x=(8√3+13)=13C0⋅(8√3)13+13C1(8√3)12(13)1+... x′=(8√3−13)13=13C0(8√3)13−13C1(8√3)12(13)1+... x−x′=2[13C1⋅(8√3)12(13)1+13C3(8√3)10⋅(13)3...] therefore, x−x′ is even integer, hence [x] is even Now, y=(7√2+9)9=9C0(7√2)9+9C1(7√2)8(9)1 +9C2(7√2)7(9)2....... y′=(7√2−9)9=9C0(7√2)9−9C1(7√2)8(9)1 +9C2(7√2)7(9)2....... y−y′=2[9C1(7√2)8(9)1+9C3(7√2)6(9)3+...] y−y′= Even integer, hence [y] is even