# Liouville number

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In number theory, a Liouville number is a real number x with the property that, for every positive integer n, there exist integers p and q with q > 1 and such that

$0<\left|x-{\frac {p}{q}}\right|<{\frac {1}{q^{n}}}.$ 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 algebraic irrational numbers. In 1844, Joseph Liouville showed that all Liouville numbers are transcendental, thus establishing the existence of transcendental numbers for the first time.

## 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 (a1, a2, …, ), such that ak ∈ {0, 1, 2, …, b − 1} ∀k ∈ {1, 2, 3, …} and there are infinitely many k with ak ≠ 0, define the number

$x=\sum _{k=1}^{\infty }{\frac {a_{k}}{b^{k!}}}$ In the special case when b = 10, and ak = 1, ∀k, the resulting number x is called Liouville's constant:

• 0.11000100000000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001...

It follows from the definition of x that its base-b representation is

$x=\left(0.a_{1}a_{2}000a_{3}00000000000000000a_{4}0000\ldots a_{n}\left[\left(nn!-1\right){\text{zeros}}\right]a_{n+1}\ldots \right)_{b}\;.$ Since this base-b representation is non-repeating 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 qn and pn as follows:

$q_{n}=b^{n!}\,;\quad p_{n}=q_{n}\sum _{k=1}^{n}{\frac {a_{k}}{b^{k!}}}=\sum _{k=1}^{n}{a_{k}}{b^{n!-k!}}\;.$ Then

{\begin{aligned}0<\left|x-{\frac {p_{n}}{q_{n}}}\right|&=\left|x-{\frac {q_{n}\sum _{k=1}^{n}{\frac {a_{k}}{b^{k!}}}}{q_{n}}}\right|=\left|x-\sum _{k=1}^{n}{\frac {a_{k}}{b^{k!}}}\right|=\left|\sum _{k=1}^{\infty }{\frac {a_{k}}{b^{k!}}}-\sum _{k=1}^{n}{\frac {a_{k}}{b^{k!}}}\right|=\left|\left(\sum _{k=1}^{n}{\frac {a_{k}}{b^{k!}}}+\sum _{k=n+1}^{\infty }{\frac {a_{k}}{b^{k!}}}\right)-\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}}}+...={\frac {b-1}{b^{(n+1)!}b^{0}}}+{\frac {b-1}{b^{(n+1)!}b^{1}}}+{\frac {b-1}{b^{(n+1)!}b^{2}}}+...={\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, we conclude that any such x is a Liouville number.

### Notes on the proof

1. The inequality $\sum _{k=n+1}^{\infty }{\frac {a_{k}}{b^{k!}}}\leq \sum _{k=n}^{\infty }{\frac {b-1}{b^{k!}}}$ follows from the fact that "il existe" k, ak ∈ {0, 1, 2, …, b−1}. Therefore, at most, ak = b-1. The largest possible sum would occur if the sequence of integers, (a1, a2, …), were (b-1, b-1, ...) where ak = b-1, ∀ k. $\sum _{k=n+1}^{\infty }{\frac {a_{k}}{b^{k!}}}$ will thus be less than, or equal to, this largest possible sum.
2. The strong inequality {\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 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 $\sum _{k=0}^{\infty }{\frac {1}{b^{k}}}={\frac {b}{b-1}}$ (the geometric series formula); therefore, if we can find an inequality from $\sum _{k=n+1}^{\infty }{\frac {a_{k}}{b^{k!}}}$ that introduces a series with (b-1) in the numerator, and if we can work to further reduce the denominator term $b^{k!}$ to $b^{k}$ , as well as shifting the series indices from 0 to $\infty$ , then we will be able to eliminate both series and (b-1) terms, getting us closer to a fraction of the form ${\frac {1}{b^{(exponent)*n}}}$ , which is the end-goal of the proof. We further this motivation here by selecting now from the sum $\sum _{k=n+1}^{\infty }{\frac {b-1}{b^{k!}}}$ a partial sum. Observe that, for any term in $\sum _{k=n+1}^{\infty }{\frac {b-1}{b^{k!}}}$ , since b ≥ 2, then ${\frac {b-1}{b^{k!}}}<{\frac {b-1}{b^{k}}}$ , ∀ k (except for when n=1). Therefore, {\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, we select a partial sum from within $\sum _{k=n+1}^{\infty }{\frac {b-1}{b^{k}}}$ (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 $b^{(n+1)!}=b^{(n+1)!}b^{0}$ .
3. For the final inequality ${\frac {b}{b^{(n+1)!}}}\leq {\frac {b^{n!}}{b^{(n+1)!}}}$ , we have chosen this particular inequality (true because b ≥ 2, where equality follows if and only if n=1) because we wish to manipulate ${\frac {b}{b^{(n+1)!}}}$ into something of the form ${\frac {1}{b^{(exponent)*n}}}$ . 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 $q_{n}=b^{n!}$ .

