Multiplicatively monotone Euclidean norm admits unique Euclidean division for exact divisor

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Suppose R is an commutative unital ring with a Euclidean norm N that is multiplicatively monotone: N(ab) \ge \max \{ N(a), N(b) \} whenever ab \ne 0.

Suppose a = bc for b \ne 0. Then, we cannot write:

a = bq + r

with r \ne 0 and N(r) < N(b).

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Given: An integral domain R with multiplicatively monotone Euclidean norm N. b \ne 0 and a = bc.

To prove: We cannot write a = bq + r with N(r) < N(b).

Proof: Suppose a = bq + r with N(r) < N(b). Then, since a = bc, we get:

bc = bq + r \implies b(c - q) = r.

Take N both sides, to get:

N(b(c-q)) = N(r).

Since N is multiplicatively monotone, we get:

N(b(c-q)) \ge \max \{ N(b), N(c - q) \}.

On the other hand, we have, by assumption:

N(r) < N(b).

This is a contradiction, completing the proof.