Spivak Calculus Chapter 3 Problem 19-(a)-(i) I have some trouble understanding the given solution of this problem. Furthermore I am not sure whether my own solution is correct. Problem 19-(a)-(i), Chapter 3 (Spivak; Calculus; 3rd edition):

Prove that there do not exist functions $f$ and $g$ with either of the following properties:
(i) $f(x)+g(x)=xy$ for all $x$ and $y$.

I tried to work out a solution and came up with the following:

*

*Let $y=1$, then $f(x)+g(1)=x$ for all $x$.

*Let $y=0$, then $f(x)+g(0)=0$ for all $x$.

Now, for example, let $x=2$. Then with 1. $f(2)+g(1)=2$ and with 2. $f(2)+g(0)=0$, thus $g(1)-g(0)=2$.
Let $x=3$. Then with 1. $f(3)+g(1)=3$ and with 2. $f(3)+g(0)=0$, thus $g(1)-g(0)=3$ also, which is already a contradiction to the statement that $g$ is a function.
Using the general equations 1. and 2., it follows that $g(1)-g(0)=x$ for all $x$, which is impossible for any function $g$. The same can be shown analogously for $f$.
Is this solution false? If so, why?
Spivak's answer:

19. (a) (i) If $f(x)+g(y)=xy$ for all $x$ and $y$, then, in particular, $$f(x)+g(0)=0$$ for all $x$. So $f(x)=-g(0)$ for all $x$, and $$-g(0)+g(y)=xy$$ for all $y$; setting $x=0$ we obtain $g(y)=g(0)$. So we must have $$0=-g(0)+g(0)=xy$$ for all $x$ and $y$, which is absurd.

Why it is possible to use the equation $f(x)=-g(0)$, where $y=0$, for the subsequent step $-g(0)+g(y)=xy$, which is said to be true for all $y$? (And similar in the last step where $x=0$.) In other words, why is $-g(0)+g(y)=xy$ true for all $y$, although $f(x)$ can only be replaced by $-g(0)$ when $y=0$?
 A: The equation $f(x)+g(0)=0$ for all $x$ establishes that $f(x)=-g(0)$ for all $x$. It does not establish that $f(x)=-g(0)$ for all $x$ and for $y=0$. $f$ is not a function of $y$ so the value of $f$ is independent of the value of $y$.
A: Your solution may be simplified by noting that the system $$f(x) + g(1) = x \\ f(x) + g(0) = 0$$ implies that the difference $$f(x) + g(1) - f(x) - g(0) = g(1) - g(0) = x$$ for all $x$; therefore, $g$ is not well-defined.  At this point, I would consider the reasoning valid.
As for Spivak's solution, he first establishes that $f(x) + g(0) = 0$ for all $x$ by choosing $y = 0$.  That means that for any choice of $x$, we must have $f(x) = -g(0)$.  this implies $f$ is a constant function.  Upon substituting back into the original functional equation, $xy = f(x) + g(y) = -g(0) + g(y)$ for all $y$.  This is allowed because $f(x)$ is constant for any $x$.  At this point, we have already run into a problem because $g(0)$ is a constant, $g(y)$ doesn't depend on $x$, yet their difference $g(y) - g(0) = xy$.  How can the LHS not depend on $x$, yet the RHS does?  To formalize this in terms of a logical contradiction, Spivak takes the argument to completion:  he then chooses $x = 0$, which then implies $$g(y) = g(0)$$ for all $y$.  Then he finally returns to the functional equation $$f(x) + g(y) = xy,$$ armed with both $f(x) = -g(0)$ and $g(y) = g(0)$, and concludes $$xy = f(x) + g(y) = -g(0) + g(0) = 0$$ for all $x$ and $y$, but this is absurd since selecting $x = y = 1$ implies $1 = 0$.
The entire proof relies on a strategy that you yourself employed:  to take the functional equation, set one variable to a particular value, and show that the resulting equation for the other variable, which is still valid for all values of that variable, results in a contradiction.
