Using only the axioms of a field, prove that the additive identity in $\Bbb R$ is unique

Question

Using only the axioms of a field, prove that the additive identity in $\Bbb R$ is unique.

My work:

(1) Assume that the additive identity is not unique. So, let there be an $a$ and $b$ such that $x+a=x$ and $x+b=x$

(2) Take the additive inverse of x and add it to both sides in both equations. So, we have $x+a+x^{-1}=x+x^{-1}$ . And we have $x+b+x^{-1}=x+x^{-1}$

(3) Using the commutative and associative law we see that $a=0$ and $b=0$

(4) Therefore by the transitive property, we must have $a=b$

Are all of these steps valid or am I using more than just the axioms of a field?

Answer 1

I have answered my own question. I believe that this proof holds upon further inspection.

Answer 2

The truth is you don't even need the full field axioms. You can prove this just using the weaker axioms that the real numbers form a group under addition.

Def: Let (G,*) be an ordered pair where G is a nonempty set and let :G x G ----> G is a binary operation on G such that: (i) (Closure Axiom) For every a,b in G, ab is in G; (ii) (Associativity) For every a,b,c in G, (a*b)c= a(bc). (iii) (Identity) There exists a unique element e in G such that for every x in G, ex =xe =x; (iv) (Inverse) For every x in G, there exists a unique element y such xy=y*x =e.

Then (G,*) is a group. We usually just refer to the set G when referring to a specific group.

It's pretty easy to prove (R,+) is a group. Indeed,comparing the axiom systems,we can see a field is simply a group under addition with an additional associative and commutative binary operation of multiplication with a nonzero multiplicative inverse.(Ok, that's not a completely precise definition, but it's more then good enough for our purposes here.)

So just assume a and b are additive identities. So a = a + b =b + a =b. And that's it!

Answer 3

What you have written down is wrong most obviously by writing $x^{-1}$ for the additive inverse; whatever you call the inverse, that expression should not be used, as it is reserved for the multiplicative inverse.

However there is a deeper problem: there can be no axiom stating the existence of an additive inverse before one has uniqueness of the additive identity. You want to say that for every $a$ there exists $b$ with $a+b=0$, but that requires $0$ to designate the additive identity. Interestingly the definitions as reproduced in Wikipedia completely ignore uniqueness problems, and immediately stick names on neutral elements and inverse elements even though the axioms only speak about their existence. While maybe acceptable for exposition purposes, that is not a good way to formally set up axioms.

The main problem is uniqueness for neutral elements, since these elements have to be designated in order to formulate further axioms. One way out is to show beforehand that a two-sided neutral element for any operation is always unique; then as soon as the existence of a neutral element is proclaimed, this uniqueness can be invoked to justify introducing a symbol for it (which implicitly assumes uniqueness). The proof of this uniqueness is easy, which is why one usually does not make much fuss about it:

Let $\#$ be a binary operation, and $a,b$ two two-sided neutral elements for $\#$, that is, for any $x$ one has $a\#x=x=x\#a$ and $b\#x=x=x\#b$. Then $a=b$, since $a=a\#b=b$ (and "=" is transitive).

Note that commutativity ensures that in a field the neutral elements are automatically two-sided when they exist.

Still, in a formal approach it is not nice to need a proof based on part of the axioms, however trivial, in order to even be able to state the remaining axioms. Therefore the formally right thing to do is to introduce symbols $0$ and $1$, alongside with symbols for addition and multiplication, into the language of fields before even stating any axioms. The axioms then do not require existence of neutral elements, they just state properties valid for those symbols. In this case questions of their uniqueness do not even arise. (One can still prove, by the above argument, that there are no "look-alike neutral elements" in the field, but it is not something that needs to be proven urgently.)

In any case, using the existence of additive inverses to prove the uniqueness of the additive neutral element does not really make sense in any set-up.