# Show that $0 < \frac{y-x}{1+xy} < 1$.

Let $E$ a finite set of real, with at least 5 elements.

I remember that my teacher proved that there exist two of its elements $x<y$, such that $0 < \dfrac{y-x}{1+xy} < 1$.

Unfortunately I can not find my notes and (inevitably..) I do not remember how to prove this result.

• – Ben Nov 15 '14 at 7:39

Use the formula $$\tan (a-b) = \frac{\tan a - \tan b}{1+\tan a \tan b}$$
Take $\arctan$ of each element, you get elements of the set $$(-\pi/2,\pi/2) =(-\pi/2,-\pi/4] \cup(-\pi/4,0] \cup (0, \pi/4] \cup (\pi/4, \pi/2).$$
There are $5$ elements, so at least two of them have are in the same interval and then $$0<\arctan y - \arctan x <\frac\pi 4.$$(0 is excluded, because the elements are different and $\arctan$ is into). $$\frac{y-x}{1+xy}= \tan (\arctan y - \arctan x)\in(0,1).$$