Simple proof...?

Question

if f(x+y) = f(x) + f(y), show that f(0) = 0 for all real pairs of x and y.

gabrielwyl, x nor y has to be 0, only x+y has to be 0.

I mean, x nor y doesn't have to be 0.

ConradD, your understanding of f(x+y) is not quite accurate. f(x+y) means that x+y is always coupled in the function. For example, (x+y)^2, e^(x+y), and so on. Furthermore, I said if f(x+y) = f(x) + f(y), which means that there are cases where this equation does not hold true.

pilot, you have to show that f(0) = 0, you can't use that statement to do anything else.

Kudos Aurora, but f(-x) = -f(x) only if the function is odd. If the function is even, then
f(-x) = f(x). What then?

Answer 1

If f(x+y) = f(x) + f(y)

Then f(0) = f(x) + f(-x) (since x+y=0)
= f(x) - f(x)
= 0 for all real pairs of x and y.

is this correct?

Answer 2

gabriel wyl is correct.
if f(x+y) = f(x) + f(y) for all x,y then f(0) = 0.

suppose f(0) is not equal 0

take x = y = 0 ( yes this is just a special case but that doesnt matter because you stated that it should hold for every x,y , if it only holds for x,y with x+y=0 what you stated later in your comments than it certainly also holds for x=0,y=0 since x+y=0)

then f(0) = 2f(0) this is ridiculous if we assume that f(0) != 0, thus the assumtion f(0) != 0 is false.
proof completed.

since f(0) = 0, it follows from f(x + (-x)) = f(x) + f(-x) = 0 that f(-x) = -f(x)

Answer 3

f(x+y) = f(x) +f(y)
For x = 0, y = 0, f(x+y) = f(0) +f(0)
f(0) = 2 (f(0)
f(0) = 0

Answer 4

If f(x+y)=0, and f(0)=0, that means that x+y=0 and y=0-x, if y=0-x, substitute that back into your equation, so you have x+(0-x)=x-x=0

Answer 5

This conjecture is false because if f(x)=x+5 for instance and f(y)=y+2 for instance then f(0+0)=7. It only takes 1 counterexample to disprove that.

Answer 6

Yes.

Answer 7

Quit doing your homework online and READ YOUR TEXTBOOK!

Answer 8

What?