Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 14H (Logic and Set Theory)

For ordinals, the inductive definition of ordinal addition isThe synthetic definition takes a well-order of type $\alpha$, followed by a disjoint well-order of type $\beta$, with every point of the first preceding the second. They agree by transfinite induction on $\beta$: the empty second order does nothing; adding its last point takes the successor; at a limit ordinal, the concatenated order is the union of its initial concatenations, whose types have the displayed supremum.

The first assertion is false: $\boxed{1+\omega =\omega <\omega +1}$. Thus ordinal addition is not commutative.

False. Put $\alpha =\beta =1$ and $\gamma =\omega$. ThenThe two ordinals differ. The order of the factors matters in the distributive law for ordinal multiplication.

True. The valid distributive law for ordinal multiplication isThe synthetic definition of ordinal multiplication takes a sequence of copies of the left factor, indexed by the right factor. A sequence indexed first by $\beta$ and then by $\gamma$ is exactly the concatenation of the two corresponding sequences. Taking order types gives the identity. Equivalently, use transfinite induction on $\gamma$, with the successor rule and the supremum rule at limit ordinals.

True. If $\alpha \beta =\beta \alpha$, repeated use of associativity givesNo unlicensed commutativity of ordinal multiplication is being used: only the assumed commuting pair is swapped.

True. A useful fact is that squaring is strictly increasing on ordinals. If $A<B$, monotonicity in both arguments gives $A^2\le BA$, and strict increase in the right argument for the positive left factor $B$ gives $BA<B^2$. Hence $A^2<B^2$.

Suppose, for a contradiction, that $\alpha \beta <\beta \alpha$. Again using monotonicity and associativity of ordinal multiplication,This contradicts the hypothesized equality of the outer terms. Reversing the roles excludes $\beta \alpha <\alpha \beta$. If either factor is zero, they already commute. Therefore the commuting squares of ordinals property isThe proof uses monotonicity, which follows directly from the ordered-copy definition; it does not assume a cancellation law for ordinals.