Question

The function f(x;y), defined on pairs of real numbers, satisfies the conditions f(a;a) = 0, f(a; f(b;c)) = f(a;b)+c for any a, b, c. Find f(3.5 ; 7).

Answer 1

Let $f(a,a)=0$ be the condition (1) and $f(a,f(b,c))=f(a,b)+c$ be the condition (2).

We have

$f(a,b)+b=f(a,f(b,b))=f(a,0)=f(a,f(a,a))=f(a,a)+a=a$, that is $f(a,b)=a-b$

So we obtain that if $f$ satisfies the conditions (1) and (2), then $f(a,b)$ can be only $a-b$

Check whether $f(a,b)=a-b$ satisfies the conditions (1) and (2).

We have $f(a,a)=a-a=0$, so $f(a,b)=a-b$ satisfies the condition (1).

Next, $f(a,f(b,c))=f(a,b-c)=a-(b-c)=(a-b)+c=f(a,b)+c$, so $f(a,b)=a-b$ satisfies the condition (2).

We obtain that $f$ satisfies the conditions (1) and (2) if and only if $f(a,b)=a-b$, then $f(3.5;7)=3.5-7=-3.5$

Answer: $-3.5$