[Guest]

Given a set of 7 distinct real numbers, prove that, there always exists two numbers x and y in this set that satisfies the following inequality:

0 < (x - y)/(1 + xy) < 1/Sqrt(3)

[Guest]

ok , here is my solution

[set of 7 distinct real numbers S, x,y E S]

To Prove : 0< (x-y)/(1+xy) < 1/sqrt(3)

Proof:

let there be 7 angles xi's i-> [1,7] // i.e, x1,x2,.....

xi's are angles xi -> [-pi/2,pi/2] ie [-90 degree, 90 degree]

let their corresponding tans inverse i.e tan(zi)'s be xi's // tan(z1)=x1,tan(z2)=x2,.....

//here tan(zi) -> [-infinity,+infinity]

We know the range of tan is -infinity to +infinity //hence constrain of distinct real numbers satisfied

if we divide the interval of [-90 degrees , 90 degree] into 6 sub-intervals

|--- |---|---|---|---|---|

-90 -60 -30 0 30 60 90

I II III IV V VI

if we transform this

we have 7 angles and six sub-intervals,

from pigeon hole principle we can infer that two angles will be in same interval , let them be X and Y.

we can see that difference between X and Y can't be greater than 30 degrees.

let x=Tan(X) and y=Tan(Y)

hence we have 0 < X-Y < pi/6 // pi = 180 degree

//Since tan is monotonically increasing function we can apply tan to this inequality

hence Tan(0) < Tan(X-Y) < Tan(pi/6)

==> 0 < Tan(X)-Tan(y)/(1+Tan(X)*Tan(Y) < 1/sqrt(3)

==> 0 < (x - y) /(1+xy) < 1/sqrt(3) //Since Tan(X)=x and Tan(Y) = y

Edited October 21, 2009 by m3ssi3r

[Guest]

ok , here is my solution

[set of 7 distinct real numbers S, x,y E S]

To Prove : 0< (x-y)/(1+xy) < 1/sqrt(3)

Proof:

let there be 7 angles xi's i-> [1,7] // i.e, x1,x2,.....

xi's are angles xi -> [-pi/2,pi/2] ie [-90 degree, 90 degree]

let their corresponding tans inverse i.e tan(zi)'s be xi's // tan(z1)=x1,tan(z2)=x2,.....

//here tan(zi) -> [-infinity,+infinity]

We know the range of tan is -infinity to +infinity //hence constrain of distinct real numbers satisfied

if we divide the interval of [-90 degrees , 90 degree] into 6 sub-intervals

|--- |---|---|---|---|---|

-90 -60 -30 0 30 60 90

I II III IV V VI

if we transform this

we have 7 angles and six sub-intervals,

from pigeon hole principle we can infer that two angles will be in same interval , let them be X and Y.

we can see that difference between X and Y can't be greater than 30 degrees.

let x=Tan(X) and y=Tan(Y)

hence we have 0 < X-Y < pi/6 // pi = 180 degree

//Since tan is monotonically increasing function we can apply tan to this inequality

hence Tan(0) < Tan(X-Y) < Tan(pi/6)

==> 0 < Tan(X)-Tan(y)/(1+Tan(X)*Tan(Y) < 1/sqrt(3)

==> 0 < (x - y) /(1+xy) < 1/sqrt(3) //Since Tan(X)=x and Tan(Y) = y

That is very nice. But please use spoilers.

[Guest]

m3ssi3r, welcome to the forum! You got the solution.

In general, please put the solutions/hints within spoilers to allow others to try the puzzle independently.