Let $A = \{{x \in \mathbb{R} \mid x \cdot \pi \in \mathbb{Z}\}}$, $B = \{{a \in \mathbb{R^+}\mid \exists k \in \mathbb{N^+}, \exists q \in \mathbb{Q^+} : a^q = k \cdot q\}}$
Is $|A| = |B|$ ?
Attempt:
If we looking at the $A$ group we can see that $x\in \mathbb{R}$ such that $x\cdot\pi \in \mathbb{Z}$ that's mean $x$ must be equal to $\dfrac{n}{\pi}$.
Therefore, let $x=\dfrac{n}{\pi}$ , $n\in \mathbb{Z}$ for group $A$.
I defined function $$f:A\rightarrow \mathbb{Z}$$ $$f(x)=x\cdot\pi$$ The function is one to one, this means that $|A|\leq|\mathbb{Z}|$
