How does one prove that there exists a number $z$ such that $z+z = 1$ from the axioms of a complete ordered field?
The attempt so far:
Define set $A$ containing all $x$ such that $x+x < 1$. Define set $B$ containing all y such that $1 < y+y$. Now by the axiom of continuity, there exists an element $z$ such that any $x$ from $A$ is smaller than or equal to $z$, and any $y$ from $B$ is larger than or equal to $z$.
But then How do I prove that $z$ cannot belong to $A$ nor $B$?