MHB Can Two Numbers from a Set of Thirteen Satisfy This Mathematical Inequality?

[anemone]

Show that among any thirteen real and distinct numbers, it is possible to choose two, says, $a$ and $b$ such that $0<\dfrac{a-b}{1+ab}<2-\sqrt{3}$.

 

[kaliprasad]

Show that among any thirteen real and distinct numbers, it is possible to choose any two, says, $a$ and $b$ such that $0<\dfrac{a-b}{1+ab}<2-\sqrt{3}$.

It cannot any 2 because chosen a and b have to be specific here , Here is the solution

Spoiler

write the numbers has $\tan\ x$. Now x shall be between $-90^0$ to $90^0$. divide -90 to 90 degrees of 12 intervals of 15 degrees each. Now as there are 13 numbers 2 numbers have to in one interval.
say the angles are x, y and tan x = a , tan y = b and a > b
$\dfrac{a-b}{1+ab}= \tan( x - y) \lt tan \ 15 \lt 2-\sqrt{3}$

 

[anemone]

It cannot any 2 because chosen a and b have to be specific here,

Hey kaliprasad, I thought I've already replied to this challenge days ago, so... sorry for the late reply!

Yes, you're right, the wording of the problem sounds wrong, I'll change it so that it becomes a valid and sound problem, thanks for pointing this out...:)

Here is the solution

Spoiler

write the numbers has $\tan\ x$. Now x shall be between $-90^0$ to $90^0$. divide -90 to 90 degrees of 12 intervals of 15 degrees each. Now as there are 13 numbers 2 numbers have to in one interval.
say the angles are x, y and tan x = a , tan y = b and a > b
$\dfrac{a-b}{1+ab}= \tan( x - y) \lt tan \ 15 \lt 2-\sqrt{3}$

Needless to say, you're another math prodigy at our site! Thanks for your neat proof and thanks for participating, also, glad to see you around at the challenge problems sub forum, kali!