# 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?

• Could you list down the other axioms? For instance, is there any axiom that allows left/right cancellation? That should help you prove uniqueness. Dec 29, 2013 at 14:29
• Here is the whole description of the field axioms in the book. Dec 29, 2013 at 23:31
• Along with the set R of real numbers we assume the existence of two operations, called "addition" and "multiplication", such that for every pair of real numbers x and y the sum x+y and the product xy are real numbers uniquely determined by x and y satisfying the following axioms. [Axiom 1] x+y=y+x [Axiom 2] x+(y+z)=(x+y)+z, x(yz)=(xy)z [Axiom 3] x(y+z)=xy+xz [Axiom 4] Given any two real numbers x and y, there exists a real number z such that x+z=y. This z is denoted by y-x; the number x-x is denoted by 0.(It can be proved that 0 is independent of x.) We write -x for 0-x. Dec 29, 2013 at 23:46
• [Axiom 5] There exists at least one real number x=/=0. If x and y are two real numbers with x=/=0, then there exists a real number z such that xz=y. This z is denoted by y/x; the number x/x is denoted by 1 and can be shown to be independent of x. We write x^-1 for 1/x if x=/=0. The Axioms 1~5 are the whole field axioms in the book. I think it is different from the field axioms we usually see in other textbooks, where we normally set there is a 0 first and for each element x, we have 0+x=x. However, in this book, it equivalently describes that for each x, there is a 0(x) such that x+0(x)=x. Dec 30, 2013 at 0:06

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, $0_x=0_{x+y}=0_{y+x}=0_y$.

• I really appreciate this impressive answer. However, from this step (y+x)+0(x)=y+(x+0(x))=y+x to this step 0(x+y)=0(x), I think we need to apply the cancellation law, (y+x)+0(x)=y+x+0(x+y) => 0(x+y)=0(x), but in this phase we still cannot prove the law. Dec 30, 2013 at 0:21
• It is not cancellation but the uniqueness of $0_{x+y}$ that is used here. Dec 30, 2013 at 2:07
• Thank you very much!! Now, I totally understand. Dec 30, 2013 at 15:36

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.

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 $x$.

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.