Answer
A is the “absolute value” relation on R: For all real numbers
x and y,x A y ⇔ |x| = |y|.
Reflexive: for all x ∈ Z xAx
true since |x| = |x| for every x
Symmetric: for all x,y ∈ Z if xAy then yAx
xAy <=> |x| = |y| >> |y| = |x| (commutive law) = yAx (by def of A)
Transitive: for all x,y, and z ∈ Z if (xAy and yAz) then xAz
xAy <=> |x| = |y| >> |y| = |x| commutive law--(1)
yAz <=> |y| = |z| -- (2)
**sub 1 in 2 |x| = |z|
xAz (By definition of A) [that's what we needed to show]
Since A is reflective symmetric and transitive, it's an equivalence relation
x = 0 , then 0 is the only value with absolute value = 0
or x doesn't equal 0, then {x,-x} are the value with their absluoate = x
hence the equvillance classes are
{0},{x,-x} for all x $\in$ R
Work Step by Step
A is the “absolute value” relation on R: For all real numbers
x and y,x A y ⇔ |x| = |y|.
Reflexive: for all x ∈ Z xAx
true since |x| = |x| for every x
Symmetric: for all x,y ∈ Z if xAy then yAx
xAy <=> |x| = |y| >> |y| = |x| (commutive law) = yAx (by def of A)
Transitive: for all x,y, and z ∈ Z if (xAy and yAz) then xAz
xAy <=> |x| = |y| >> |y| = |x| commutive law--(1)
yAz <=> |y| = |z| -- (2)
**sub 1 in 2 |x| = |z|
xAz (By definition of A) [that's what we needed to show]
Since A is reflective symmetric and transitive, it's an equivalence relation
x = 0 , then 0 is the only value with absolute value = 0
or x doesn't equal 0, then {x,-x} are the value with their absluoate = x
hence the equvillance classes are
{0},{x,-x} for all x $\in$ R