 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-2/22f/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Assume

u (0) = 0

. Applying midpoint preservation to

2 v

and

0

gives

u (v) = \frac{u ( 2 v ) + u ( 0 )}{2} = \frac{1}{2} u (2 v),

u (2 v) = 2 u (v)

. Consequently,

u (v + w) = u (2 \frac{v + w}{2}) = 2 u (\frac{v + w}{2}) = u (v) + u (w) .

Thus

u

is additive. It follows successively that

u (n v) = n u (v) (n \in Z), u (q v) = q u (v) (q \in Q) .

An isometry is continuous. For any

λ \in R

, choose rationals

q_{j} \to λ

. Then

u (λ v) = lim_{j \to \infty} u (q_{j} v) = lim_{j \to \infty} q_{j} u (v) = λ u (v) .

Together with additivity, this proves that

u

is real-linear. Hence the origin-fixing case of the Mazur-Ulam theorem gives

u (a v + b w) = a u (v) + b u (w)

for all

a, b \in R

and

v, w \in V

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