In number theory, a Liouville number is a real number x {\displaystyle x} with the property that, for every positive integer n {\displaystyle n} , there exists a pair of integers ( p , q ) {\displaystyle (p,q)} with q > 1 {\displaystyle q>1} such that
0 < | x − p q | < 1 q n . {\displaystyle 0<\left|x-{\frac {p}{q}}\right|<{\frac {1}{q^{n}}}.}
The inequality implies that Liouville numbers possess an excellent sequence of rational number approximations. In 1844, Joseph Liouville proved a bound showing that there is a limit to how well algebraic numbers can be approximated by rational numbers, and he defined Liouville numbers specifically so that they would have rational approximations better than the ones allowed by this bound. Liouville also exhibited examples of Liouville numbers thereby establishing the existence of transcendental numbers for the first time. One of these examples is Liouville's constant
L = 0.110001000000000000000001 … , {\displaystyle L=0.110001000000000000000001\ldots ,}
in which the n {\displaystyle n} th digit after the decimal point is 1 if n {\displaystyle n} is the factorial of a positive integer and 0 otherwise. It is known that π and e, although transcendental, are not Liouville numbers.
The existence of Liouville numbers (Liouville's constant) Liouville numbers can be shown to exist by an explicit construction. For any integer b ≥ 2 {\displaystyle b\geq 2} and any sequence of integers a 1 , a 2 , … {\displaystyle a_{1},a_{2},\dots } such that a k ∈ { 0 , 1 , 2 , … , b − 1 } {\displaystyle a_{k}\in \{0,1,2,\ldots ,b-1\}} for all k {\displaystyle k} and a k ≠ 0 {\displaystyle a_{k}\neq 0} for infinitely many k {\displaystyle k} , define the number
x = ∑ k = 1 ∞ a k b k ! {\displaystyle x=\sum _{k=1}^{\infty }{\frac {a_{k}}{b^{k!}}}} . In the special case when b = 10 {\displaystyle b=10} , and a k = 1 {\displaystyle a_{k}=1} for all k {\displaystyle k} , the resulting number x {\displaystyle x} is called Liouville's constant:
L = 0. 11 000 1 00000000000000000 1 … {\displaystyle L=0.{\color {red}11}000{\color {red}1}00000000000000000{\color {red}1}\ldots }
It follows from the definition of x {\displaystyle x} that its base- b {\displaystyle b} representation is
x = ( 0. a 1 a 2 000 a 3 00000000000000000 a 4 … ) b {\displaystyle x=(0.a_{1}a_{2}000a_{3}00000000000000000a_{4}\ldots )_{b}}
where the n {\displaystyle n} th nonzero digit is in the n ! {\displaystyle n!} -th place. Since this base- b {\displaystyle b} representation is non-repeating, it follows that x {\displaystyle x} is not a rational number. Therefore, for any rational number p / q {\displaystyle p/q} , | x − p / q | > 0 {\displaystyle |x-p/q|>0} . Now, for any integer n ≥ 1 {\displaystyle n\geq 1} , p n {\displaystyle p_{n}} and q n {\displaystyle q_{n}} can be defined as follows:
q n = b n ! ; p n = q n ∑ k = 1 n a k b k ! = ∑ k = 1 n a k b n ! − k ! {\displaystyle q_{n}=b^{n!}\,;\quad p_{n}=q_{n}\sum _{k=1}^{n}{\frac {a_{k}}{b^{k!}}}=\sum _{k=1}^{n}a_{k}b^{n!-k!}} . Then
