# For any 7 different real numbers, there are among them two numbers x and y such that $0<\frac{x-y}{1+xy} < √3$

For any 7 different real numbers, there are among them two numbers x and y such that $\frac{x-y}{1+xy}$ is greater than zeron and less than the square root of 3.

I find this fact quite amazing for many reasons. I want to know more about it; the proof for it would be cool, or perhaps some information on who postulated it; i'm particularly interested on how could he have ever come up with the conjecture. I'm also looking for an example of a set of 6 different reals such that no pair of them with the above property can be found. (If it exists, but does it?)

• This is a well known exercise the hint of the proof is let $x_1,\cdots,x_7$ be $7$ numbers and take $x_i=tan(y_i)$ for every $x_i$ Apr 24, 2015 at 22:54
• I hadn't seen this one, and I must say that it's a complete "why didn't I think of that" moment. Apr 24, 2015 at 23:22
• Now that is interesting! Endorsement. Apr 24, 2015 at 23:48

Given $x_1,x_2,\ldots,x_7$, set $x_i = \tan(t_i)$, where $t_i \in \left(-\dfrac{\pi}2, \dfrac{\pi}2\right)$. Divide $\left(-\dfrac{\pi}2, \dfrac{\pi}2\right)$ into $6$ equal intervals of length $\dfrac{\pi}6$. By PHP, we have two of the $t_i$'s to lie in the same interval, i.e., we have $$-\dfrac{\pi}6 < t_i - t_j < \dfrac{\pi}6$$ Now taking $\tan$ on both sides and I trust you can finish from here.
• $\tan \pi/6 = 1/\sqrt 3$, so doesn't that actually show that $0 < (x-y)/(1+xy) < 1/\sqrt 3$ for some $x, y$? Apr 24, 2015 at 23:36
• @MartinR Yes, true. We can get a much better bound of $1/\sqrt3$ as opposed to $\sqrt3$. Apr 24, 2015 at 23:51
• Or the other way around: 4 real numbers are sufficient for the bound $\sqrt 3$. Apr 24, 2015 at 23:53
Let the $7$ different numbers be $y_1,y_2,\cdots ,y_7$. Define $7$ angles measured in degrees $x_1, x_2,\cdots, x_7$ such that: $x_i =\tan^{-1} y_i$, then $x_i \neq x_j$, if $i \neq j$, and $x_i \in \left(-90^{\circ}, 90^{\circ}\right)$. Divide this interval into $6$ equal subintervals: $(-90^{\circ}, -60^{\circ}], (-60^{\circ}, -30^{\circ}], \cdots, (60^{\circ}, 90^{\circ})$. Thus by pigeon hole principle, there must be at least $2$ x's that are in the same interval, say $x_i > x_j$, and this means $0 < x_i - x_j < 30^{\circ} \to 0 < \tan(x_i-x_j) < \tan (30^{\circ}) \to 0 < \dfrac{y_i-y_j}{1+y_iy_j} < \sqrt{3}$
To answer the last part of your question: Let $t_i (1 \le i \le 6)$ be any six numbers in $(-\pi/2,\pi/2)$ separated by more than $\pi/6$ (for instance, $-5\pi/11,-3\pi/11,-\pi/11,\pi/11,3\pi/11,5\pi/11$). Then take $x_i=\tan(t_i)$.