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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 $B$?

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In an ordered field you have $1>0$, hence $1+1>0$. Now, this is enough to conclude that $1+1 \neq 0$, so that there exists its multiplicative inverse $a=(1+1)^{-1}$ since we are in a field. To prove that $a+a=1$, simply use distributive property: $$1=a(1+1)=a+a$$

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  • $\begingroup$ then how do you prove that two of those inverses add up to 1? $\endgroup$ – rr01 Mar 1 '17 at 8:35
  • $\begingroup$ @rr01 Perform the calculation $\frac{1}{1+1}+\frac{1}{1+1}$, and you'll see. $\endgroup$ – Arthur Mar 1 '17 at 8:38

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