## Task 3 - 340633

Already 5000 years ago, there was a highly developed knowledge of fraction arithmetic in Egypt. Unit fractions were preferred; these are fractions whose numerator is 1 and whose denominator is a natural number.

(a) Give one way to represent the fractions $\frac{2}{7}$ and $\frac{3}{13}$ as the sum of at least two different unit fractions!

The number of summands is not specified; no justification is required.

(b) Represent the fraction $\frac{1}{36}$ as the sum of at least two unit fractions such that one of the summands is as large as possible! Explain why no larger summand is possible!

(c) Solve the same problem for $\frac{1}{n}$ instead of $\frac{1}{36}$, where $n$ is any natural number greater than 1!
Examples for (a) are:

$$
\frac{2}{7}=\frac{1}{4}+\frac{1}{28} \quad ; \quad \frac{3}{13}=\frac{1}{6}+\frac{1}{26}+\frac{1}{39}
$$

(b),(c) It holds that

$$
\begin{aligned}
& \frac{1}{36}-\frac{1}{37}=\frac{37-36}{36 \cdot 37}=\frac{1}{3332}, \quad \text { hence } \quad \frac{1}{36}=\frac{1}{37}+\frac{1}{3332} \\
& \frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n \cdot(n+1)}=\frac{1}{n(n+1)}, \quad \text { hence } \quad \frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}
\end{aligned}
$$

Here, the summand $\frac{1}{37}$ (or the summand $\frac{1}{n+1}$) has the required property that no larger unit fraction can be a summand in such a representation.

Proof:

Among the natural numbers, 37 (or $n+1$) is the next largest to 36 (or to $n$), i.e.: Among the natural numbers that are greater than 36 (or greater than $n$), there is no smaller one than 37 (or $n+1$).

From this it follows:

Among the unit fractions that (appear as summands in the desired representation and therefore) must be smaller than $\frac{1}{36}$ (or $\frac{1}{n}$), there is no larger one than $\frac{1}{37}$ (or $\frac{1}{n+1}$).