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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 possible values of $f(0)$?

A.  Any real number
B.  Any positive real number
C.  $0$ and $1$ only
D.  $1$ only
E.  $0$ only

The answer for this problem is E. For the following problem to find the answer do you have to plug in 0 to prove the function?

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Does the equation plug in like this f(0+y)=f(0)+f(y)... this is where I am lost. – Little Jon Feb 16 '13 at 21:42
It's even easier. Just make $x=y=0$. – 1015 Feb 16 '13 at 21:43
@LittleJon: Yes, it is enough: $f(y)=f(0)+f(y)$, so substract $f(y)$ to get $f(0)=0$. – Berci Feb 16 '13 at 21:44
@LittleJon: If you are interested, you can read more about that functional equation over here:'s_functional_equation – Salech Alhasov Feb 17 '13 at 1:04
up vote 6 down vote accepted

Substitution of values alone will only confirm that $f(0) = 0$ is a value equal to $f(0)$, e. g. when $x = y = 0$.

Substitution of $x = y = 0$, e.g., gives us:

$$f(0)=f(0+0)=f(0)+f(0) \implies 0=f(0)$$

Now, we check whether $f(0)$ must be $0$ and no other value:

Suppose that $f(0) = a$ where $a\in \mathbb{R}$.

Then $$a = f(0) = f(0 + 0) = f(0) + f(0) = a + a =2a $$

$$a = 2a\implies a =0$$ is the only possible value of $f(0)$.

So $0$ is the one and only value satisfying of $f(0)$

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Is this clear, now, Little Jon? – amWhy Feb 16 '13 at 21:56
So a+a=2a was a demonstration and a=0 is one of the following numbers that could plug into the function. – Little Jon Feb 16 '13 at 21:57
Substitution yields that $f(0) = 0$, that 0 is a correct answer; we just want to rule out other possible answers. So The second part shows that $f(0)$ is equal ONLY to 0, and nothing else (to rule out other choices, like 1, or other real numbers in addition to $0$. It demonstrates that $f(0) = a = 0$ is the only number that works. – amWhy Feb 16 '13 at 22:01
Yes I understand thanks for the explanation leading to the answer! – Little Jon Feb 16 '13 at 22:04

Yes, use it for $x=y=0$: $$f(0)=f(0+0)=f(0)+f(0) \implies 0=f(0)$$

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Suppose that $f(0) = c$ where $c$ is any real number. Then \begin{align*} c &= f(0) \\ &= f(0 + 0) \\ &= f(0) + f(0) \\ &= c + c \\ &=2c \end{align*}

This tells us that $c = 2c$ which means $c=0$ is the only possible value.

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Just to add to the collection above, if the vector space is $V$:

$$f(v)=f(v+0)=f(v)+f(0)\Longrightarrow f(0)=0$$

for any $v\in V$

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