GRE8767 Q37
Posted: 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.
Thanks,
David
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