a) Give a direct proof of $\aleph_{\alpha} +\aleph_{\alpha} = \aleph_{\alpha}$ by expressing $\omega_{\alpha}$ as a disjoint union of two sets of cardinality $\aleph_{\alpha}$.
b) Give a direct proof of $n \cdot \aleph_{\alpha} = \aleph_{\alpha}$ by constructing a one-to-one mapping of $\omega_{\alpha}$ onto $n \times \omega_{\alpha}$, where $n$ is a positive natural number.
Remark: The exercises are taken from the book 'Introduction to Set Theory' by Hrbacek and Jech Chapter $7$ Section $2$
Attempt:
a) Note that cardinal addition $\kappa + \lambda $ is defined as $|A \cup B|$ where $|A|=\kappa$, $|B|=\lambda$ and $A \cap B=\emptyset$. Therefore, $\aleph_{\alpha} +\aleph_{\alpha} = \aleph_{\alpha}$ is the same as $|\omega_{\alpha} \cup \omega_{\alpha}|=|\omega_{\alpha}|$. But my problem is how to define a map by expressing $\omega_{\alpha}$ as a disjoint union of two sets of cardinality $\aleph_{\alpha}$ as $\omega_{\alpha}$ is not disjoint with itself.
b) Let $|C|=n$. I want to get a bijection $g:C \times \omega_{\alpha} \rightarrow \omega_{\alpha}$. My idea of the map is as follow: Arrange the pair $(1,1),(2,2),...,(n,n)$ as one column in a matrix. Then the second column is $(1,n+1),(2,n+2),...,(n,2n)$. So there will be a bijection between $C \times \omega_{\alpha}$ and $\omega_{\alpha}$. But I don't know how to formally define such function.
You misunderstand the first part, $\aleph_\alpha+\aleph_\alpha$ is $|2\times\omega_\alpha|$ which is the disjoint union of $\{0\}\times\omega_\alpha$ and $\{1\}\times\omega_\alpha$. And to show that there is a bijection between that disjoint union and $\omega_\alpha$ you just need to notice that you can define a notion of "even" and "odd" ordinals, and then use the same proof (essentially) that $\Bbb N$ and $\Bbb Z$ have the same cardinality.
The second part follows almost trivially by induction from the first.