Why to prove: If $z$ and $a$ are elements in $\mathbb R$ with $z + a = a$, then $z=0$?

Question

I do not understand how this is a theorem in my textbook requiring a proof when it seems to me that this is simply the axiom of additive identity A.K.A existence of a 0 element. Wouldn't using axioms to prove another axiom contradict the results of Godel? Or at the very least be considered circular reasoning ?

Answer 1

I doubt that your textbook states that $z+a=a\implies z=0$ is an axiom. It is easy to prove, though:\begin{align}z&=z+0\\&=z+\bigl(a+(-a)\bigr)\\&=(z+a)+(-a)\\&=a+(-a)\\&=0.\end{align}Note that this uses not only the existence of $0$ but also the existence of an inverse.

And, no, there is no theorem by Gödel preventing the use of an axiom to prove another axiom.

Answer 2

The addition axioms say:

1) There exists an number $0$ so that $a+0=0+a=a $ for all $a $

But there is nothing in that axiom that says $0$ is unique or that there can't be several $a+b=a+c=a+d=a $ where $b,c,d$ might all be different. Or that those numbers might be unique to $a $ and not be true for $e+b $, $e+c $ etc.

BTW we could have defined an axiom that $a+b=c+b\implies a=c $ and then had the additive inverse axiom become something to be proven.

Apostol'Apostol's analysis text has an axiom i'I've never seen elsewhere that given any $a,z $ there exist a $b $ so that $a+b=z $ and subtraction is defined $b=z-a $. From that one axiom they define they have to prove that if $a+z=a $ and $b+q=b $ then $z=q $ regardless of what $a,b$ are. In other words, they have to prove $0$ exists. They also use one axiom to prove the existence of inverses as well as the existence of $0$.