Exercise 13. A set of $n$ non-zero and distinct real numbers is said to be regular if, when these numbers are written on the board in ascending order, the difference between two adjacent numbers is always the same, regardless of the two adjacent numbers chosen. For example, the set $\{4,18,-3,11\}$ is regular because if the numbers are written in ascending order, $18-11=11-4=4-(-3)$. A set $A$ of non-zero and distinct real numbers is said to be super-regular if it is regular and the set formed by the inverses of the numbers in set $A$ is also regular.

What is the largest integer $n$ such that there exists a super-regular set of $n$ non-zero and distinct real numbers?
## Solution to Exercise 13

Answer: $n=4$

In this exercise, we aim to demonstrate that 4 is the largest integer satisfying the property stated in the problem. The solution necessarily includes two parts. First, we show that if a super-regular set has size $k$, then $k \leqslant 4$; this step is called the analysis. Second, we show that there indeed exists a super-regular set of size 4; this step is called the construction or synthesis.

In the following, we call the common difference of a regular set the difference between the largest number and the second largest number in that set.

## Analysis:

We start by showing that there does not exist a super-regular set of size 3 whose elements are all of the same sign.

By contradiction, we assume a super-regular set $A=\{a, b, c\}$ such that $a0$, meaning all three elements are strictly positive. We set $r=b-a>0$ as the common difference of $A$. Then, $c-b=r$, so $c=a+2r$ and $b=a+r$.

We set $B=\left\{\frac{1}{a}, \frac{1}{a+r}, \frac{1}{a+2r}\right\}$ as the set composed of the inverses of the elements of $A$. We verify that $\frac{1}{a+2r}0$ and $a1 / b>1 / c$. Moreover, the inequality $1 / a>1 / b>1 / c$ is false precisely when the reals are not of the same sign: if $a=-1, b=1, c=3$, it does not hold.

Therefore, many students used $b-a=c-b$ and $1 / b-1 / a=1 / c-1 / b$ to obtain $a=c$, and received 2 points if the inequality $1 / a>1 / b>1 / c$ was mentioned, 1 point otherwise, as they missed the difficulty of the problem.

In fact, it was necessary to use the previous reasoning for a super-regular set of any size and show that having three consecutive elements of the same sign was impossible. Very few students understood this part of the problem, and those who did were often able to complete the exercise.

Note that many students confuse inverse and opposite: the opposite of 2 is -2, its inverse is $1 / 2$.