 Solution

The shuffle bound gives $\alpha \le \beta \#\beta$. By part (d), writeNatural addition doubles corresponding Cantor coefficients, soOn the other hand, ordinary ordinal addition absorbs each intermediate lower-order remainder, givingConsequently $\alpha <\beta +\beta +\beta$.

The stronger inequality with two copies is false. Take $\beta =\omega$, $\alpha =\omega +\omega$, let $X$ be the first block and $Y$ the second. Both have order type $\omega$, but $\alpha =\beta +\beta$.