If $f(x+y)=f(x)*f(y)$ and $f(0)$ is a non 0, what is $f(0)?$

Question

Question

If $f(x+y)=f(x)f(y)$ and $f(0)\ne0$, what is $f(0)?$

My thought process

I've decided to set $f(0)=f(x)f(y)$

and also as a side note i put $f(0)=f(0)f(0)$ incase if X and Y were both 0

But that's all I have gotten so far and I needed help with this

Answer 1

It is assumed that $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbb{R}$.

So you can choose, as a particular case, $x=0$ and $y=0$.

Then you get : $f(0) = f(0)^2$, that is $f(0)\left(1-f(0)\right)=0$.

Since it is assumed that $f(0)\neq0$, then you can divide both sides of the last equality by $f(0)$, and you get $1-f(0)=0$. Finally $f(0)=1$.

Note also that, knowing that, you can now choose $y=-x$ and get $f(0)=f(x)f(-x)$, so that :

$$\forall x\in\mathbb{R},\,f(-x)=\frac{1}{f(x)}$$

I suggest that, starting from that point you try to see what else can be derived concerning $f$ ...

Answer 2

Let $t=f(0)$. Then $t^2=t$. This is equivalent to

$$t(t-1)=0,$$

hence $t=0$ or $t=1$.