One question about additive identity arising from Apostol's field axiom in the Mathematical Analysis 2nd edition In the begining of this book, the field axiom 4 talks about "Given any two real numbers x and y, there exists a real number z such that x+z=y and this z is denoted by y-x." Therefore, for each x, we can denote an element 0 by x-x. Apostol further describes that this 0 can be proved to be independent of the choice of the element x. Now, here is my question: how can we prove the independence of 0. I think it might be difficult or tricky to me because I tried to suppose there are 0(1) and 0(2) for x(1) and x(2), respectively, and to prove 0(1)=0(2), but I still cannot work it out. Can anyone help me?
 A: Let $0_x$ denote the (unique? I shall assume this follows from the other axioms, though you didn't mention it) solution of $x+0_x=x$. Then (assuming that the associativity of addition is among the other axioms)
$(y+x)+0_x=y+(x+0_x)=y+x$ which implies $0_{x+y}=0_x$. By the same argument $0_{y+x}=0_y$. And hence, assuming that commutativity of addition is among the other axioms, $0_x=0_{x+y}=0_{y+x}=0_y$.
A: I just found a way to prove the independence of $0$ without using the uniqueness. Given an $x$ in $R$ there exists $0_x$ such that $x+0_x=x$. For any real number $y$ in $R$, by the Axiom4 itself, there is a number $z$ such that $x+z=y$. Thus, $0_x+y=0_x+(x+z)=(0_x+x)+z=x+z=y$. Hence this $0$ is independent of the choice of $x$ and the uniqueness of $0$ therefore holds. 
A: We need to show if $x,y$ are real, z is unique if $x+z=y$.
Let $0_x^1$ be the real number s.t. $x+0_x^1=x$ (it exists by the axiom).
Let $(-x)_1$ be the real number s.t. $x+(-x)_1=0_x^1$.
Note $0_x^1 ,(-x)_1$ may not be unique (for now).
For any real number s, there exists a $w$ s.t. $s=x+w$ by the axiom. So $$0_x^1+s=0_x^1+x+w=x+w=s \ for \  all\ s.$$If $x+z'=y$ too, then $$z=z+0_x^1=z+x+(-x)_1=y+(-x)_1$$And$$z'=z'+0_x^1=z'+x+(-x)_1=y+(-x)_1=z$$
As it is given that x+y is unique for all real x,y.  Now we can say $0_x^1 = 0_x$.
Since $0_x+s =s $ for any $s$, $0_x=0_s$, so $0$ is independent of $x$.
A: To say that $0$ depends on the choice of $x$ is to say that $x-x\neq y-y$. Suppose that $$x-x\neq y-y$$
then 
$$x+y \neq y+x$$
But by associativity of addition, $$x+y = y+x$$
We have thus reached a contradiction.
