
theorem ff:
not sqrt 2 is Element of FAdj(F_Rat,{3-CRoot(2),zeta})
proof
set F = FAdj(F_Rat,{3-CRoot(2),zeta}), K = FAdj(F_Rat,{2-CRoot(2),zeta});
now assume sqrt 2 is Element of FAdj(F_Rat,{3-CRoot(2),zeta}); then
   reconsider a = sqrt 2 as Element of F;
   {a} is Subset of F; then
FAdj(F_Rat,{3-CRoot(2),zeta})
     = FAdj(F,{2-CRoot(2)}) by FIELD_7:2,FIELD_7:3
    .= FAdj(F_Rat,{3-CRoot(2),zeta}\/{2-CRoot(2)}) by FIELD_7:34
    .= FAdj(F_Rat,{3-CRoot(2),zeta,2-CRoot(2)}) by ENUMSET1:3; then
C: 6 = deg(FAdj(F_Rat,{3-CRoot(2),zeta,2-CRoot(2)}),K) * 4
       by grad4,grad6,FIELD_7:30;
     now let n be Integer, k be Nat;
       assume C3: k <> 0 & 3/2 = n/k;
       then (3/2) * k = (n/k) * k .= n * k* k" .= n by C3,XCMPLX_1:203;
       then C4: 3 * k = n * 2;
       C5: 3 = 2 * 1 + 1;
       now assume 2 > k;
         then k is trivial by NAT_2:29;
         then per cases by NAT_2:def 1;
         suppose k = 0;
           hence contradiction by C3;
           end;
         suppose k = 1;
           hence contradiction by C5,C4;
           end;
         end;
       hence 2 <= k;
       end; then
   denominator(3/2) = 2 by RAT_1:def 3;
hence contradiction by C,RAT_1:21;
end;
hence thesis;
end;
