In number theory, a Liouville number is a real number x with the property that, for every positive integer n, there exist infinitely many pairs of integers (p, q) with q > 1 such that
Liouville numbers are "almost rational", and can thus be approximated "quite closely" by sequences of rational numbers. They are precisely the transcendental numbers that can be more closely approximated by rational numbers than any algebraic irrational number. In 1844, Joseph Liouville showed that all Liouville numbers are transcendental, thus establishing the existence of transcendental numbers for the first time.
π and e are not Liouville numbers.^{[1]}
The existence of Liouville numbers (Liouville's constant)
Here we show that Liouville numbers exist by exhibiting a construction that produces such numbers.
For any integer b ≥ 2, and any sequence of integers (a_{1}, a_{2}, …, ), such that a_{k} ∈ {0, 1, 2, …, b − 1} for all k ∈ {1, 2, 3, …} and there are infinitely many k with a_{k} ≠ 0, define the number
In the special case when b = 10, and a_{k} = 1, for all k, the resulting number x is called Liouville's constant:
 L = 0.11000100000000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001...
It follows from the definition of x that its baseb representation is
where the nth term is in the (n!)th decimal place.
Since this baseb representation is nonrepeating it follows that x cannot be rational. Therefore, for any rational number p/q, we have x − p/q  > 0.
Now, for any integer n ≥ 1, define q_{n} and p_{n} as follows:
Then
Therefore, we conclude that any such x is a Liouville number.
Notes on the proof
 The inequality follows from the fact that there exists k, a_{k} ∈ {0, 1, 2, …, b−1}. Therefore, at most, a_{k} = b1. The largest possible sum would occur if the sequence of integers, (a_{1}, a_{2}, …), were (b1, b1, ...) where a_{k} = b1, for all k. will thus be less than, or equal to, this largest possible sum.
 The strong inequality follows from our motivation to eliminate the series by way of reducing it to a series for which we know a formula. In the proof so far, the purpose for introducing the inequality in 1. comes from intuition that (the geometric series formula); therefore, if we can find an inequality from that introduces a series with (b1) in the numerator, and if we can work to further reduce the denominator term to , as well as shifting the series indices from 0 to , then we will be able to eliminate both series and (b1) terms, getting us closer to a fraction of the form , which is the endgoal of the proof. We further this motivation here by selecting now from the sum a partial sum. Observe that, for any term in , since b ≥ 2, then , for all k (except for when n=1). Therefore, (since, even if n=1, all subsequent terms are smaller). In order to manipulate the indices so that k starts at 0, we select a partial sum from within (also less than the total value since it's a partial sum from a series whose terms are all positive). We will choose the partial sum formed by starting at k = (n+1)! which follows from our motivation to write a new series with k=0, namely by noticing that .
 For the final inequality , we have chosen this particular inequality (true because b ≥ 2, where equality follows if and only if n=1) because we wish to manipulate into something of the form . This particular inequality allows us to eliminate (n+1)! and the numerator, using the property that (n+1)!  n! = (n!)n, thus putting the denominator in ideal form for the substitution .
Irrationality
Here we will show that the number x = c/d, where c and d are integers and d > 0, cannot satisfy the inequalities that define a Liouville number. Since every rational number can be represented as such c/d, we will have proven that no Liouville number can be rational.
More specifically, we show that for any positive integer n large enough that 2^{n − 1} > d > 0 (that is, for any integer n > 1 + log_{2}(d ) ) no pair of integers (p, q ) exists that simultaneously satisfies the two inequalities
From this the claimed conclusion follows.
Let p and q be any integers with q > 1. Then we have,
If cq − dp  = 0, we would have
meaning that such pair of integers (p, q ) would violate the first inequality in the definition of a Liouville number, irrespective of any choice of n.
If, on the other hand, cq − dp  > 0, then, since cq − dp is an integer, we can assert the sharper inequality cq − dp  ≥ 1. From this it follows that
Now for any integer n > 1 + log_{2}(d ), the last inequality above implies
Therefore, in the case cq − dp  > 0 such pair of integers (p, q ) would violate the second inequality in the definition of a Liouville number, for some positive integer n.
We conclude that there is no pair of integers (p, q ), with q >1, that would qualify such an x = c/d as a Liouville number.
Hence a Liouville number, if it exists, cannot be rational.
(The section on Liouville's constant proves that Liouville numbers exist by exhibiting the construction of one. The proof given in this section implies that this number must be irrational.)
Uncountability
Consider, for example, the number
 3.1400010000000000000000050000....
