Hi,

This questionessentiallydepends on the domain and codomain of f.

For example, let's assume f is a function from to , we have f(0)=f(0-0)=f(0)f(0), so f(0)=1.

Then f(0)=f(3-3)=f(3)f(3), so f(3) =1 or f(3)=-1.

Now we have to prove that both choices are possible:

For the first case define f(x)=1 for all x in . This function satisfies f(x-y)=f(x)f(y), is never zero and f(3)=1.

For the second case define f(x)=1 for x even, f(x)=-1 for x odd. This function satisfies f(x-y)=f(x)f(y), is never zero and f(3)=-1.

Now let's assume f is a function from to .

We get f(0)=1 the same way as above.

For every x in we have f(0)=f(x-x)=f(x)f(x), so f(x)=1 or f(x)=-1 for every x in .

Next, f(-x)=f(0-x)=f(0)f(x)=f(x) for every x in .

We still can define f(x)=1 for all x, hence satisfying the case f(3)=1.

But case f(3)=-1 can't be satisfied now: look at f(1.5). As noted above it is either 1 or -1. In both cases we get: -1=f(3)=f(1.5-(-1.5))=f(1.5)f(-1.5)=f(1.5)f(1.5)=1, a contradiction.

So we cannot have f(3)=-1 for f .