Answer
Not Reflexive
Symmetric
Not Transative
Work Step by Step
Define a relation I on R as follows: For all real numbers
x and y, x I y ⇔ x − y is irrational.
Reflexive: for all x ∈ Z xAx
false since x-x = 0 (note that 0 is rational = 0/1)
Symmetric: for all x,y ∈ Z if xIy then yIx
xIy <=> x-y = irrational number X
-x+y = -X (muliplie by -1)
y - x = -X (commutive law, adding -Sign wont change the fact that X is irrational)
hence yIx by definition of I
Transitive: for all x,y, and z ∈ Z if (xIy and yIz) then xIz
let x = $\sqrt{2}$, y = 1, z = $\sqrt{2}$
xIy <=> x-y = $\sqrt{2}$ - 1 , irrational
yIz <=> y-z = 1 - $\sqrt{2}$ , irrational
but xIz <=> $\sqrt{2}$ - $\sqrt{2}$ = rational{hence false}
