A calculator performs five operations: addition, subtraction, multiplication, division, and square root extraction. Find the formula by which the smallest of two arbitrary numbers $a$ and $b$ can be determined on this calculator.
Answer: $((a+b)-\sqrt{(a-b) \cdot(a-b)}) / 2$. Indeed, denoting the smaller of the numbers $a$ and $b$ by $x$, and the larger by $y$, we get $y-x=|a-b|=\sqrt{(a-b)^{2}}, y+x=a+b$, hence,

$$
x=\frac{(y+x)-(y-x)}{2}=\frac{a+b-\sqrt{(a-b)^{2}}}{2}
$$