= Solution
\b[True.] For $\beta>0$, multiplication $\gamma\mapsto\beta\gamma$ is strictly increasing and <continuous> at <limit ordinals>, and $\beta\gamma\ge\gamma$. Thus there is a greatest $\gamma$ with $\beta\gamma\le\alpha$: the set is bounded, and its supremum still satisfies the inequality by continuity if it is a limit. The remaining ordered interval in $\alpha$ has an <ordinal> <order type> $\delta$, so $\alpha=\beta\gamma+\delta$. If $\delta\ge\beta$, then $\beta(\gamma+1)\le\alpha$, contradicting maximality. Hence $\delta<\beta$.
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