GRE8767 Q37

Forum for the GRE subject test in mathematics.
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Joined: Wed Oct 01, 2008 9:16 pm

GRE8767 Q37

Postby kaiserguy » Tue Oct 14, 2008 8:44 pm

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.


Posts: 10
Joined: Mon Oct 13, 2008 10:36 pm

Postby Kastro » Tue Oct 14, 2008 9:37 pm

You're on the right track, and from (4) we can certainly arrive at f(x) + f(-y) = f(-x) + f(y). This is not all that condition (3) says, however, and we can conclude nothing about f(xy) from this alone.

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.

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