3.14(3 zeros)1(17 zeros)5(95 zeros)9(599 zeros)2(4319 zeros)6...
where the digits are zero except in positions n! where the digit equals the nth digit following the decimal point in the decimal expansion of π.
As shown in the section on the existence of Liouville numbers, this number, as well as any other nonterminating decimal with its nonzero digits similarly situated, satisfies the definition of a Liouville number. Since the set of all sequences of nonnull digits has the cardinality of the continuum, the same thing occurs with the set of all Liouville numbers.
Moreover, the Liouville numbers form a dense subset of the set of real numbers.
Liouville numbers and measure
From the point of view of measure theory, the set of all Liouville numbers L is small. More precisely, its Lebesgue measure, λ(L), is zero. The proof given follows some ideas by John C. Oxtoby.^{[2]}^{:8}
For positive integers n > 2 and q ≥ 2 set:
we have
Observe that for each positive integer n ≥ 2 and m ≥ 1, we also have
Since
and n > 2 we have
Now
and it follows that for each positive integer m, L ∩ (−m, m) has Lebesgue measure zero. Consequently, so has L.
In contrast, the Lebesgue measure of the set of all real transcendental numbers is infinite (since the set of algebraic numbers is a null set).
Structure of the set of Liouville numbers
For each positive integer n, set
The set of all Liouville numbers can thus be written as
Each U_{n} is an open set; as its closure contains all rationals (the from each punctured interval), it is also a dense subset of real line. Since it is the intersection of countably many such open dense sets, L is comeagre, that is to say, it is a dense G_{δ} set.
Irrationality measure
The LiouvilleRoth irrationality measure (irrationality exponent, approximation exponent, or Liouville–Roth constant) of a real number x is a measure of how "closely" it can be approximated by rationals. Generalizing the definition of Liouville numbers, instead of allowing any n in the power of q, we find the largest possible value for μ such that is satisfied by an infinite number of integer pairs (p, q) with q > 0. This maximum value of μ is defined to be the irrationality measure of x.^{[3]}^{:246} For any value μ less than this upper bound, the infinite set of all rationals p/q satisfying the above inequality yield an approximation of x. Conversely, if μ is greater than the upper bound, then there are at most finitely many (p, q) with q > 0 that satisfy the inequality; thus, the opposite inequality holds for all larger values of q. In other words, given the irrationality measure μ of a real number x, whenever a rational approximation x ≅ p/q, p,q ∈ N yields n + 1 exact decimal digits, we have
for any ε>0, except for at most a finite number of "lucky" pairs (p, q).
Almost all numbers have an irrationality measure equal to 2.^{[3]}^{:246}
Below is a table of known upper and lower bounds for the irrationality measures of certain numbers.
Number  Irrationality Measure  Simple continued Fraction  Notes  

Lower bound  Upper bound  
Rational number where and  1  Finite continued fraction.  Every rational number has an irrationality measure of exactly 1.
Examples include 1, 2 and 0.5  
Algebraic number  2  Infinite continued fraction. Periodic if quadratic irrational.  By the ThueSiegelRoth theorem the irrationlity measure of any algebraic number is exactly 2. Examples include square roots like and and the golden ratio .  
2  Infinite continued fraction.  If the elements of the continued fraction expansion of an irrational number satisfy for positive and , the irrationality measure .
Examples include or where the continued fractions behave predictable: and  
2  
2  
The Trott constants in any base for simple continued fractions^{[4]}  1  2  It is not known whether the continued fraction expansions of the Trott constants are finite or infinite.  The Trott constants are precisely those constants whose decimal expansion equal their continued fraction expansion. In base 10 we have: .
Since in any base the number of digits is exactly itself, no term in the simple continued fraction can exceed and the terms are therefore bounded. This results in an irrationality measure of at most 2. 
^{[4]}^{[5]}  2  2.49846...  Infinite continued fraction.  , is a harmonic series. 
^{[4]}^{[6]}  2  2.93832...  , is a logarithm.  
^{[4]}^{[6]}  2  3.76338...  ,  
^{[4]}^{[7]}  2  3.57455...  
^{[4]}^{[8]}  2  5.11620...  
^{[4]}  2  5.51389...  
and ^{[4]}^{[9]}  2  5.09541...  and

and are linearly dependent over . 
^{[4]}^{[10]}  2  7.10320...  It has been proven that if the series (where n is in radians) converges, then 's irrationality measure is at most 2.5.^{[11]}^{[12]}  
^{[13]}  2  6.09675...  Of the form  
^{[14]}  2  4.788...  
^{[14]}  2  6.24...  
^{[14]}  2  4.076...  
^{[14]}  2  4.595...  
