Sum of two particular irrationals is irrational 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?
 A: 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. 