## Irrationality

An equivalent definition to the one given above is that for any positive integer n, there exists an infinite number of pairs of integers (pq) obeying the above inequality.

Now 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 2n − 1 > d > 0 (that is, for any integer n > 1 + log2(d ) ) no pair of integers (pq ) exists that simultaneously satisfies the two inequalities

$0<\left|x-{\frac {p}{q}}\right|<{\frac {1}{q^{n}}}\,.$ From this the claimed conclusion follows.

Let p and q be any integers with q > 1. Then we have,

$\left|x-{\frac {p}{q}}\right|=\left|{\frac {c}{d}}-{\frac {p}{q}}\right|={\frac {|cq-dp|}{dq}}$ If |cq − dp | = 0, we would have

$\left|x-{\frac {p}{q}}\right|={\frac {|cq-dp|}{dq}}=0\,,$ meaning that such pair of integers (pq ) 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

$\left|x-{\frac {p}{q}}\right|={\frac {|cq-dp|}{dq}}\geq {\frac {1}{dq}}$ Now for any integer n > 1 + log2(d ), the last inequality above implies

$\left|x-{\frac {p}{q}}\right|\geq {\frac {1}{dq}}>{\frac {1}{2^{n-1}q}}\geq {\frac {1}{q^{n}}}\,.$ Therefore, in the case |cq − dp | > 0 such pair of integers (pq ) 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 (pq ), 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 non-terminating decimal with its non-zero digits similarly situated, satisfies the definition of a Liouville number. Since the set of all sequences of non-null 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 is zero. The proof given follows some ideas by John C. Oxtoby.:8

For positive integers n > 2 and q ≥ 2 set:

$V_{n,q}=\bigcup \limits _{p=-\infty }^{\infty }\left({\frac {p}{q}}-{\frac {1}{q^{n}}},{\frac {p}{q}}+{\frac {1}{q^{n}}}\right)$ we have

$L\subseteq \bigcup _{q=2}^{\infty }V_{n,q}.$ Observe that for each positive integer n ≥ 2 and m ≥ 1, we also have

$L\cap (-m,m)\subseteq \bigcup \limits _{q=2}^{\infty }V_{n,q}\cap (-m,m)\subseteq \bigcup \limits _{q=2}^{\infty }\bigcup \limits _{p=-mq}^{mq}\left({\frac {p}{q}}-{\frac {1}{q^{n}}},{\frac {p}{q}}+{\frac {1}{q^{n}}}\right).$ Since

$\left|\left({\frac {p}{q}}+{\frac {1}{q^{n}}}\right)-\left({\frac {p}{q}}-{\frac {1}{q^{n}}}\right)\right|={\frac {2}{q^{n}}}$ and n > 2 we have

