Is $6.12345678910111213141516171819202122\ldots$ transcendental?

My son was busily memorizing digits of $\pi$ when he asked if any power of $\pi$ was an integer. I told him: $\pi$ is transcendental, so no non-zero integer power can be an integer.

After tiring of memorizing $\pi$, he resolved to discover a new irrational whose expansion is easier to memorize. He invented (probably re-invented) the number $J$:

$$J = 6.12345678910111213141516171819202122\ldots$$

which clearly lets you name as many digits as you like pretty easily. He asked me if $J$ is transcendental just like $\pi$, and I said it must be but I didn't know for sure. Is there an easy way to determine this?

I can show that $\pi$ is transcendental (using Lindemann-Weierstrass) but it doesn't work for arbitrary numbers like $J$, I don't think.

Solution 1:

This is a transcendental number, in fact one of the best known ones, it is $6+$ Champernowne's number.

Kurt Mahler was first to show that the number is transcendental, a proof can be found on his "Lectures on Diophantine approximations", available through Project Euclid. The argument (as typical in this area) consists in analyzing the rate at which rational numbers can approximate the constant (see the section on "Approximation by rational numbers: Liouville to Roth" in the Wikipedia entry for Transcendence theory).

An excellent book to learn about proofs of transcendence is "Making transcendence transparent: an intuitive approach to classical transcendental number theory", by Edward Burger and Robert Tubbs.

Solution 2:

This is just the Champernowne constant plus $6$. Since the Champernowne constant is transcendental, so is this number (as $6$ is, of course, algebraic).