0 < | x − p n q n | = | ∑ k = 1 ∞ a k b k ! − ∑ k = 1 n a k b k ! | = ∑ k = n + 1 ∞ a k b k ! ≤ ∑ k = n + 1 ∞ b − 1 b k ! < ∑ k = ( n + 1 ) ! ∞ b − 1 b k = b − 1 b ( n + 1 ) ! + b − 1 b ( n + 1 ) ! + 1 + b − 1 b ( n + 1 ) ! + 2 + ⋯ = b − 1 b ( n + 1 ) ! b 0 + b − 1 b ( n + 1 ) ! b 1 + b − 1 b ( n + 1 ) ! b 2 + ⋯ = b − 1 b ( n + 1 ) ! ∑ k = 0 ∞ 1 b k = b − 1 b ( n + 1 ) ! ⋅ b b − 1 = b b ( n + 1 ) ! ≤ b n ! b ( n + 1 ) ! = 1 b ( n + 1 ) ! − n ! = 1 b ( n + 1 ) n ! − n ! = 1 b n ( n ! ) + n ! − n ! = 1 b ( n ! ) n = 1 q n n {\displaystyle {\begin{aligned}0<\left|x-{\frac {p_{n}}{q_{n}}}\right|&=\left|\sum _{k=1}^{\infty }{\frac {a_{k}}{b^{k!}}}-\sum _{k=1}^{n}{\frac {a_{k}}{b^{k!}}}\right|=\sum _{k=n+1}^{\infty }{\frac {a_{k}}{b^{k!}}}\\[6pt]&\leq \sum _{k=n+1}^{\infty }{\frac {b-1}{b^{k!}}}<\sum _{k=(n+1)!}^{\infty }{\frac {b-1}{b^{k}}}={\frac {b-1}{b^{(n+1)!}}}+{\frac {b-1}{b^{(n+1)!+1}}}+{\frac {b-1}{b^{(n+1)!+2}}}+\cdots \\[6pt]&={\frac {b-1}{b^{(n+1)!}b^{0}}}+{\frac {b-1}{b^{(n+1)!}b^{1}}}+{\frac {b-1}{b^{(n+1)!}b^{2}}}+\cdots ={\frac {b-1}{b^{(n+1)!}}}\sum _{k=0}^{\infty }{\frac {1}{b^{k}}}\\[6pt]&={\frac {b-1}{b^{(n+1)!}}}\cdot {\frac {b}{b-1}}={\frac {b}{b^{(n+1)!}}}\leq {\frac {b^{n!}}{b^{(n+1)!}}}={\frac {1}{b^{(n+1)!-n!}}}={\frac {1}{b^{(n+1)n!-n!}}}={\frac {1}{b^{n(n!)+n!-n!}}}={\frac {1}{b^{(n!)n}}}={\frac {1}{q_{n}^{n}}}\end{aligned}}}
Therefore, any such x {\displaystyle x} is a Liouville number.
Notes on the proof The inequality
∑ k = n + 1 ∞ a k b k ! ≤ ∑ k = n + 1 ∞ b − 1 b k ! {\displaystyle \sum _{k=n+1}^{\infty }{\frac {a_{k}}{b^{k!}}}\leq \sum _{k=n+1}^{\infty }{\frac {b-1}{b^{k!}}}}
follows since ak ∈ {0, 1, 2, ..., b−1} for all k, so at most ak = b−1. The largest possible sum would occur if the sequence of integers (a1, a2, ...) were (b−1, b−1, ...), i.e. ak = b−1, for all k. ∑ k = n + 1 ∞ a k b k ! {\displaystyle \sum _{k=n+1}^{\infty }{\frac {a_{k}}{b^{k!}}}} will thus be less than or equal to this largest possible sum.
The strong inequality ∑ k = n + 1 ∞ b − 1 b k ! < ∑ k = ( n + 1 ) ! ∞ b − 1 b k {\displaystyle {\begin{aligned}\sum _{k=n+1}^{\infty }{\frac {b-1}{b^{k!}}}<\sum _{k=(n+1)!}^{\infty }{\frac {b-1}{b^{k}}}\end{aligned}}} follows from the motivation to eliminate the series by way of reducing it to a series for which a formula is known. In the proof so far, the purpose for introducing the inequality in #1 comes from intuition that ∑ k = 0 ∞ 1 b k = b b − 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{b^{k}}}={\frac {b}{b-1}}} (the geometric series formula); therefore, if an inequality can be found from ∑ k = n + 1 ∞ a k b k ! {\displaystyle \sum _{k=n+1}^{\infty }{\frac {a_{k}}{b^{k!}}}} that introduces a series with (b−1) in the numerator, and if the denominator term can be further reduced from b k ! {\displaystyle b^{k!}} to b k {\displaystyle b^{k}} , as well as shifting the series indices from 0 to ∞ {\displaystyle \infty } , then both series and (b−1) terms will be eliminated, getting closer to a fraction of the form 1 b exponent × n {\displaystyle {\frac {1}{b^{{\text{exponent}}\times n}}}} , which is the end-goal of the proof. This motivation is increased here by selecting now from the sum ∑ k = n + 1 ∞ b − 1 b k ! {\displaystyle \sum _{k=n+1}^{\infty }{\frac {b-1}{b^{k!}}}} a partial sum. Observe that, for any term in ∑ k = n + 1 ∞ b − 1 b k ! {\displaystyle \sum _{k=n+1}^{\infty }{\frac {b-1}{b^{k!}}}} , since b ≥ 2, then b − 1 b k ! < b − 1 b k {\displaystyle {\frac {b-1}{b^{k!