# Please make x and y and z

This puzzle continues "Please make 1 and 2 and 4".

A mathematical expression is feasible, if it obeys the following rules:

• Any real number may be used
• One may use brackets "(" and ")" to structure the expression, and to make it well-defined
• The allowed mathematical operations are addition ($+$), subtraction ($-$), multiplication ($*$).
• Furthermore there is the special operation plus-minus ($\pm$).

Every feasible expression encodes the following set of numbers: every occurrence of $\pm$ in the expression is replaced once by $+$ and once by $-$, in all possible combinations.

Question: Is it true that for any choice of three real numbers $x,y,z$, there does exist a feasible expression that encodes the set $\{x,y,z\}$?

$x + (\frac{1}{2}\pm\frac{1}{2})*(y-x + (\frac{1}{2}\pm\frac{1}{2})*(z-y))$
Each $(\frac12\pm\frac12)$ is either 0 or 1. If the first is 0, we have x. If it is 1, we have x + y - x = y if the second is 0, or x + y - x + z - y = z if the second is 1.
Start from $x$ and add $0$ or $1$ times either $y-x$ or $z-x$ so:
$$x+(\dfrac12\pm\dfrac12)\times(\dfrac{(z+y-2x)}{2}\pm\dfrac{z-y}{2})$$