7. let s be the subset of q defined by s = {req:r > 0 and r <2| show that s is a non-empty subset of q bounded above but sup s does not belong to q. s is no-emp

7. Let S be the subset of Q defined by S = {rEQ:r > 0 and r <2| Show that S is a non-empty subset of Q bounded above but sup S does not belong to Q. S is no-empty, since 1 E S. S is bounded above, since 2 is an upper bound of the set. If possible, let sup S e Q and sup S = u. Then u > 0 and u E By the law of trichotomy, exactly one of the following holds u>2,u =2,u2< 2. Case 1. Let u2> 2. Then u2 - 2> 0. Let us take another rational number r ._4+3u Then r >0. 3+2u ... 1) T=u3+2u 4 3+2u 0. Therefore 0 0 and r > 2 .. ...(ii) (1i) shows that r is an upper bound of S and (i) shows that u is not the supremum of S. This is a contradiction to the assumption that u = sup S. Therefore u 2. Case II. u = 2. We have seen that there exists no rational number r such that r = 2. Therefore u # 2. Case III. u2<2. Let us take again the rational number r= Then r > 0 and T- =4U 0. Therefore 0< u 0. Therefore r > 0 andr<2... .. (iv) (iv) shows that r E S, From (iii) it follows that u belongs to S and u is less than an element Tof S. Therefore u is not the supremum of S, a contradiction. Therefore u2. None of the three possibilities provided by the law of trichotomy can hold. Hence our assumption that sup S is a rational number 1s wrong. Therefore no rational number can be the supremum of S.