I know its a really basic question. I'm new to this subject and I'm having some troubles understanding how aritmetic of ordinals works. So, a detailed explanation could be really helpful. Find the smallest ordinal $ \gamma $ that satisfies
$ \omega+\gamma=\gamma $
Also, I have another basic question. Is it true that for every finite natural number $ n $ it follows that: $ n\omega=\omega n $
Thanks in advance.
The least such $\gamma$ is $\omega^2$. For contradiction assume there is a $\gamma<\omega^2$ such that $\omega+\gamma=\gamma$. The sequence $(0,\omega,\omega2,\omega3,\omega4,\ldots)$ has limit $\omega^2$, so $\gamma$ must fall somewhere within those ordinals in terms of size, i.e. there must be some $n\in\mathbb N$ where $\omega n\le \gamma<\omega(1+n)$. With a bit of induction it can be verified that $\omega+\omega n=\omega(1+n)$, and since ordinal addition is increasing in the right argument, we have $\gamma<\omega(1+n)=\omega+\omega n\le \omega+\gamma$, which contradicts our assumption that $\gamma=\omega+\gamma$.