Prove $f(x)f(y) = f(x + y)$ with $f(0) = 1$ implies $f(x) \neq 0$ for all $x$

Question

Prove that if $f$ is a function that satisfies $f(x)f(y) = f(x + y)$ for all real $x, y$ with $f(0) = 1$, then $f(x) \neq 0$ for all $x$.

I have the following solution given:

Letting $y = -x$ in $f(x)f(y) = f(x + y)$ gives $f(x)f(-x) = f(0) = 1$. Hence, we cannot have $f(x) = 0$ for any $x$ since $f(x)f(-x) \neq 0$.

I don't understand this solution. Why are we allowed to fix $y$ as $-x$. I understand that $f(x)f(-x)$ can never equal $0$, but doesn't this lose generality? Can someone please clarify?

Answer 1

You want to show that for every $x$ we have $f(x)\not=0$

If on the contrary, there exists an $x$ for which $f(x)=0$, then we have $$1=f(0)=f(x+(-x))=f(x)f(-x)=0\times f(-x)=0$$

That is $1=0$ which is not possible.

Thus such an $x$ does not exist.

Answer 2

Assume there is an $x$ with $f(x)=0$;

$0=f(x)f(y)=f(x+y)$.

$z:=x+y$, $y \in \mathbb{R}$ arbitrary;

$f(z)=0$, for all real $z$.

But $f(0)=1$, a contradiction.