Past exam of the mathematics course of the University of Cambridge 2005 iii Paper 27 8 Solution Created 2026-10-03 Updated 2026-10-14
Fix an ordinal . The Generalized Cantor normal form asserts that every nonzero ordinal has a unique expressionProducts, sums and powers here are ordinal multiplication, ordinal addition and ordinal exponentiation; the coefficients need not be finite if is infinite. The case is Cantor normal form, and a finite base on finite ordinals gives ordinary positional notation. Zero has the empty expansion.
We first prove the ordinal division algorithm. For , the function is strictly increasing, continuous at limit ordinals and unbounded. Hence there is a greatest with : the first with cannot be a limit by continuity and must equal . The interval after the initial segment has a unique order type , so . Maximality of gives . These facts also prove uniqueness of .
Similarly is strictly increasing, continuous at limits and unbounded, so there is a unique largest with . Divide by to obtainThe quotient is positive, and makes . If , repeat on ; its leading exponent is strictly smaller than . This process terminates, since an infinite run would give an infinite descending sequence of ordinals. It constructs the required form.
For uniqueness, a finite expression with largest exponent and coefficients below has value below . Prove this from the end backwards: a tail with largest exponent is below , andThus the displayed leading exponent is the uniquely determined largest with . The ordinal division algorithm uniquely determines its coefficient and the remaining tail, and recursion proves uniqueness of the whole expression. No commutative rearrangement of ordinal sums is used.