I am studying the field axioms for real numbers from the book 'Tom m Apostol Calculus' and i am wondering if the following can be proved using the field axioms only. a=a and b=(b).
the proofs that i have attempted are as follows
a+0=0+a ( associative law) a=0+a but '0+a' is not stated as a.
the proof for
(a+0)+0= a+(0+0) [ commutative law]
(a)+0=a+(0)
but (0) need not be equal to 0.
It's not an axiom of field, since it's just a matter of notation. It's a rule of brackets that $(x)=x$.
In fact addition is a binary operation on your field $K$, defined as a map $+: K \times K \longrightarrow K$ with $(a,b) \mapsto +(a,b)$ satisfying some axioms.
The axiom $a+0 = 0+a = a$ is rewritten as $+(a,0) = +(0,a) = a$.
So, for example, $$a+(0+0) = +(a, +(0,0)) = +(a, 0) = a$$