Answer
Reflexive
Symmetric
Transitive
Work Step by Step
For all m, n ∈ Z, m F n ⇔ 5 | (m − n).
Reflexive: it's reflexive if a F a is true for all a $\in$ Z
hence a F a = 5 | (a - a) = 5| 0 which is true.
Symmetric: for all x,y $\in$ Z, if xCy then yCx
Suppose x,y are particular but arbitrary chosen integers that satisfy the relation xFy, we need to show that yFx.
xCy = 5| x - y
= x - y = 5*P , P $\in$ Z (Definition of divides)
-x+y = -(5*P) (Multiplying both sides by -1)
y-x = 5*{-P} (commutive Law And Taking -p common factor from the right side)
note that -p $\in$ Z hence by definition of divides y-x|5, hence by definition of F
yFx. [THAT"S WHAT WE NEED TO SHOW]
Transitive : for all x,y and z $\in$ Z if (xCy and yCz) then (xCz)
xCy = 5| (x-y) > x-y = 5k , (def of divides) ...(1)
yCx = 5 | (y - z) > y-z = 5j, (def of divides) ... (2) , k,j $\in$ Z
adding (1) and (2)
x-z = 5k + 5j = 5(k+j) k+j $\in$ Z since the sum of 2 integers is an integer
x-z = 5*{int} by def of divides 5| x - z
by def of C -> xCz [that's what we need to show]
