Ancient Hindu mathematicians used the following rational approximation for the number $\sqrt{2}$:

$$
t=1+\frac{1}{3}+\frac{1}{3 \cdot 4}-\frac{1}{3 \cdot 4 \cdot 34}
$$

What is the maximum percentage error of the approximation? - How many leading digits of $t$ in decimal form match the decimal form of the approximate fraction of $\sqrt{2}$?
I. After bringing to a common denominator and simplifying,

$$
t=\frac{577}{408}, \text { from which } t^{2}=\frac{332929}{166464}=2+\frac{1}{166464}
$$

thus $t$ is an upper approximate value of $\sqrt{2}$.

The difference between the approximate value and the number to be approximated is $t-\sqrt{2}$, and this difference relative to $\sqrt{2}$ is

$$
p=100 \cdot \frac{t-\sqrt{2}}{\sqrt{2}}=100 \frac{t^{2}-2}{\sqrt{2}(t+\sqrt{2})}=\frac{100}{166464(t \sqrt{2}+2)}
$$

percent. Substituting $\sqrt{2}$ for $t$ in the denominator decreases the denominator, increasing the quotient; finally, with further increase, we can choose a simple upper bound for $p$:

$$
p<\frac{100}{166464 \cdot 4}<0.00016 \%
$$

II. From this, by reversing, we get that the value of the above difference is

$$
t-\sqrt{2}=\sqrt{2} \cdot \frac{p}{100}<1.5 \cdot 0.0000016=0.0000024
$$

In each of the 3 increases performed, we increased cautiously, so it is foreseeable that in the decimal representation, the difference between the approximate values of $t$ and $\sqrt{2}$ will already appear in the 6th decimal place, but it may also appear earlier. Calculating

$$
t=1.414215 \ldots, \sqrt{2}=1.414213 \ldots
$$

thus they agree in the first 6 significant digits.

Peter Kafka (Pannonhalma, Benedekine Order II. Class)