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A very interesting function equation $f(a,b)=f(a,c)+f(c,b)$ implies $f(a,b)=g(a)-g(b)$.

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Function equation: $f(a,b)=f(a,c)+f(c,b)$ for all positive reals $a>c>b\geq 0$.

My solution:

$f(a,c)=f(a,b)-f(c,b)$,

Let $b=0$,

$f(a,c)=f(a,0)-f(c,0)$.

Define $g(a)=f(a,0)$. We get the answer.

It seems too simple; we don't even need the function be continuous or other conditions. Am I correct?


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