Exercise 3. Let E be a finite set of strictly positive real numbers such that, for any strictly positive real number $x$, E contains as many elements strictly greater than $x$ as elements strictly less than $\frac{1}{x}$. Determine the product of all elements of $E$.
## Solution.

Let $x_{0} \in E$ such that $x_{0}>1$. Then, by hypothesis,

- the sets $E \cap]-\infty, \frac{1}{x_{0}}[$ and $E \cap] x_{0},+\infty[$ have cardinals of the same parity,
- the sets $E \cap]-\infty, x_{0}[$ and $E \cap] \frac{1}{x_{0}},+\infty[$ have cardinals of the same parity.

Consequently, the two sets

$$
\begin{aligned}
\left.E \cap] \frac{1}{x_{0}}, x_{0}\right] & =(E \cap] \frac{1}{x_{0}},+\infty[) \backslash(E \cap] x_{0},+\infty[) \\
E \cap\left[\frac{1}{x_{0}}, x_{0}[\right. & =(E \cap]-\infty, x_{0}[) \backslash(E \cap]-\infty, \frac{1}{x_{0}}[)
\end{aligned}
$$

have cardinals of the same parity. These sets are composed of the same elements, except possibly $x_{0}$ and $\frac{1}{x_{0}}$: the first set contains $x_{0}$ but not $\frac{1}{x_{0}}$; thus, the second set contains one and only one element of $\left\{\frac{1}{x_{0}}, x_{0}\right\}$ but cannot contain $x_{0}$: it therefore contains $\frac{1}{x_{0}}$.

We have thus established that if $x_{0} \in E$, then $\frac{1}{x_{0}} \in E$. The product of all elements of $E$ is therefore 1.

![](https://cdn.mathpix.com/cropped/2024_06_04_6f8c9f47cc79216a1a37g-04.jpg?height=89&width=187&top_left_y=1159&top_left_x=969)

## Common Exercises