)  c/q
n
with integer coefficients c 0 , c 1 , ... , c n .
 α  (p/q) (1/c)
g(p/q)
.
Theorem: Each Liouville number is transcendental.
Proof: To get a contradiction suppose that a Liouville number α is a root of a degree n polynomial
f(x)  Z[x] . The number α is irrational and by the preceding theorem
 c  0  p, q  Z with q  0 and α  (p/q)  c/q
n
.
Choose a natural number m with 1  c 2 m . Since α is a Liouville number
 s, t  Z with t  1 and 0  α  (s/t)  1/ t
nm
.
Combining inequalities, c/ t n  α  (s/t)  1/ t n t m , which forces c t m  1 and contradicts 1  c 2 m  c t m .
Definition: The Lebesgue measure (or measure) of a set A of reals is
meas(A)
 glb {

k 1
len (I k ) } ,
where the glb is taken over all covers C  { I k } k  1 of A by a countable number of open intervals. Put
Review Problem 9-12. Prove these properties of measure.
(a, Monotonicity) A  B implies meas(A)  meas(B) .
(b, Countable Subadditivity) meas(  n  1 A n )  
(c) meas ( [a, b] )  b  a .
(d) The measure of any countable set is zero.
n 1
meas(A
n
)
for any countable collection of sets.
Theorem: The set T of transcendental numbers in [0,1] has measure 1, but the set L of Liouville numbers
in [0,1] has measure zero. Thus there are "many" transcendental numbers that are not Liouville numbers.
Proof that meas (T) = 1: The inclusion T  [0,1] gives meas (T)  meas ( [0,1] )  1 . For the reverse
inequality let A be the set of algebraic numbers in [0,1]. A is countable and has measure zero. Since
[0,1]  A  T , countable subadditivity implies
1  meas ( [0,1] )  meas (A)  meas (T)  meas (T) .
Proof that meas(L) = 0: For n a natural number and q an integer with q > 1,define
n
A(n, q)  { x  [0,1] :  p  Z, 0  x  (p/q)  1/q } .
For fixed n


 q  2  p   A(n, q)  { x  [0,1] :  p, q  Z, q  1 and 0  x  (p/q)
 1/q
n
}.
For each n the last set contains L. Four steps will show meas(L) = 0.
(1) If A(n, q)   then 0  p  q . Why? If x is in A(n,q) then x  (p/q)  1/q n  1/q  1/2 for some p.
But if p > q or p   1 then x  (p/q)  1/2 for every x in [0,1].
(2) For each n and q > 1,

meas
q
  p   A(n, q)   4 /q n 1 .
Why? Using point (1)
can be covered by the open intervals ( (p  1)/q n , (p  1)/q n ) ,
 p   A(n, q)   p  0 A(n, q)
  p   A(n, q)    qp  0 (2/q n )  2 (q  1)/q n  (4 q)/q n .
0  p  q , so meas


(3) For each n > 2, meas  q  2  p   A(n, q)  4/(n  2) . Why? By countable subadditivity and (2)
meas


q2


 p   A(n, q) 


 q  2 meas

 q  2 4 /q
  p   A(n, q) 
n 1
4

1
x
1 n
dx  4/(n  2)
(4) Finally, L   q  2  p   A(n, q) and for each n > 2
meas(L)  meas


q2


 p   A(n, q)  4/(n  2)
.
The last can be true for all n > 2 only if meas (L) = 0.
The table summarizes a few facts about the sizes of several subsets on [0,1]. All five sets are dense in
[0,1] and thus have content 1.
Subset of [0,1]
Cardinality
Measure
Rationals
Irrationals
Algebraic Numbers
Transcendental Numbers
Liouville Numbers
Countable
Uncountable
Countable
Uncountable
Uncountable
0
1
0
1
0
More on Liouville Numbers
Theorem: If z is a Liouville number and r is a non-zero rational then rz is a Liouville number.
Proof: Write r = a/b with a and b integers. Assume b > 0. Given a natural number n, choose a natural
number m > n so that a b n 1  2 m  n . Applying the Liouville definition there are integers p and q with
q  2 and 0  z  (p/q)  1/q
m
. Multiplying the last inequality by r , 0  r z  (a p/b q)  a /(b q m ) .
Combining that inequality with a b n 1  2 m  n  q m  n gives 0  r z  (a p/b q)  a /(b q m )  1/(bq) n .
Corollary: Every interval (a,b) contains a Liouville number.
Proof: For z > 0 a Liouville number, the interval (a/z, b/z) contains a non-zero rational number.
Theorem: If each term of the sequence (a k ) is one of the integers 1 through 9, then z 

 k 1 a k 10
 k!
is a Liouville number. Further, the numbers of this form are all district.
Proof: Given a natural number n let q  10 n ! and p  
z  (p/q) 

 k  n  1 a k 10
 k!


 k  (n  1)! 9 (10
k
n
a
k 1 k
)  10
10
1  (n  1)!
n ! k !
 10
q
 k 1 a k 10
 n (n !)
n
 1/q
n
 k!
.
, and
the first equality shows 0  z  (p/q) .
For the other assertion let w  

b
k 1 k
10
 k!
be determined by a sequence (b k ) with a k  b k for some
k. Let m be the least index k with a k  b k . Then
z  w  ( a m  b m ) 10
 m!

 a m  b m 10
 m!

 a m  b m 10
 m!

 10
 m!

 k!

 k!
 k  m  1 ( a k  b k ) 10
 k  m  1 ( a k  b k ) 10

 k  m 1

 k  m  1 (8 ) 10
 i  ( m  1)! (8 ) 10

 m!

 10
 m!
 ( 80 / 9 ) 10
 ( m  1)!
 k!
 k!

 10
a k  b k 10
i
0
Corollary: The set of Liouville numbers is uncountable.
Proof: A Cantor diagonalization argument proves that there are uncountably many numbers of form
z

 k 1 a k 10
 k!
with each a k  { 1, 2, 3, ... , 9} .
Theorem: If f(x)  Z[x] and α is a Liouville number then f(α ) is a Liouville number.
Proof: Write f(x)  f(α )  (x  α) g(x) . Since f(x) is a polynomial { x : x  α and f(x)  f(α ) } is finite.
Fix delta with
0  δ  min{ x  α : x  α and f(x)  f(α ) } .
Let r be the degree of f(x) and set
M  max { g(x) : α  x  δ } .
To see that f(α ) is a Liouville number let n be a natural number. Choose a natural number m  n r so
that 1  δ 2 m and M 2 n r  2 m . Since α is a Liouville number
 p, q  Z with q  2 and α  (p/q)  1/q
m
.
There are now just four steps
(1) g(p/q)  M and f(p/q)  f(α ) . Why? The estimate α  (p/q)  1/q m  1/2 m  δ implies
g(p/q)
(2)
M
g(p/q)
g(p/q)
(3)
and
q
f(p/q)  f(α ) .
mn r
 M 2
0  α  (p/q)
0  f(α )  f(p/q)
. Why? Using the defining inequality M  2 m  n r and (1) to estimate g(p/q),
mn r
 1/q
q
nr
mn r
.
. Why? Combining (2) with the original inequality estimating α  (p/q) ,
 α  (p/q)
g(p/q)
q
mn r
/q
m
r n
 1/(q ) .
(The leading inequality follows from (1); f(α )  f(p/q)  0 .)
(4)
f(p/q)  (an integer)/q
f(p/q)

 k  0 c k (p/q)
r
r
. Why? Write
k

 k 0 ck p
r
k
f(x) 
q
rk
 k 0 ck x
/q
r
r
k
with integer coefficients c 0 , c1 , ... , c r .