NDA Math Sets Relations and Functions

(x,y)∊S id x divides y.
For x∊Z,x divides x.
Hence,(x,x)∊s,∀x∊Z
Hence,S is reflexive....(1)

If (x,y)∊S⇒x divides y⇒y=Kx
Where K is an integer.
⇒x=

1

K

y,where

1

K

may not be an integer.
Hence,y does not divide x.
Hence,(y,x)∉S
Hence,S is not symmetric...(2)

If (x,y)∊S and (y,z)∊S,then
y=K1x and z=K2y(K1K2∊Z)
⇒z=K2K1x
Hence,x divides z.
Hence,(x,z)∊S
Hence ,the given relation is transitive....(3)
From (1)(2)(3) the given relation is only reflexive and transitive