 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-2/13i/b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

No such ordinal exists. For

α > 1

α^{2} = αα \geq α \cdot 2 = α + α > α

. Since ordinal exponentiation is nondecreasing in the exponent for base

α > 1

α^{ω} \geq α^{2} > α

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