MathDB

81

Part of 1989 IMO Longlists

Problems(1)

Prove that f(a + b) = max{f(a), f(b)}

Source: IMO Longlist 1989, Problem 81

9/18/2008
A real-valued function f f on Q \mathbb{Q} satisfies the following conditions for arbitrary α,β∈Q: \alpha, \beta \in \mathbb{Q}:
(i) f(0) \equal{} 0, (ii) f(α)>0 if α≠0, f(\alpha) > 0 \text{ if } \alpha \neq 0, (iii) f(\alpha \cdot \beta) \equal{} f(\alpha)f(\beta), (iv) f(\alpha \plus{} \beta) \leq f(\alpha) \plus{} f(\beta), (v) f(m)≤1989 f(m) \leq 1989 ∀m∈Z. \forall m \in \mathbb{Z}.
Prove that f(\alpha \plus{} \beta) \equal{} \max\{f(\alpha), f(\beta)\} \text{ if } f(\alpha) \neq f(\beta).
functionalgebra unsolvedalgebra