{\begin{aligned}m(L\cap (-m,\,m))&\leq \sum _{q=2}^{\infty }\sum _{p=-mq}^{mq}{\frac {2}{q^{n}}}=\sum _{q=2}^{\infty }{\frac {2(2mq+1)}{q^{n}}}\\[6pt]&\leq (4m+1)\sum _{q=2}^{\infty }{\frac {1}{q^{n-1}}}\leq (4m+1)\int _{1}^{\infty }{\frac {dq}{q^{n-1}}}\leq {\frac {4m+1}{n-2}}.\end{aligned}} Now

$\lim _{n\to \infty }{\frac {4m+1}{n-2}}=0$ 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 T of all real transcendental numbers is infinite (since T is the complement of a null set).

In fact, the Hausdorff dimension of L is zero, which implies that the Hausdorff measure of L is zero for all dimension d > 0. Hausdorff dimension of L under other dimension functions has also been investigated.

## Structure of the set of Liouville numbers

For each positive integer n, set

$U_{n}=\bigcup \limits _{q=2}^{\infty }\bigcup \limits _{p=-\infty }^{\infty }\left\{x\in \mathbf {R} :0<\left|x-{\frac {p}{q}}\right|<{\frac {1}{q^{n}}}\right\}=\bigcup \limits _{q=2}^{\infty }\bigcup \limits _{p=-\infty }^{\infty }\left({\frac {p}{q}}-{\frac {1}{q^{n}}},{\frac {p}{q}}+{\frac {1}{q^{n}}}\right)\setminus \left\{{\frac {p}{q}}\right\}$ The set of all Liouville numbers can thus be written as

$L=\bigcap \limits _{n=1}^{\infty }U_{n}=\bigcap \limits _{n\in \mathbb {Z} ^{+}}\bigcup \limits _{q\in \mathbb {Z} _{\geqslant 2}}\bigcup \limits _{p\in \mathbb {Z} }\left(\left({\frac {p}{q}}-{\frac {1}{q^{n}}},{\frac {p}{q}}+{\frac {1}{q^{n}}}\right)\setminus \left\{{\frac {p}{q}}\right\}\right).$ Each Un is an open set; as its closure contains all rationals (the $\displaystyle p/q$ 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 Gδ set.

Along with the above remarks about measure, it shows that the set of Liouville numbers and its complement decompose the reals into two sets, one of which is meagre, and the other of Lebesgue measure zero.

## Irrationality measure

The irrationality measure (or irrationality exponent or 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 least upper bound of the set of real numbers μ such that

$0<\left|x-{\frac {p}{q}}\right|<{\frac {1}{q^{\mu }}}$ is satisfied by an infinite number of integer pairs (p, q) with q > 0. This least upper bound is defined to be the irrationality measure of x.: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

${\frac {1}{10^{n}}}\geq \left|x-{\frac {p}{q}}\right|\geq {\frac {1}{q^{\mu +\varepsilon }}}$ for any ε>0, except for at most a finite number of "lucky" pairs (p, q).

For a rational number α the irrationality measure is μ(α) = 1.:246 The Thue–Siegel–Roth theorem states that if α is an algebraic number, real but not rational, then μ(α) = 2.:248

Almost all numbers have an irrationality measure equal to 2.:246

Transcendental numbers have irrationality measure 2 or greater. For example, the transcendental number e has μ(e) = 2.:185 The irrationality measure of π is at most 7.60630853: μ(log 2)<3.57455391 and μ(log 3)<5.125.

The Liouville numbers are precisely those numbers having infinite irrationality measure.:248

### Irrationality base

The irrationality base is a weaker measure of irrationality introduced by J.Sondow and is regarded as an irrationality measure for Liouville numbers. It is defined as follows:

Let $\alpha$ be an irrational number. If there exists a real number $\beta \geq 1$ with the property that for any $\varepsilon >0$ , there is a positive integer $q(\varepsilon )$ such that

$\left|\alpha -{\frac {p}{q}}\right|>{\frac {1}{(\beta +\varepsilon )^{q}}}{\text{ for all integers }}p,q{\text{ with }}q\geq q(\varepsilon )$ ,

then $\beta$ is called the irrationality base of $\alpha$ and is represented as $\beta (\alpha )$ If no such $\beta$ exists, then $\alpha$ is called a super Liouville number.

