Is the relation $ on set Z+ defined by x$y iff (x+y)/xy >= 0 an equivalence relation?

Answer 1

reflexive:

n$n --> (n+n)/(n*n) = 2n/n^2 = 2/n >0 since n>0

yes it is reflexive

symmetric:

given: x$y ---> (x+y)/(xy) >= 0

(y + x)/(yx) >=0 by symmetric property of integers

---> y$x

yes, it is symmetric

transitive:

given x$y and y$z

(x+y)/(xy)>=0 and (y+z)/(yz) >= 0 for x>0, y>0, and z>0

then x+z>0 and xz>0 , since x>0 and z>0

(x+z)/((xz) >0

x$z

yes it is transitive.

equivalence relation proven