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 2 k! is a Liouville number. k 1
n n! k! n k! Proof: Given a natural number n let q 2 n! and p 2 q 2 . k 1 k 1 z (p/q) 2 k! 2 k 21 (n 1)! 2 n (n !) 1/q n , and k n 1 k (n 1)! 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 the integer cq dp is non-zero and cq dp 1 . This gives
1/q n 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) .
n Proof: Write f(x) c x k . Using Problem 1-2 k 0 k k k k 1 k 2 2 k 3 k 2 k 1 x a (x a) (x a x a x ... a x a ) (x a) g k (x) , n n n f(x) f(a) c (x k a k ) c (x a) g (x) (x a) c g (x) . k 0 k k 0 k k k 0 k k
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 . 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.
(1) g(p/q) 1/c . Why? α (p/q) c/q n c 1 implies α 1 p/q α 1 and thus g(p/q) M 1/c .
(2) f (p/q) 0 . Why? If f(p/q) = 0 then p/q would be a rational root of f(x). Since α is n irrational, α p/q and thus p/q α i for some i. But then α (p/q) c/q c α α i , an impossibility.
n (3) f(p/q) 1/q n . Why? Write f(x) c x k with integer coefficients c , c , ... , c . k 0 k 0 1 n
n n f(p/q) c (p/q) k c p k q n k /q n k 0 k k 0 k
n Since , the integer w c p k q n k is non-zero, w 1 and f(p/q) w /q n 1 /q n . k 0 k
(4) Finally 1/q n f(p/q) f(α ) f(p/q) α (p/q) g(p/q) α (p/q) (1/c) contradicting the assumption α (p/q) c/q n .
Theorem: Each Liouville number is transcendental.
Proof: To get a contradiction suppose that a Liouville number is a root of a degree n polynomial f(x) Z[x] . The number 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 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 { len (I ) } , k 1 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( A ) meas(A ) for any countable collection of sets. n 1 n n 1 n (c) meas ([a, b] ) b a . (d) The measure of any countable set is zero.
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 A(n, q) { x [0,1] : p Z, 0 x (p/q) 1/q n } . For fixed n n q 2 p A(n, q) { x [0,1] : p, q Z, q 1 and 0 x (p/q) 1/q } . 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].
n 1 (2) For each n and q > 1, meas p A(n, q) 4 /q . Why? Using point (1) q n n p A(n, q) p 0 A(n, q) can be covered by the open intervals ( (p 1)/q , (p 1)/q ) , q , so meas A(n, q) (2/q n ) 2 (q 1)/q n (4 q)/q n . p p 0
(3) For each n > 2, meas q 2 p A(n, q) 4/(n 2) . Why? By countable subadditivity and (2)
meas A(n, q) meas A(n, q) q 2 p q 2 p
4 /q n 1 4 x1 n dx 4/(n 2) q 2 1
(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 Countable 0 Irrationals Uncountable 1 Algebraic Numbers Countable 0 Transcendental Numbers Uncountable 1 Liouville Numbers Uncountable 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.
k! Theorem: If each term of the sequence (a ) is one of the integers 1 through 9, then z a 10 k k 1 k is a Liouville number. Further, the numbers of this form are all district.
n n! k! n k! Proof: Given a natural number n let q 10 n! and p a 10 q a 10 . k 1 k k 1 k z (p/q) a 10 k! 9 (10 k ) 101 (n 1)! 10 n (n !) 1/q n , and k n 1 k k (n 1)! the first equality shows 0 z (p/q) .
For the other assertion let w b 10 k! be determined by a sequence (b ) with a b for some k 1 k k k k k. Let m be the least index k with . Then
z w ( a b )10 m! ( a b )10 k! m m k m 1 k k
a b 10 m! ( a b )10 k! m m k m 1 k k
a b 10 m! a b 10 k! m m k m 1 k k 10 m! (8)10 k! k m 1 10 m! (8)10 i i (m 1)!
10 m! (80 / 9)10 (m 1)! 0
Corollary: The set of Liouville numbers is uncountable.
Proof: A Cantor diagonalization argument proves that there are uncountably many numbers of form k! z a 10 with each a {1, 2, 3, ... , 9} . k 1 k k
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 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 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 and f(p/q) f(α) .
(2) g(p/q) q m n r . Why? Using the defining inequality M 2 m n r and (1) to estimate g(p/q),
g(p/q) M 2 m n r q m n r .
(3) 0 α (p/q) 1/q n r . Why? Combining (2) with the original inequality estimating α (p/q) ,
0 f(α) f(p/q) α (p/q) g(p/q) q m n r /q m 1/(q r )n . (The leading inequality follows from (1); f(α) f(p/q) 0 .)
r k (4) f(p/q) (an integer)/q r . Why? Write f(x) c x with integer coefficients c , c , ... , c . k 0 k 0 1 r r r f(p/q) c (p/q) k c p k q r k /q r k 0 k k 0 k