^{[14]}  2  5.793...  Of the form  
^{[14]}  2  3.673...  
^{[14]}  2  3.068...  
^{[15]}^{[16]}  2  4.60105...  Of the form  
^{[16]}  2  3.94704...  
^{[16]}  2  3.76069...  
^{[16]}  2  3.66666...  
^{[16]}  2  3.60809...  
^{[16]}  2  3.56730...  
^{[16]}  2  6.64610...  Of the form  
^{[16]}  2  5.82337...  
^{[16]}  2  3.51433...  
^{[16]}  2  5.45248...  
^{[16]}  2  3.47834...  
^{[16]}  2  5.23162...  
^{[16]}  2  3.45356...  
^{[16]}  2  5.08120...  
^{[16]}  2  3.43506...  
^{[14]}  4.5586...  and  
^{[14]}  6.1382...  and  
^{[14]}  59.976...  
^{[17]}  2  4  Infinite continued fraction.  where is the th term of the ThueMorse sequence. 
Champernowne constants in base ^{[18]}  Infinite continued fraction.  Examples include  
Liouville numbers  Infinite continued fraction, not behaving predictable.  The Liouville numbers are precisely those numbers having infinite irrationality measure.^{[3]}^{:248} 
Irrationality base
The irrationality base is a weaker measure of irrationality introduced by J. Sondow,^{[19]} and is regarded as an irrationality measure for Liouville numbers. It is defined as follows:
Let be an irrational number. If there exists a real number with the property that for any , there is a positive integer such that
 ,
then is called the irrationality base of and is represented as
If no such exists, then is called a super Liouville number.
Example: The series is a super Liouville number, while the series is a Liouville number with irrationality base 2. ( represents tetration.)
Liouville numbers and transcendence
Establishing that a given number is a Liouville number provides a useful tool for proving a given number is transcendental. However, not every transcendental number is a Liouville number. The terms in the continued fraction expansion of every Liouville number are unbounded; using a counting argument, one can then show that there must be uncountably many transcendental numbers which are not Liouville. Using the explicit continued fraction expansion of e, one can show that e is an example of a transcendental number that is not Liouville. Mahler proved in 1953 that π is another such example.^{[20]}
The proof proceeds by first establishing a property of irrational algebraic numbers. This property essentially says that irrational algebraic numbers cannot be well approximated by rational numbers, where the condition for "well approximated" becomes more stringent for larger denominators. A Liouville number is irrational but does not have this property, so it can't be algebraic and must be transcendental. The following lemma is usually known as Liouville's theorem (on diophantine approximation), there being several results known as Liouville's theorem.
Below, we will show that no Liouville number can be algebraic.
Lemma: If α is an irrational number which is the root of a polynomial f of degree n > 0 with integer coefficients, then there exists a real number A > 0 such that, for all integers p, q, with q > 0,
Proof of Lemma: Let M be the maximum value of f ′(x) (the absolute value of the derivative of f) over the interval [α − 1, α + 1]. Let α_{1}, α_{2}, ..., α_{m} be the distinct roots of f which differ from α. Select some value A > 0 satisfying
Now assume that there exist some integers p, q contradicting the lemma. Then
Then p/q is in the interval [α − 1, α + 1]; and p/q is not in {α_{1}, α_{2}, ..., α_{m}}, so p/q is not a root of f; and there is no root of f between α and p/q.
By the mean value theorem, there exists an x_{0} between p/q and α such that
Since α is a root of f but p/q is not, we see that f ′(x_{0}) > 0 and we can rearrange:
Now, f is of the form c_{i} x^{i} where each c_{i} is an integer; so we can express f(p/q) as
the last inequality holding because p/q is not a root of f and the c_{i} are integers.
Thus we have that f(p/q) ≥ 1/q^{n}. Since f ′(x_{0}) ≤ M by the definition of M, and 1/M > A by the definition of A, we have that
which is a contradiction; therefore, no such p, q exist; proving the lemma.
Proof of assertion: As a consequence of this lemma, let x be a Liouville number; as noted in the article text, x is then irrational. If x is algebraic, then by the lemma, there exists some integer n and some positive real A such that for all p, q
Let r be a positive integer such that 1/(2^{r}) ≤ A. If we let m = r + n, and since x is a Liouville number, then there exist integers a, b where b > 1 such that
which contradicts the lemma. Hence, if a Liouville number exists, it cannot be algebraic, and therefore must be transcendental.
See also
References
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(help)  ^ The irrationality measure of π does not exceed 7.6304, according to Weisstein, Eric W. "Irrationality Measure". MathWorld.