Liouville's number revisited

Question

Liouville's Number is defined as $L = \sum_{n=1}^{\infty}(10^{-n!})$. Does it have other applications than just constructing a transcendental number?

(Personally, I would have defined it (as "Steven's Number" :-)) as binary: $S = \sum_{n=1}^{\infty}(2^{-n!})$, since each digit can only be "0" or "1": the corresponding power of 2 (instead of 10) included or not. Since according to Cantor most number are transcendental one can conjecture that this is also the case for Steven's Number. Can a proof for this be devised based on the proof for Liouville's number?)

I'm not a mathematician, so please type slowly! :-)

Answer 1

the number $S$ is also transcendental, for precisely the same reason as $L$: they both have rational approximations that are far too good to hold for an algebraic number. See here for more details.

If the specific number $L$ is useful for anything else, I'm not aware of it. It's really just a simple example, one of many which can be dreamed up, of a number which fails to be algebraic since it defies Liouville's Lemma in diophantine approximation.

Answer 2

Liouville's Number makes it's next appearance in Making Transcendence Transparent - Edward B. Burger, Robert Tubbs. The transcendence of $e + L$ is proved. (This might be the end of it's career though).