Ans is D
for reflexivity
(m-m) =(2k-1)
2k-1=0
k=½, which does not belong to Z
for symmetric
mOn is possible if m==n or m!=n will relate only if one is odd and another one even
if (m-n) = 2k-1 then (n-m)=2k-1
(n-m)=2k-1
-(m-n)=2k-1
so if in (m-n) we took k=x so for to hold (n-m) we just need to take k=-x
For transitivity
m-n=2k1-1, let m is odd and n is even
n-p=2k2-1 n is even and assume m is odd
m-p=2k3-1 so now m and p both are odd, so their subtraction will lead to even no . which can’t be written in 2k3-1 format where k3 belongs to z