such that 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[1] thereby establishing the existence of transcendental numbers for the first time.
[2] One of these examples is Liouville's constant in which the nth digit after the decimal point is 1 if
[3] Liouville numbers can be shown to exist by an explicit construction.
cannot satisfy the inequalities that define a Liouville number.
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
exists that simultaneously satisfies the pair of bracketing inequalities If the claim is true, then the desired conclusion follows.
would violate the first inequality in the definition of a Liouville number, irrespective of any choice of n .
would violate the second inequality in the definition of a Liouville number, for some positive integer n. Therefore, to conclude, there is no pair of integers
The proof of this assertion 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 cannot be algebraic and must be transcendental.
is an irrational root of an irreducible polynomial of degree
with integer coefficients, then there exists a real number
be a minimal polynomial with integer coefficients, such that
and we can rearrange: 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/(2r) ≤ A and define m = r + n. Since x is a Liouville number, there exist integers a, b with b > 1 such that which contradicts the lemma.
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.
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 is true of the set of all Liouville numbers.
From the point of view of measure theory, the set of all Liouville numbers
One could show even more - the set of Liouville numbers has Hausdorff dimension 0 (a property strictly stronger than having Lebesgue measure 0).
from each punctured interval), it is also a dense subset of real line.
It is defined by adapting the definition of Liouville numbers: instead of requiring the existence of a sequence of pairs
—a sequence which necessarily contains infinitely many distinct pairs—the irrationality exponent
is satisfied by an infinite number of integer pairs
satisfying the above inequality yields good approximations of