How should I interpret $y-x$ in this context?

Solution 1:

You should interpret $y-x$ as just a whole symbol when representing $z$ in the axiom 4. The first sentence does not define $-x$, the last sentence does instead.

What immediately follows from the first sentence is that there exists a real number $z$ such that $x+z=x$. We symbolically denote this $z$ by $x-x$, or alternatively $0(x)$. Notice that I didn't omit "$(x)$" for $0$. What is special about zero is that $0(x)$ is independent of $x$, that is, $0(x)=0(y)$ for $\forall x,y\in \mathbb{R}$. This follows from commutative laws and associative laws of addition, and uniquness of $0(x+y)$ (which is guaranteed in the first sentence of the axiom 4). Given this, we can safely denote $0(x)$ simply by $0$, and it immediately follows from the first sentence that there exists a real number $w$ such that $x+w=0$. While a 'straightforward' notation of $w$ is $0-x$, we denote it by $-x$ in practice. Thus the negative of $x$ is finally defined, and we can easily check that $y-x$ defined in the first sentence coincides with $y+(-x)$.

What the discussion above tells us is that $x+(-x)=0$ is trivial; it is the definition of $-x$! Note that we don't have to prove the uniquness of $-x$, because it is guaranteed by the first sentence of the axiom 4. Also, you can show that $y+(y-x)=(y+y)-x$ without knowledge of $y-x=y+(-x)$ (if you have any difficulty in proving this, please let me know).

Appendix: Proof of the uniqueness of zero

Let $x,y\in\mathbb{R}$. Associative laws of addition and definition of $0(x)$ indicate $$(x+y)+0(y)=x+(y+0(y))=x+y.$$ Commutative laws of addition is additionally required to show that $$(x+y)+0(x)=x+(y+0(x))=x+(0(x)+y)=(x+0(x))+y=x+y.$$ After all, we obtain $(x+y)+0(x)=x+y$ and $(x+y)+0(y)=x+y$, and uniqueness of $0(x+y)$ requires $0(x)=0(y)$.