Q. What is wrong with the following argument?

Let R e the real numbers

(1) "For all x,y in R, f(x)+f(y)=f(xy)

is equivalent to

(2) For all x,y in R, f(-x) +f(y)= f((-x)y)

which is equivalent to

(3) For all x,y in R, f(-x) + f(y) = f((-x)y)=f(x(-y))= f(x)+ f(-y)

From this for y=0, we make the conclusion that

(4) For all x in R, f(-x)=f(x)

Since the steps are reversible, any function with property (4) has property (1). Therefore, for all x,y in R cos x+ cos y =cos(xy)

(A) (2) does not imply (1)

(B) (3) does not imply (2)

(C) (3) does not imply (4)

(D) (4) does not imply (3)

(E) (4) is not true for f = cos

The answer is D. But I can't see how (3) implies (2). I think (4) does imply (3). We can add f(y) to both sides of (4) to get

f(x)+f(y) = f(-x) + f(y)

From (4) we know f(y)=f(-y), so we can change the first f(y) to f(-y) to arrive at f(x)+f(-y) = f(-x) + f(y) as required.

Maybe, I'm doing something stupid. Any help would be appreciated.

Thanks,

David

## GRE8767 Q37

Forget completely about the condition in (1), read the proof in reverse, and it should be clear that the middle two equalities in (3) are not necessarily true.