Tomi's Challenges - #12

$\Large \frac{a^2F(a)-a^2F(b)+2abF(a)-2abF(b)+b^2F(a)-b^2F(b)+4F(a)-4F(b)}{(a-b)x}=0$

$\Large \frac{a^2(F(a)-F(b))+2ab(F(a)-F(b))+b^2(F(a)-F(b))+4(F(a)-F(b))}{(a-b)x}=0$

$\Large \frac{(F(a)-F(b))((a+b)^2+4)}{(a-b)x}=0$

We have $a\neq b$ and $x\neq 0$ . Getting rid of the denominator, we get:

$(F(a)-F(b))((a+b)^2+4)=0$

The 2nd factor has no roots for any value of $a$ and $b$ . Taking the root of the first expression, we get $F(a)=F(b)$ .

So the solution to this equation is $((F(a)=F(b)) \wedge (a\neq b) \wedge (x\neq 0)) \vee (\emptyset \wedge (a\neq b) \wedge (x\neq 0))$ . This logical expression simplifies to the following, by the laws of algebras of sets (Note: We treat statements as sets here because they are actually sets that consist of all numbers such that the statement is true)

$(F(a)=F(b)) \wedge (a\neq b) \wedge (x\neq 0)$ .

This expression is logically equivalent to the following:

For all $a, b$ , $(\neg (F(a)=F(b) \Longrightarrow (a = b)) \wedge (x\neq 0))$ .

This means that $F$ is not injective, not bijective, but it is surjective by definition. Its argument can also never be equal to zero. We can see that the only function satisfying this condition is C. If its argument was equal to zero, the equation we just solved would be undefined and therefore $F(x)$ could be pretty much anything, which contradicts with the fact that it's uniquely determined by C.