Example: The series $\varepsilon _{2e}=1+{\frac {1}{2^{1}}}+{\frac {1}{4^{2^{1}}}}+{\frac {1}{8^{4^{2^{1}}}}}+{\frac {1}{16^{8^{4^{2^{1}}}}}}+{\frac {1}{32^{16^{8^{4^{2^{1}}}}}}}+\ldots$ is a super Liouville number.

## 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.

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,

$\left|\alpha -{\frac {p}{q}}\right|>{\frac {A}{q^{n}}}$ 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 α1, α2, ..., αm be the distinct roots of f which differ from α. Select some value A > 0 satisfying

$A<\min \left(1,{\frac {1}{M}},\left|\alpha -\alpha _{1}\right|,\left|\alpha -\alpha _{2}\right|,\ldots ,\left|\alpha -\alpha _{m}\right|\right)$ Now assume that there exist some integers p, q contradicting the lemma. Then

$\left|\alpha -{\frac {p}{q}}\right|\leq {\frac {A}{q^{n}}}\leq A<\min \left(1,{\frac {1}{M}},\left|\alpha -\alpha _{1}\right|,\left|\alpha -\alpha _{2}\right|,\ldots ,\left|\alpha -\alpha _{m}\right|\right)$ Then p/q is in the interval [α − 1, α + 1]; and p/q is not in {α1, α2, ..., αm}, 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 x0 between p/q and α such that

$f(\alpha )-f({\tfrac {p}{q}})=(\alpha -{\frac {p}{q}})\cdot f'(x_{0})$ Since α is a root of f but p/q is not, we see that |f ′(x0)| > 0 and we can rearrange:

$\left|\alpha -{\frac {p}{q}}\right|={\frac {\left|f(\alpha )-f({\tfrac {p}{q}})\right|}{|f'(x_{0})|}}=\left|{\frac {f({\tfrac {p}{q}})}{f'(x_{0})}}\right|$ Now, f is of the form $\sum _{i=0}^{n}$ ci xi where each ci is an integer; so we can express |f(p/q)| as

$\left|f\left({\frac {p}{q}}\right)\right|=\left|\sum _{i=0}^{n}c_{i}p^{i}q^{-i}\right|={\frac {1}{q^{n}}}\left|\sum _{i=0}^{n}c_{i}p^{i}q^{n-i}\right|\geq {\frac {1}{q^{n}}}$ the last inequality holding because p/q is not a root of f and the ci are integers.

Thus we have that |f(p/q)| ≥ 1/qn. Since |f ′(x0)| ≤ M by the definition of M, and 1/M > A by the definition of A, we have that

$\left|\alpha -{\frac {p}{q}}\right|=\left|{\frac {f({\tfrac {p}{q}})}{f'(x_{0})}}\right|\geq {\frac {1}{Mq^{n}}}>{\frac {A}{q^{n}}}\geq \left|\alpha -{\frac {p}{q}}\right|$ 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

$\left|x-{\frac {p}{q}}\right|>{\frac {A}{q^{n}}}$ Let r be a positive integer such that 1/(2r) ≤ A. If we let m = r + n, then, since x is a Liouville number, there exist integers a, b where b > 1 such that

$\left|x-{\frac {a}{b}}\right|<{\frac {1}{b^{m}}}={\frac {1}{b^{r+n}}}={\frac {1}{b^{r}b^{n}}}\leq {\frac {1}{2^{r}}}{\frac {1}{b^{n}}}\leq {\frac {A}{b^{n}}}$ which contradicts the lemma; therefore x is not algebraic, and is thus transcendental.