Is a homomorphic function under addition even on $\mathbb{R}$?

Question

Consider a function with the property

$$f(x +y) = f(x) +f(y)$$

Now

$$f(-y) = f(-2y + y) = f(-2y) + f(y) \tag{1}$$ $$ f(y) = f(2y + (-y)) = f(2y) + f(-y) \tag{2} $$

Substituting (2) into (1)

$$f(-y) = f(-2y) + f(2y) + f(-y)$$

$$ f(2y) = f(-2y)$$

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Edit: the last line should read $-f(2y) = f(-2y)$ if I would have done my arithmetic correctly.

Answer 1

No. In fact the function is odd, so it's not even unless it is 0.

Note that $$f(0)=f(0+0)=f(0)+f(0)$$ Hence $f(0)=0$. Now $$0=f(0)=f(y+(-y))=f(y)+f(-y)$$ Hence $$f(y)=-f(-y)$$