Calculate $\sqrt{\pi}$ to four decimal places.
Suppose the upper approximate value of $\pi$ is given with $n$ decimal places of accuracy; since the integer part of this number is a single digit, the absolute error $e$ is less than one unit of the $n+1$-th digit's order, and the relative error $e'$ is less than $\frac{1}{10^n}$. The relative error of the square root $E'$ is less than half of this. In symbols:

$$
E' < \frac{e'}{2} < \frac{1}{2 \times 10^n} < \frac{1}{10^n}
$$

Thus, the absolute error $E'$ of the square root is less than one unit of the $n$-th digit's order. But since the integer part of the square root is also a single digit, the $n$-th digit is the $(n-1)$-th decimal, so

$$
E < \frac{1}{2 \times 10^{n-1}}
$$

That is, if I want to know $\sqrt{\pi}$ to $n-1$ decimals, I need to take the upper approximate value of $\pi$ to $n$ decimals; so, if I want it to 4 decimals, I need to take $\pi$ to 5 decimal places. The upper approximate value of $\pi$ is 3.14160 and the square root of this is 1.7724, which is a lower approximate value.

(Visnya Aladár, fóreálisk. VIII. o. t. Pécs.)

The problem was also solved by: Friedmann Bernát, S.-A.-Ujhely; Galter János, Sz.-Udvarhely.