Prove the existence of $\frac{1}{2}$ from the following axioms The question is how to prove that there exists an element $z$ such that $z+z = 1$ from the following axioms (assume we are talking about set $R$):
A1: $x≠y$ implies $x < y$ or $x > y$;
A2: $x < y$ implies not $y < x$
A3: $x < y$ implies there exists $t$ such that $x < t < y$
A4: For any two sets $S$ and $T$ which is a subset of $R$, if (any $x$ from $S$ and $y$ from $T$ implies $x < y$) then (there exists a $z$ such that for all $m$ from $S$ and $n$ from $T$, $m≠z$ and $n≠z$ implies $m < z < n$)
A5: $x+(y+z)=(x+z)+y$
A6: For all $x$ and $y$, there exists $z$ such that $x = y+z$
A7: $x+z < y+t$ implies $x < y$ or $z < t$
A8: $1$ is an element of $R$
A9: $1 < 1+1$
So far following the textbook, I construct set $K$ containing all $x$ such that $x+x < 1$, and set $L$ containing all $y$ such that $1 < y+y$. Now by A4, there exists an element $z$ such that any $x$ from $K$ is smaller than or equal to $z$, and any $y$ from $L$ is larger than or equal to $z$. It is all good up to this point. Now I try to prove that z cannot belong to K nor L. 
Assume $z$ is an element of $K$. Then $z+z < 1$ and there exists an element $t$ such that $z+z < t < 1$ by A3. Define set $N$ containing all $p$ such that $p+p < t$. Then by A4, there exists $q$ such that any $p$ from $N$ is smaller than or equal to $q$, and any $y$ from $L$ is larger than or equal to $q$. But now I have trouble proving that $q ≠ z$. If $q ≠ z$ then the contradiction is immediate. 
For anyone wondering, the textbook is "Introduction to Logic and the Methodology of the Deductive Sciences," by none other than A. Tarski himself. Chapter 10, exercise 5.
 A: There is something fundamentally wrong with this question. Axiom A4 talks about sets, but that makes sense only within a meta-system that 'knows' what sets are, not within any axiomatization for $R$. In particular, within the axiomatization as stated, A4 is simply useless because there are no set specification axioms and so one cannot create anything that A4 can be applied to! Thus it makes no sense to ask for a proof of something from those axioms. One cannot just brush the issue aside and say that we can construct any set of objects that we wish, otherwise we can immediately get a contradiction via Russell's set. One can still work within the meta-system and ask whether every model (with full semantics) for the axiomatization satisfies "$\exists x\ ( x+x = 1 )$", which does have a positive answer...
$
\def\nn{\mathbb{N}}
\def\zz{\mathbb{Z}}
\def\rr{\mathbb{R}}
$


Actually it was proven in the textbook that A1,2,5,6,7 are equivalent to an abelian group with respect to the operation +. I assumed that I could use this.

Just for completeness here is a sketch of the proof, in case future readers want to see how it is done. Thanks to Rutger Moody for finding how to get commutativity, which was the major part I could not figure out despite the simplicity of the solution. After that associativity immediately follows, and then it is not hard to get existence of additive identity.
Take any $x,y,z \in R$. Let $w \in R$ such that $x = y+w$. Then $y+x = y+(y+w)$ $= (y+w)+y$ $= x+y$. Thus $z+(x+y) = z+(y+x) = z+(x+y)$.
Therefore $+$ is commutative and associative on $R$ and we can omit brackets from now on.
Let $o \in R$ such that $1 = 1+o$. Take any $x \in R$. Let $y \in R$ such that $x = 1+y$. Then $x+o = 1+y+o = 1+o+y = 1+y = x$.
Therefore $o$ is an identity for $+$ on $R$.

From this we get that $R$ is an ordered abelian group. Specifically, for any $x,y,z \in R$ such that $x < y$, we have $x \ne y$ and $\neg z > z$ by A2, and hence $x+z < y+z$ by A7 and the abelian group properties. Thus if $x+y = z$ and $x < z$, then $y > 0$, where $0$ denotes the additive identity of $R$. Therefore if $x < y$ and $y < z$, then letting $a,b \in R$ such that $x+a = y$ and $y+b = z$, we get $x+(a+b) = z$ and $a+b > 0$ since $a,b > 0$, and hence $x < z$.
Your attempt had used this property of transitivity in applying A4 to $K,M$ without justification!
But then the desired claim is not hard. For convenience let "$x \le y$" be short for "$x < y \lor x = y$".

Let $K = \{ x : x \in R \land x+x < 1 \}$.
Let $M = \{ x : x \in R \land x+x > 1 \}$.
Let $z \in R$ such that $x \le z \le y$ for any $x \in K$ and $y \in M$.
If $z+z < 1$:
  Let $t \in R$ such that $z+z+t = 1$. Then $t > 0$.
  Let $u \in R$ such that $0 < u < t$.
  Let $v \in R$ such that $u+v = t$. Then $v > 0$.
  Also $v < t$, otherwise $t = u+v > 0+t = t$.
  By A1 and symmetry we can assume that $u \le v$.
  Let $w \in R$ such that $0 < w < u$.
  Thus $(z+w)+(z+w) < z+z+u+v = 1$ and hence $z+w \in K$.
  But $z+w > z$, contradicting the definition of $z$.
Therefore $\neg z+z < 1$.
By symmetry $\neg z+z > 1$ and hence $z+z = 1$.
