Editing Liouville number (section)

== Irrationality ==
Here the proof will show that the number <math>~ x = c / d ~,</math> where {{mvar|c}} and {{mvar|d}} are integers and <math>~ d > 0 ~,</math> cannot satisfy the inequalities that define a Liouville number. Since every [[rational number]] can be represented as such<math>~ c / d ~,</math> the proof will show that '''no Liouville number can be rational'''.

More specifically, this proof shows that for any positive integer {{mvar|n}} large enough that <math>~ 2^{n - 1} > d > 0~</math> [equivalently, for any positive integer <math>~ n > 1 + \log_2(d) ~</math>)], no pair of integers <math>~(\,p,\,q\,)~</math> exists that simultaneously satisfies the pair of bracketing inequalities

:<math>0 < \left|x - \frac{\,p\,}{q}\right| < \frac{1}{\;q^n\,}~.</math>

If the claim is true, then the desired conclusion follows.

Let {{mvar|p}} and {{mvar|q}} be any integers with <math>~q > 1~.</math> Then,

:<math> \left| x - \frac{\,p\,}{q} \right| = \left| \frac{\,c\,}{d} - \frac{\,p\,}{q}  \right| = \frac{\,|c\,q - d\,p|\,}{ d\,q }</math>

If <math>  \left| c\,q - d\,p \right| = 0~,</math> then

:<math>\left| x - \frac{\,p\,}{q}\right|= \frac{\,|c\,q - d\,p|\,}{ d\,q } = 0 ~,</math>

meaning that such pair of integers <math>~(\,p,\,q\,)~</math> would violate the ''first'' inequality in the definition of a Liouville number, irrespective of any choice of {{mvar|n}}&nbsp;.

If, on the other hand, since <math>~\left| c\,q - d\,p \right| > 0 ~,</math> then, since <math>c\,q - d\,p</math>  is an integer, we can assert the sharper inequality <math>\left| c\,q - d\,p \right| \ge 1 ~.</math>  From this it follows that

:<math>\left| x - \frac{\,p\,}{q}\right|= \frac{\,| c\,q - d\,p |\,}{d\,q} \ge \frac{1}{\,d\,q\,}</math>

Now for any integer <math>~n > 1 + \log_2(d)~,</math> the last inequality above implies

:<math>\left| x - \frac{\,p\,}{q} \right| \ge \frac{1}{\,d\,q\,} > \frac{1}{\,2^{n-1}q\,} \ge \frac{1}{\;q^n\,} ~.</math>

Therefore, in the case <math>~ \left| c\,q - d\,p \right| > 0 ~</math> such pair of integers <math>~(\,p,\,q\,)~</math> would violate the ''second'' inequality in the definition of a Liouville number, for some positive integer {{mvar|n}}.

Therefore, to conclude, there is no pair of integers <math>~(\,p,\,q\,)~,</math> with <math>~ q > 1 ~,</math> that would qualify such an <math>~ x = c / d ~,</math> as a Liouville number.

Hence a Liouville number cannot be rational.