Formalizing a proof regarding real numbers. The Question:
Let $ x,y,a,z \in \mathbb{R}$ 
$ x \lt y $
$ 0 \lt a \lt 1 $
Prove:
If $ z=ax + (1-a)y$
Then $ x \lt z \lt y $
My take on the problem:
$z$ smaller than $y$:
$ z=ax + (1-a)y = ax +y -ay = a(x-y) + y $
$ x \lt y $, therefore $x-y \lt 0 $
$a(x-y) \lt 0 $ 
$ a(x-y) + y \lt y$ 
$ z \lt y $
$z$ bigger than $x$:
Since $x-y$ is the distance between $x$ and $y$, Multiplied by something smaller than $1$ but bigger than $0$ (that is, $a$). Subtracting something that is smaller than the distance from $y$ will get you a number greater than $x$.
I drew a number line and reached to that conclusion, But how do I state that in a mathematical way?
 A: Why aren't you using the same method as for $z<y$? Set $b=1-a$, so $0<b<1$; then
$$
z=ax+(1-a)y=(1-b)x+by=x+b(y-x)>x
$$
because $b>0$ and $y-x>0$.
A: We are given $z=ax+(1-a)y$. In other words, $z = x + (1-a)(y-x)$, or $$x = z + (1-a)(x-y).$$
Since the two terms on the right are each positive, as $x<y$ and $a<1$, this means $x>z$.
A: What you are effectively saying is that $x < y + a(x-y)$. To prove this, start with $x = y + (x - y)$; we now need only to show that $x - y < a(x - y)$. From the assumption $0 < a < 1$, multiply by $x - y$, noting that since $x - y < 0$, the signs of the inequalities will be reversed:
$$0 > a(x - y) > x - y$$
Thus $x = y + (x - y) < x + a(x - y) = z$, as required.

That said, there is a symmetry to the problem, so it would be much more coherent to make the same argument as you did for the first half, rather than starting from scratch with a new explanation.
A: You know that $x = ax + (1-a)x$, 
Since $$ax + (1-a)x < ax + (1-a)y \implies x < z$$
A: $$\left.
\begin{array}{c} 
   x \lt y \\ 0 \lt a \lt 1 \\ z= ax + (1-a)y
\end{array}
\right \} \implies x < z < y$$
\begin{align}
   z-x 
      &= ax + (1-a)y - x \\
      &= (a-1)x + (1-a)y \\
      &= (y-x)(1-a) \\
      &> 0 \\
      &\implies x < z\\
   y-z 
      &= y - (ax + (1-a)y) \\
      &= y - ax - y + ay \\
      &= a(y-x) \\
      &> 0 \\
      &\implies z < y
\end{align}
BTW, for $\vec X, \vec Z \in \mathbb R^n$ the function
$f : \mathbb R \to \overleftrightarrow{XZ}$ defined by 
$f(t) = (1-t)\vec X + t \vec Z$ is a bijective map onto the line 
$\overleftrightarrow{XZ}$ with the properties


*
  
*$f(0)=\vec X$ and $f(1)= \vec Y$ 
  
*$\|f(s) - f(t)\| = |s-t|$ 
  
* 
      $\vec B$ is between $f(a)=\vec A$ and $f(c)=\vec C $
      if and only if for some $b$ between $a$ and $c$, 
      $f(b) = \vec B$.
  

