Answer
For all m, n ∈ Z, m F n ⇔ 4 | (m − n)
Reflexive: it's reflexive if a F a is true for all a ∈ Z
hence a F a = 4 | (a - a) = 4| 0 which is true.
Symmetric: for all x,y ∈ 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 = 4| x - y
= x - y = 4*P , P ∈ Z (Definition of divides)
-x+y = -(4*P) (Multiplying both sides by -1)
y-x = 4*{-P} (commutive Law And Taking -p common factor from the right side)
note that -p ∈ Z hence by definition of divides y-x|4, hence by definition of F
yFx. [THAT"S WHAT WE NEED TO SHOW]
Transitive : for all x,y and z ∈ Z if (xCy and yCz) then (xCz)
xCy = 4| (x-y) > x-y = 4k , (def of divides) ...(1)
yCx = 4 | (y - z) > y-z = 4j, (def of divides) ... (2) , k,j ∈ Z
adding (1) and (2)
x-z = 4k + 4j = 4(k+j) k+j ∈ Z since the sum of 2 integers is an integer
x-z = 4*{int} by def of divides 4| x - z
by def of C -> xCz [that's what we need to show]
For all m, n ∈ Z, m F n ⇔ 4 | (m − n).
[a] = { x $\in$ z | x = 4k +a}
equivalent classes are :
{ x $\in$ z | x = 4k } for some integer k
{ x $\in$ z | x = 4k + 1 } for some integer k
{ x $\in$ z | x = 4k + 2 } for some integer k
{ x $\in$ z | x = 4k + 3 } for some integer k
Work Step by Step
For all m, n ∈ Z, m F n ⇔ 4 | (m − n)
Reflexive: it's reflexive if a F a is true for all a ∈ Z
hence a F a = 4 | (a - a) = 4| 0 which is true.
Symmetric: for all x,y ∈ 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 = 4| x - y
= x - y = 4*P , P ∈ Z (Definition of divides)
-x+y = -(4*P) (Multiplying both sides by -1)
y-x = 4*{-P} (commutive Law And Taking -p common factor from the right side)
note that -p ∈ Z hence by definition of divides y-x|4, hence by definition of F
yFx. [THAT"S WHAT WE NEED TO SHOW]
Transitive : for all x,y and z ∈ Z if (xCy and yCz) then (xCz)
xCy = 4| (x-y) > x-y = 4k , (def of divides) ...(1)
yCx = 4 | (y - z) > y-z = 4j, (def of divides) ... (2) , k,j ∈ Z
adding (1) and (2)
x-z = 4k + 4j = 4(k+j) k+j ∈ Z since the sum of 2 integers is an integer
x-z = 4*{int} by def of divides 4| x - z
by def of C -> xCz [that's what we need to show]
For all m, n ∈ Z, m F n ⇔ 4 | (m − n).
[a] = { x $\in$ z | x = 4k +a}
equivalent classes are :
{ x $\in$ z | x = 4k } for some integer k
{ x $\in$ z | x = 4k + 1 } for some integer k
{ x $\in$ z | x = 4k + 2 } for some integer k
{ x $\in$ z | x = 4k + 3 } for some integer k