}}}<{\frac {b-1}{b^{k}}}} , for all k (except for when n=1). Therefore, ∑ k = n + 1 ∞ b − 1 b k ! < ∑ k = n + 1 ∞ b − 1 b k {\displaystyle {\begin{aligned}\sum _{k=n+1}^{\infty }{\frac {b-1}{b^{k!}}}<\sum _{k=n+1}^{\infty }{\frac {b-1}{b^{k}}}\end{aligned}}} (since, even if n=1, all subsequent terms are smaller). In order to manipulate the indices so that k starts at 0, partial sum will be selected from within ∑ k = n + 1 ∞ b − 1 b k {\displaystyle \sum _{k=n+1}^{\infty }{\frac {b-1}{b^{k}}}} (also less than the total value since it is a partial sum from a series whose terms are all positive). Choose the partial sum formed by starting at k = (n+1)! which follows from the motivation to write a new series with k=0, namely by noticing that b ( n + 1 ) ! = b ( n + 1 ) ! b 0 {\displaystyle b^{(n+1)!}=b^{(n+1)!}b^{0}} . For the final inequality b b ( n + 1 ) ! ≤ b n ! b ( n + 1 ) ! {\displaystyle {\frac {b}{b^{(n+1)!}}}\leq {\frac {b^{n!}}{b^{(n+1)!}}}} , this particular inequality has been chosen (true because b ≥ 2, where equality follows if and only if n=1) because of the wish to manipulate b b ( n + 1 ) ! {\displaystyle {\frac {b}{b^{(n+1)!}}}} into something of the form 1 b exponent × n {\displaystyle {\frac {1}{b^{{\text{exponent}}\times n}}}} . This particular inequality allows the elimination of (n+1)! and the numerator, using the property that (n+1)! − n! = (n!)n, thus putting the denominator in ideal form for the substitution q n = b n ! {\displaystyle q_{n}=b^{n!}} .
Irrationality Here the proof will show that the number x = c d , {\displaystyle ~x={\frac {c}{d}}~,} where c and d are integers and d > 0 , {\displaystyle ~d>0~,} cannot satisfy the inequalities that define a Liouville number. Since every rational number can be represented as such c / d , {\displaystyle ~c/d~,} the proof will show that no Liouville number can be rational. More specifically, this proof shows that for any positive integer n large enough that 2 n − 1 > d > 0 {\displaystyle ~2^{n-1}>d>0~} [equivalently, for any positive integer n > 1 + log 2 ( d ) {\displaystyle ~n>1+\log _{2}(d)~} )], no pair of integers ( p , q ) {\displaystyle ~(\,p,\,q\,)~} exists that simultaneously satisfies the pair of bracketing inequalities
0 < | x − p q | < 1 q n . {\displaystyle 0<\left|x-{\frac {\,p\,}{q}}\right|<{\frac {1}{\;q^{n}\,}}~.}
If the claim is true, then the desired conclusion follows. Let p and q be any integers with q > 1 . {\displaystyle ~q>1~.} Then,
| x − p q | = | c d − p q | = | c q − d p | d q {\displaystyle \left|x-{\frac {\,p\,}{q}}\right|=\left|{\frac {\,c\,}{d}}-{\frac {\,p\,}{q}}\right|={\frac {\,|c\,q-d\,p|\,}{d\,q}}}
If | c q − d p | = 0 , {\displaystyle \left|c\,q-d\,p\right|=0~,} then
| x − p q | = | c q − d p | d q = 0 , {\displaystyle \left|x-{\frac {\,p\,}{q}}\right|={\frac {\,|c\,q-d\,p|\,}{d\,q}}=0~,}
meaning that such pair of integers ( p , q ) {\displaystyle ~(\,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, | c q − d p | > 0 {\displaystyle ~\left|c\,q-d\,p\right|>0~} , then, since c q − d p {\displaystyle c\,q-d\,p} is an integer, we can assert the sharper inequality | c q − d p | ≥ 1 . {\displaystyle \left|c\,q-d\,p\right|\geq 1~.} From this it follows that
| x − p q | = | c q − d p | d q ≥ 1 d q {\displaystyle \left|x-{\frac {\,p\,}{q}}\right|={\frac {\,|c\,q-d\,p|\,}{d\,q}}\geq {\frac {1}{\,d\,q\,}}}