Beginner Functional Equation

Algebra Level 3

A function f ⁣ : Q → Q f \colon \mathbb Q \to \mathbb Q satisfies f ( x + y ) = f ( x ) + f ( y ) f(x+y)=f(x)+f(y) .
The deduction that f ( x ) = a x f(x)=ax for some a ∈ Q a\in\mathbb Q is:

Correct Wrong

展豪 張 by 展豪 張 , 22, Hong Kong

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1 solution

展豪 張
Mar 10, 2016

For m , n ∈ Q m,n \in \mathbb Q ,
f ( m ) = f ( 1 + 1 + ⋯ + 1 ) = f ( 1 ) + f ( 1 ) + ⋯ + f ( 1 ) = m f ( 1 ) f(m)=f(1+1+\cdots +1)=f(1)+f(1)+\cdots +f(1)=mf(1)
n f ( m n ) = f ( m n ) + f ( m n ) + ⋯ + f ( m n ) = f ( m n + m n + ⋯ + m n ) = f ( m ) = m f ( 1 ) \displaystyle nf(\frac mn)=f(\frac mn)+f(\frac mn)+\cdots +f(\frac mn)=f(\frac mn+\frac mn+\cdots +\frac mn)=f(m)=mf(1)
∴ f ( m n ) = m n f ( 1 ) \displaystyle \therefore f(\frac mn)=\frac mnf(1)
The a a in the question is f ( 1 ) f(1)

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