Recently the following question came up in tutorial as I had a student looking at ordinal arithmetic. Basically, you want to show that if an ordinal is divisible on the right by 2 and by 3, then it is divisible on the right by 6. It caught me off guard, because how often are we called on to prove non-immediate facts about ordinal arithmetic? Here is the solution I finally came up with.
Lemma: Suppose \(\gamma\) and \(\delta\) are ordinals. Then
\[\gamma+\delta+\gamma =\delta+\gamma+\delta \Longleftrightarrow \gamma=\delta.\]
Proof. We work on the non-trivial implication, so suppose by way of contradiction that \(\gamma<\delta\) but
(*) \(\gamma+\delta+\gamma = \delta+\gamma+\delta.\)
Our first observation is
(**) \(\delta+\gamma<\gamma+\delta.\)
If not, then \(\gamma+\delta\leq\delta+\gamma\) and using our assumption that \(\gamma<\delta\) we obtain
\[(\gamma+\delta)+\gamma\leq(\delta+\gamma)+\gamma<(\delta+\gamma)+\delta,\]
contradicting (*).
From (**) we have
\[\gamma+(\delta+\gamma)<\gamma+(\gamma+\delta)\leq \delta+\gamma+\delta\]
contradicting (*) again. The same argument shows that assuming \(\delta<\gamma\) leads to a contradiction, and so the lemma holds. \(_\blacksquare\)
Problem: Suppose \(\epsilon\) is divisible on the right by both 2 and 3. Show \(\epsilon\) is divisible on the right by 6.
Solution. We may assume that \(\epsilon\) is non-zero (or else the conclusion is immediate), so suppose \(\alpha\) and \(\beta\) are non-zero ordinals for which
\[\epsilon = \alpha+\alpha = \beta+\beta+\beta.\]
Clearly \(\beta<\alpha\) and so we can fix a non-zero ordinal \(\gamma\) such that
\[\alpha=\beta+\gamma.\]
Combining these together, we find
\[\beta+\gamma+\beta+\gamma = \beta+\beta+\beta\]
and so
(1) \[\gamma+\beta+\gamma = \beta+\beta.\]
We now claim that \(\gamma\) is strictly less than \(\beta\). If not and \(\beta\leq\gamma\) then from our previous work we obtain
\[\gamma+\beta+\gamma\leq\gamma+\beta\]
and therefore
\[\beta+\gamma\leq\beta.\]
But this is absurd because \(\gamma\) is non-zero. Thus \(\gamma<\beta\) as claimed.
Give this, there is a non-zero ordinal \(\delta\) such that
\[\beta=\gamma+\delta.\]
Again, by (1) we have
\[\gamma+\beta+\gamma=\gamma+\gamma+\delta+\gamma=\gamma+\delta+\gamma+\delta\]
and so
\(\gamma+\delta+\gamma = \delta + \gamma+\delta.\)
Now we apply our lemma to conclude that \(\gamma=\delta\). In particular,
\(\beta = \gamma+\gamma\)
and
\(\alpha = \beta+\gamma = \gamma+\gamma+\gamma\)
Putting all this together, we obtain
\[\epsilon = \alpha + \alpha = \gamma\cdot 6 = \beta+\beta+\beta\]
and so \(\epsilon\) is divisible on the right by 6.