DE/DB = sinX /(2pi*X/2pi) in radians. = sin(X) / X = f(x) f(x) i a monotonously decresing function between 0 and pi. f(0)=1 and f(pi/2)=2/pi, f(pi)=0. There are infinitely -many rational numbers between 1 and 2/pi()~0.6 => there are infinitely-many rational values for DE/DB. -EOP

A*sinA/PI = a/b

ReplyDeletewhere A is the angle.

The answer depends on the radius.

If r=1 then I am not sure about the answer, but I don't think there are a and b integers.

But say if r=sqrt(2)/PI, then it DE/|DB| can be a rational number.

thank you very much for publishing this question, great fox

ReplyDeleteThis comment has been removed by the author.

ReplyDelete0 < angle(DCB)=DB <2PI

ReplyDeletein this distance there is infinite rational numbers can be shown x/y (x, y integers)

let DB=x/y

angle(DCB)=DB=x/y (radian)

DE=rsin(x/y)

DE/DB=rsin(x/y)*(y/x)

r, y/x are rational numbers then sin(x/y) must be rational number where 0 < x/y < 2PI

for example

sin(PI/2)=1 but PI/2 is not a rational number

sin(PI/6)=1/2 but PI/6 is not a rational num

my math knowledge is not enough to solve it

DE/DB = sinX /(2pi*X/2pi) in radians.

ReplyDelete= sin(X) / X = f(x)

f(x) i a monotonously decresing function between 0 and pi.

f(0)=1 and f(pi/2)=2/pi, f(pi)=0.

There are infinitely -many rational numbers between 1 and 2/pi()~0.6

=> there are infinitely-many rational values for DE/DB.

-EOP

Thank you very much for this solution, Actually i meaned if arc length (DB)=X is a rational number then can sin(X) be a rational number? (0< x< 2pi)

ReplyDeleteOR

Let DE/DB=sin(x)/x be a rational number

sin(x)-x*DE/DB=0 : Is solution of this equation a rational number?

example

sin(x)/x=0.5

sin(x)-0.5*x=0 by solving this equation

x=1,8954.....