Sum of two particular irrationals is irrational

Question

I am trying to show that a rational number say, $Q$, plus two specific irrational numbers, $\frac{\pi}{4}$ and $\ln(2)$, will produce an irrational number.

I realize that if I can show $\frac{\pi}{4} \pm \ln(2)$ is irrational then $Q$ plus that result will also be irrational.

I am feeling stuck because I know the irrationals aren't closed under addition and I don't think $\frac{\pi}{4} \pm \ln(2)$ will produce a rational, but how can I rigorously show this?

My question explicitly is: Does a rational number $Q + \frac{\pi}{4} + \ln(2)$ produces a rational number and does Q + $\frac{\pi}{4} - \ln(2)$ produces a rational number?

Answer 1

Transcendence of $\pi/4\pm \log2$

It is possible to answer your question from this paper that I linked: http://www.math.tifr.res.in/~saradha/tij.pdf

Theorem 1: If $T$ is defined by $$ T=\sum_{n=0}^{\infty}\frac{(-1)^n(\alpha n + \beta)}{(qn+s_1)(qn+s_2)} $$ where $\alpha, \beta \in \overline{\mathbb{Q}}$, $s_1, s_2\in\mathbb{Z}$, $|\alpha|+|\beta|>0$, and the cyclotomic polynomial $\Phi_{2q}$ is irreducible over $\mathbb{Q}(\alpha,\beta)$, $qn+s_1$, $qn+s_2$ do not vanish when $n\geq 0$, and assume $\alpha\neq 0$ if $s_1\equiv s_2$ mod $q$. Then $T$ is transcendental.

We apply the theorem for $\pi/4\pm \log 2$. Note that from Taylor series, $$ \pi/4= 1-\frac13+\frac15-+\cdots, \ \ \log2= 1 - \frac12 + \frac13 -+ \cdots, $$ Thus, $$ \pi/4+\log 2 = \sum_{n=0}^{\infty} (-1)^n \left(\frac1{2n+1} + \frac1{n+1} \right)=\sum_{n=0}^{\infty}(-1)^n \frac{6n+4}{(2n+1)(2n+2)}, $$ $$ \pi/4-\log 2 = \sum_{n=0}^{\infty}(-1)^n \left(\frac1{2n+1} - \frac1{n+1} \right)=\sum_{n=0}^{\infty} (-1)^n \frac{-2n}{(2n+1)(2n+2)}. $$

Therefore, by Theorem 1, the numbers $\pi/4\pm \log 2$ are transcendental, hence irrational.

Addendum

By applying Theorem 2 of the paper and How find this sum $\sum\limits_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)(3n+3)}$, we obtain that $$ \sum_{n=0}^{\infty}\dfrac{1}{(3n+1)(3n+2)(3n+3)}=\frac14\left(\frac{\pi}{\sqrt 3} - \log 3\right) $$ is transcendental.

A General Result

As @Ted Shifrin notes, a general result of this sort can be proved by Baker's theorem (which is a generalization of Gelfond-Schneider):

[Baker's Theorem]

If $\alpha_1, \ldots, \alpha_n$ are algebraic numbers, not $0$ or $1$. If $\log\alpha_1, \ldots, \log\alpha_n$ are linearly independent over $\mathbb{Q}$, then $1, \log\alpha_1, \ldots, \log\alpha_n$ are linearly independent over $\overline{\mathbb{Q}}$.

Using $\log(-1)=i\pi$ and the Fundamental Theorem of Arithmetic, we see that $\log(-1), \log 2, \log 3$ are linearly independent over $\mathbb{Q}$. Thus, by Baker's theorem, $1, \log(-1), \log 2, \log 3$ are linearly independent over $\overline{\mathbb{Q}}$. This implies that if $\beta_1, \beta_2, \beta_3$ are algebraic and not all zero, then $$ \beta_1 \pi+\beta_2 \log 2 + \beta_3 \log 3 $$ is transcendental.