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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-1/6/c/solution

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Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 1 / 6 / c / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

There is an exception to the printed assertion: for

β = - α

, the inner product is negative but

α + β = 0

is not a root. This already occurs in the rank-one root system

R = {α, - α}

.

For the intended assertion, assume

β \neq = - α

. By the preceding list, at least one of

n_{α, β}, n_{β, α}

equals

- 1

. If the first does, the reflection axiom gives

s_{α} (β) = β - n_{α, β} α = β + α \in R .

If the second does, use

s_{β} (α) = α + β

instead. Thus the precise reusable conclusion is

(α, β) < 0, β \neq = - α ⟹ α + β \in R .

This proves that obtuse nonopposite roots have a root sum directly from the reflection and integrality axioms, without appealing to the classification of root systems.

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