SOLUTION: please help me with this If the function f satisfies the equation f(x+y)=f(x)+f(y) for every pair of real numbers x and y, what are the posible values of f(0)? any real number a

Question 256094: please help me with this
If the function f satisfies the equation f(x+y)=f(x)+f(y) for every pair of real numbers x and y, what are the posible values of f(0)?
any real number
any positive real nuber
0 and 1 only
1 only
0 only
and why?
Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
please help me with this
If the function f satisfies the equation f(x+y)=f(x)+f(y) for every pair of real numbers x and y, what are the posible values of f(0)?
any real number
any positive real nuber
0 and 1 only
1 only
0 only
and why?
Real Numbers are all numbers that can be expressed on an infinitely long number line.
Real Numbers include negative numbers ( such as -100, -33, -1 ) , zero (0) , positive numbers ( such as 1, 33, 100 ), rational numbers ( integer a / integer b where b is not 0 ) and irrational numbers ( such as pi, e, sqrt(2) ).
f(x+y)=f(x)+f(y)
example functions f(x)=x^2+x+1
f(y)=y-1
f(x)+f(y)=x^2+x+1+y+1
The two functions f(x) and g(x) can be added to make a new function h(x) where h(x)=f(x) + g(x). It is sometimes written as (f + g)(x).
here is another example:
Let f(x) = x + 3 and g(x) = 2x - 5
then (f + g)(x) = (x + 3) + (2x - 5)
so (f + g)(x) = 3x - 2
going back to original problem:
f(x+y)=f(x)+f(y)
f(0)=?
suggests x+y=0
example functions
f(x)=x^2+x+1
f(x+y)=(x+y)^2+(x+y)+1
f(0) with the above would result in 1
f(y)=y-1
f(x+y)=x+y-1
f(0) with the above would result in -1
f(x+y) = f(x) + f(y)
f(0) =f(0) + f (0)
we want f(x) + f(y) to equal 0
that means f(x) has to equal -1 * f(y) and vice versa
or f(x) and f(y) must be 0
We can only have zero only to satisfy this

RELATED QUESTIONS
The function f : R --> R satisfies f(x) f(y) = f(x + y) + xy for all real numbers x and... (answered by ikleyn)

If {{{ f(x+y) = f(x) + f(y) + xy + 1 }}} for all real numbers x and y and {{{ f(1)=1 }}}, (answered by Fombitz)

Please help me with this: The roots of a quadratic function y = f(x) are x = 2 and x =... (answered by Cromlix)

For all positive numbers a and b, a function f satisfies the equation f (a b) = f(a) +... (answered by richard1234)

Let f be a real-valued function such that f(x-f(y)) = f(x) - xf(y) for any real numbers x (answered by Edwin McCravy)

For all real numbers {{{x}}}, the function {{{f(x)}}} satisfies {{{2f(x) + f(1-x) =... (answered by Edwin McCravy)

I need some help with this one...can anyone help me out? Thanks! Directions: Let the (answered by praseenakos@yahoo.com)

Let f be a function such that f(x+y) = x + f(y) + f(y^2) - y^2 for any two real numbers (answered by Edwin McCravy,mccravyedwin)

The function f : \mathbb{R} \rightarrow \mathbb{R} satisfies f(x) f(y) - f(xy) = -2x -... (answered by CPhill,ikleyn,greenestamps)