A problem on ordinal arithmetic.

Question

Let $0 < \alpha \leq \beta$ be ordinals. Could anyone advise me on how to prove there exists $\delta $ and $\xi$ such that $\xi < \alpha$ and $\alpha \ . \delta + \xi = \beta?$(ordinal operations).

Thank you for your attention.

Answer 1

Use transfinite induction on $\beta$ to show first that $\alpha\cdot\beta\ge\beta$. This shows that there is a least $\gamma$ such that $\alpha\cdot\gamma\ge\beta$. Argue that either you have equality, or else $\gamma$ has to be a successor. In the latter case, if $\gamma=\delta+1$, we then have $\alpha\cdot\delta<\beta<\alpha\cdot\delta+\alpha$, and conclude: Consider the least $\xi$ such that $\alpha\cdot\delta+\xi\ge\beta$, and argue that we have equality, and therefore $\xi<\alpha$.

You can analyze the argument above a bit, and see that in fact we have also established uniqueness of the pair $(\delta,\xi)$.