f(xy) = f(x) * f(y)

As usual in functional equations, one could first try to subtitute some values for the variables. Since we have some information about $f(0)$ and want to know the value of $f(2)$ , we want them both to appear. So we substitute $x->0,y->2$ and obtain $f(0)=f(0)\cdot f(2)$ . $f(0)$ cancels out since it is not 0 and so we obtain $f(2) = 1$

f(0)=f(0.2)=f(0).f(2) since f(0) does not equal to 0 then we can divide both sides with f(0), resulting f(2)=1.

f(xy)=f(x).f(y) Putting x=0,y=0,we have f(0^2)=f(0).f(0) f(0)=[f(0)]^2 Either f(0)=0 or f(0)=1 But f(0) is not equal to 0 f(0)=1 Now, Putting x=0,y=2 f(0*2)=f(0).f(2) f(0)=f(0).f(2) Therefore f(2)=1

by taking x=2 and y=0 we get f(2 0) = f(2) f(0) which implies f(0)=f(2)*f(0) since f(0) is not zero we get f(2)=1

First, we substitute x=0, y=0 into the equation, which gives $f(0) = f(0) \cdot f(0)$ . This expression factorizes into $f(0) [ f(0)-1] = 0$ . Hence $f(0) = 0$ or $1$ . However, the challenge states that $f(0)\neq 0$ , thus we must have $f(0)=1$ .

Now, substitute in $x=0, y = 2$ into the equation, which gives $f(0) = f(0) \cdot f(2)$ . Since $f(0)=1$ , this simplifies to $1 = 1 \cdot f(2)$ , or that $f(2) = 1$ .

Note: Using the same method, the last paragraph actually shows that $f(t)=1$ for all values of $t$ . Specifically, substitute in $x=0, y=t$ into the equation (why can we do this?), which gives $f(0) = f(0) \cdot f(t)$ , which simplifies to $f(t) = 1$ .

$x = 2 , y = 0$

$f(2\times 0) = f(2)\times f(0) \Rightarrow f(0) = f(2)\times f(0) \Rightarrow f(2) = 1$

It is given f(xy)= f(x) . f(y) and that f(0) not equal to 1.

Hence we see ,

$f( 1\times0)$ = f(1) . f(0) and $f( 2\ times 0)$ = f(2). f(0) Hence we can conclude 2 things,

First , f(1) = f(2) as f(1). f(0) = f(2) . f(0) and that f(1) = 1 as it acts as a multiplicative identity with f(0)

Similarly we get f(x) =f(x+1) and then principle of induction allows that f(2) = 1 as f(1) = 0.