Here we have both $\alpha$ and $\beta$ are ordinals. I need to prove that $ α + β = β\Leftrightarrow αω ≤ β$.
It seems that maybe the forward direction can be done by induction, but could someone please show me how should I do it? May I please ask how to prove this statement?
Recall that if $\eta\leq\delta$, then there is a unique $\gamma\leq\delta$ such that $\eta+\gamma=\delta$.
Now, in the one direction, $\alpha+\beta=\alpha+(\alpha\cdot\omega+\gamma)=(\alpha+\alpha\cdot\omega)+\gamma = \dots$
In the other direction, recall that $\alpha\cdot\omega=\sup\{\alpha\cdot n\mid n<\omega\}$, so by induction if $\alpha+\beta=\beta$, we get that $\alpha\cdot n+\beta=\beta$ for all $n$, therefore $\alpha\cdot n\leq\beta$ for all $n$. What does that tell you about $\alpha\cdot\omega$?