Editing Liouville number (section)

== Liouville numbers and transcendence ==
'''No Liouville number is algebraic.''' The proof of this assertion proceeds by first establishing a property of [[irrational number|irrational]] [[algebraic number]]s. 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. The following [[lemma (mathematics)|lemma]] is usually known as '''Liouville's theorem (on diophantine approximation)''', there being several results known as [[Liouville's theorem (disambiguation)|Liouville's theorem]]<!--intentional link to DAB page-->.

'''Lemma:''' If <math>\alpha</math> is an irrational root of an irreducible polynomial of degree <math>n>1</math> with integer coefficients, then there exists a real number <math>A>0</math> such that for all integers <math>p,q</math> with <math>q>0</math>,
:<math>\left|\alpha-\frac{p}{q}\right|>\frac{A}{q^n}</math>

'''Proof of Lemma:''' Let <math>f(x)=\sum_{k\,=\,0}^na_kx^k</math> be a [[Minimal polynomial (field theory)|minimal polynomial]] with integer coefficients, such that <math>f(\alpha)=0</math>.

By the [[fundamental theorem of algebra]], <math>f</math> has at most <math>n</math> distinct roots.<br>
Therefore, there exists <math>\delta_1>0</math> such that for all <math>0<|x-\alpha|<\delta_1</math> we get <math>f(x)\ne0</math>.

Since <math>f</math> is a minimal polynomial of <math>\alpha</math> we get <math>f'\!(\alpha)\ne0</math>, and also <math>f'</math> is [[Continuous function|continuous]].<br>
Therefore, by the [[extreme value theorem]] there exists <math>\delta_2>0</math> and <math>M>0</math> such that for all <math>|x-\alpha|<\delta_2</math> we get <math>0<|f'\!(x)|\le M</math>.

Both conditions are satisfied for <math>\delta=\min\{\delta_1,\delta_2\}</math>.

Now let <math>\tfrac{p}{q}\in(\alpha-\delta,\alpha+\delta)</math> be a rational number. [[Without loss of generality]] we may assume that <math>\tfrac{p}{q}<\alpha</math>. By the [[mean value theorem]], there exists <math>x_0\in\left(\tfrac{p}{q},\alpha\right)</math> such that
:<math>f'\!(x_0)=\frac{f(\alpha)-f\bigl(\frac{p}{q}\bigr)}{\alpha-\frac{p}{q}}</math>
Since <math>f(\alpha)=0</math> and <math>f\bigl(\tfrac{p}{q}\bigr)\ne0</math>, both sides of that equality are nonzero. In particular <math>|f'\!(x_0)|>0</math> and we can rearrange:
:<math>\begin{align}\left|\alpha-\frac{p}{q}\right|&=\frac{\left|f(\alpha)-f\bigl(\frac{p}{q}\bigr)\right|}{|f'\!(x_0)|}=\frac{\left|f\bigl(\frac{p}{q}\bigr)\right|}{|f'\!(x_0)|}\\[5pt]&=\frac{1}{|f'\!(x_0)|}\left|\,\sum_{k\,=\,0}^na_kp^kq^{-k}\,\right|\\[5pt]&=\frac{1}{|f'\!(x_0)|\,q^n}\,\underbrace{\left|\,\sum_{k\,=\,0}^na_kp^kq^{n-k}\,\right|}_{\ge\,1}\\&\ge\frac{1}{Mq^n}>\frac{A}{q^n}\quad:\!0<A<\min\!\left\{\delta\,,\frac{1}{M}\right\}\end{align}</math>

'''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''

: <math>  \left| x - \frac{p}{q}  \right|> \frac{A}{q^{n}} </math>

Let ''r'' be a positive integer such that 1/(2<sup>''r''</sup>) ≤ ''A'' and define ''m'' = ''r'' + ''n''. Since ''x'' is a Liouville number, there exist integers ''a'', ''b'' with ''b'' > 1 such that

: <math>\left|x-\frac ab\right|<\frac1{b^m}=\frac1{b^{r+n}}=\frac1{b^rb^n} \le \frac1{2^r}\frac1{b^n} \le \frac A{b^n}, </math>

which contradicts the lemma. Hence a Liouville number cannot be algebraic, and therefore must be transcendental.

Establishing that a given number is a Liouville number proves that it is transcendental. However, not every transcendental number is a Liouville number. The terms in the [[simple continued fraction|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 (mathematical constant)|''e'']], one can show that ''e'' is an example of a transcendental number that is not Liouville. [[Kurt Mahler|Mahler]] proved in 1953 that [[pi|{{pi}}]] is another such example.<ref>Kurt Mahler, "On the approximation of π", ''Nederl. Akad. Wetensch. Proc. Ser. A.'', t.&nbsp;56 (1953), p.&nbsp;342–366.</ref>