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A binary operation is commutative if $a\ast b=b\ast a$ for every pair in its domain. A particular pair may commute even when the operation is not commutative. For example, ordinal addition and ordinal multiplication are not generally commutative.

Two ordinals commute under ordinal multiplication if and only if their squares commute. One implication follows from associativity. For the converse, squaring is strictly increasing: $A<B$ implies $A^2\le BA<B^2$. If $\alpha \beta <\beta \alpha$, monotonicity givesa contradiction to commuting squares. Swap the factors to exclude the reverse inequality; zero factors cause no exception.

Ordinal multiplication is distributive in its right argument:Indeed a sequence of copies of $\alpha$ indexed by the concatenation of $\beta$ and $\gamma$ is the concatenation of the two sequences of copies. The other distributive law fails: $(1+1)\omega =\omega$, while $\omega +\omega >\omega$.

Let nonzero ordinals $\xi ,\eta$ have leading exponents $\rho ,\sigma$. If $\eta$ is finite, then $\xi \eta$ has leading exponent $\rho$. If $\eta$ is infinite, so that $\sigma >0$, then $\xi \eta$ has leading exponent $\rho +\sigma$. This follows by multiplying the leading terms in Cantor normal form and using continuity of ordinal multiplication at limit ordinals.

The recursive definition of ordinal addition is $\alpha +0=\alpha$, $\alpha +(\beta +1)=(\alpha +\beta )+1$, and $\alpha +\lambda ={\sup }_{\beta <\lambda }(\alpha +\beta )$ for a nonzero limit ordinal $\lambda$. The synthetic definition is the order type of the ordered disjoint union of a copy of $\alpha$ followed by a copy of $\beta$.

For the synthetic operation, adding the empty order does nothing, adding a successor appends one final point, and concatenation with a limit order is the increasing union of the initial concatenations with all its proper initial segments. Their order types have supremum ${\sup }_{\beta <\lambda }(\alpha +\beta )$. Thus the synthetic operation satisfies the recursive equations. Transfinite induction gives uniqueness of the recursively defined operation, proving equivalence.

The recursive definitions of ordinal multiplication and ordinal exponentiation areThe last equation in each line is for nonzero limit $\lambda$. In particular the convention $0^0=1$ belongs to this recursive definition.

No such ordinal exists. For $\alpha >1$, ordinal multiplication gives ${\alpha }^2=\alpha \alpha \ge \alpha \cdot 2=\alpha +\alpha >\alpha$. Since ordinal exponentiation is nondecreasing in the exponent for base $\alpha >1$, ${\alpha }^{\omega }\ge {\alpha }^2>\alpha$.

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.

Ordinal exponentiation is defined by transfinite recursion:when $\lambda$ is a nonzero limit ordinal. If $\alpha \ge 1$, each successor step and each supremum show that $\beta \mapsto {\alpha }^{\beta }$ is nondecreasing. If $\alpha \ge 2$, transfinite induction strengthens this to strict increase: ${\alpha }^{\beta }<{\alpha }^{\beta +1}$, and the value at a limit exceeds every earlier value.

Strict increase can disappear when the base changes. For example, with $\omega <\omega +1<\omega +2$,because for every positive finite $n$, both finite powers have leading term ${\omega }^n$, and taking the supremum gives ${\omega }^{\omega }$.

We next proveApply transfinite induction to $\gamma$. The zero and successor cases follow directly from the recursive definitions and the associativity of ordinal multiplication. At a limit $\lambda$, continuity of ordinal multiplication in its right argument givesThe superficially similar identity $(\alpha \beta {)}^{\gamma }={\alpha }^{\gamma }{\beta }^{\gamma }$ is false: for $\alpha =\omega$, $\beta =2$, and $\gamma =2$,

Finally,For the first equivalence, bases zero and one fail, while strict increase embeds every countable ordinal into the powers of any base $\alpha \ge 2$. For the second, $\alpha \ge {\omega }_2$ immediately gives ${\alpha }^{{\omega }_1}\ge \alpha$. Conversely, if $\alpha <{\omega }_2$, then $|\alpha |\le {\aleph }_1$. Every exponent below the first uncountable ordinal ${\omega }_1$ is countable, so each ${\alpha }^{\beta }$ has cardinality at most ${\aleph }_1$; the supremum of ${\aleph }_1$ such ordinals still has cardinality at most ${\aleph }_1$ and is therefore below the second uncountable ordinal ${\omega }_2$.

Always true. Under the synthetic definition of ordinal multiplication, $\alpha \beta$ is the order type of $\beta$ consecutive copies of $\alpha$. Grouping the copies first into $\gamma$ blocks of $\beta$ copies, or first into single blocks of type $\alpha \beta$, gives the same lexicographic well-order. Hence

True. Left distributivity over a sum in the right argument,follows by transfinite induction on $\gamma$ from the recursive definition of ordinal multiplication; the successor and limit steps are exactly its defining clauses.