if $A=\{2n-1:n\in\mathbb{N}\}$, $B=\{2n:n\in\mathbb{N}\}$, $C=\{4n:n\in\mathbb{N}\}$, $\alpha(n)=2n$, and$$\beta(n)=\begin{cases}
(n+1)/2 & \text{ if } n= odd, \
n/2 & \text{ if } n=even,
\end{cases}$$then $\alpha(A)=\{4n-2:n\in\mathbb{N}\}$, $\alpha(B)=\{4n:n\in\mathbb{N}\}$, and $\alpha(C)=\{8n:n\in\mathbb{N}\}$.am i rite?

Question

if $A=\{2n-1:n\in\mathbb{N}\}$, $B=\{2n:n\in\mathbb{N}\}$, $C=\{4n:n\in\mathbb{N}\}$, $\alpha(n)=2n$, and$$\beta(n)=\begin{cases}
(n+1)/2 & \text{ if } n= odd, \ 
n/2 & \text{ if } n=even,
\end{cases}$$then $\alpha(A)=\{4n-2:n\in\mathbb{N}\}$, $\alpha(B)=\{4n:n\in\mathbb{N}\}$, and $\alpha(C)=\{8n:n\in\mathbb{N}\}$.am i rite?

anonymous · Answer

btw with $\mathbb{N}$ i mean $\{1,2,...\}$

anonymous · Answer

some people like to $\mathbb{N}\cup\{0\}$ lol

anonymous · Answer

sure looks right to me.
i take it 
$$\beta(n)$$ is not important here

anonymous · Answer

oh yes theres part 2 lemme write it here btw thx so much

anonymous · Answer

$\beta(A)=\beta(B)=\mathbb{N}$ and $\beta(C)=B$?

anonymous · Answer

mr. satellite?

anonymous · Answer

$$\beta(B)=\mathbb N$$ for sure

anonymous · Answer

as is 
$$\beta(A)$$

anonymous · Answer

yes you have them all!

anonymous · Answer

thank you!!!