100 followers problem!

$\large \text{(A)}$

Consider $\alpha = \sqrt{2}-1$ and $\beta = 2-\sqrt{2}$ , then both $\alpha$ and $\beta$ are irrational but $\alpha + \beta = 1$ is rational (note that $0 < \alpha, \beta < 1$ is also true in this example).

Given one counterexample, statement (A) is definitely not true for all cases.

$\large \text{(B)}$

Since $|c| < 1$ , so we have

$\displaystyle \sum_{i=0}^{\infty} \alpha c^i = \alpha \sum_{i=0}^{\infty} c^i = \alpha \left( \dfrac{1}{1-c} \right)$

Now, $\dfrac{1}{1-c} \in \mathbb{Q}$ and $\dfrac{1}{1-c} \neq 0$ for any $0<c<1$ and $c \in \mathbb{Q}$ . Therefore, since $\alpha$ is irrational, so $\alpha \left( \dfrac{1}{1-c} \right)$ must be irrational.

$\large \text{(C)}$

The integration can be solved as

$\begin{aligned} \displaystyle \int_0^\pi \left( \beta \cos(x) + c \right) \,dx &= \beta \int_0^\pi \cos(x) \,dx + c \int_0^\pi \,dx \\ &= \beta \left[ \sin(x) \right] {\huge{|}}_{0}^{\pi} + c \left[ x \right] {\huge{|}}_{0}^{\pi} \\ &= c \pi \end{aligned}$

We know that $\pi$ is irrational and $c \in \mathbb{Q}$ and $c \neq 0$ , thus $c \pi$ must be irrational.

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Hence options (B) and (C) follow.