How to prove for all $x$ and $y$, there exists a $z$ such that $x+z=y$ with only using field axioms for real numbers?

Question

How would I prove that for all $x$ and $y$, there exists a $z$ such that $x+z=y$?

Answer 1

Maybe this can help.

Let $F$ be a field and let $x,y\in F$. We know that $(F,+)$ is an abelian group. Therefore, $-x\in F$. Take $z=y+(-x)$. Then $z\in F$ and applying commutative and associative properties under $+$, we get $$x+z=x+[y+(-x)]=x+[-x+y]=[x+(-x)]+y=0+y=y,$$ where $0$ is the additive identity in $F$.

Answer 2

• Solve for $z$.
• Show that formula for $z$ can be computed from field operations.
• Prove that the formula for $z$ is a solution using the field axioms.