Answer
Reflexive
Symmetric
Transitive
Work Step by Step
Define a relation Q on R as follows: For all real numbers
x and y, x Q y ⇔ x − y are rational.
Reflexive: for all x ∈ Z xAx
true since x-x = 0 (note that 0 is rational = 0/1)
Symmetric: for all x,y ∈ Z if xQy then yQx
xQy <=> x-y = $\frac{a}{b}$ b ≠ 0
-x+y = $\frac{-a}{b}$ (muliplie by -1)
y - x = $\frac{-a}{b}$ (commutive , b ≠ 0 hence rational)
yQx by definition of Q
Transitive: for all x,y, and z ∈ Z if (xQy and yQz) then xQz
xQy <=> x-y = $\frac{a}{b}$ , b ≠ 0
yQz <=> y-z = $\frac{c}{d}$ , d ≠ 0
add (1) and (2)
x-z = $\frac{ad+cb}{bd}$ , bd ≠ 0 (x-z is rational)
xQz (By definition of Q) [that's what we needed to show]