www.ConcurrentInverse.com
                                   2005 October 11
                                   (c) 2005 Wm.C.Corwin





The more general case will now be considered:
[k+l[5]] = Q + [h+f[5]] where k,l,h, and f a re rationals.

It the above is possible we will separate the rational and irrational parts
after eliminating the radicals.

k + l[5] = {Q} + 2Q[h+f[5]] + h + f[5]
k-{Q}-h + (l-f)[5] = 2Q[h+f[5]]
{k-{Q}-h} + 5{l-f} + 2(k-{Q}-h)(l-f)[5] = 4{Q}h + 4{Q}f[5]

{k-{Q}-h} -4{Q}h + 5{l-f} = 0           2(k-{Q}-h)(l-f) -4{Q}f =0
{k-h} -2{Q}(k+h) + {{Q}} + 5{l-f} =0       2(k-h-{Q})(l-f) -4{Q}f =0

Separating out the rational and irrational parts:
{k-h}-2{Q}(k+h)+{{Q}}+5{l-f}= 0 and  2(k-h)(l-f)-2{Q}l-2{Q}f =0

                                    k - h = {Q}(l+f)/(l-f)


        2{Q}(k+h) = {k-h}+{{Q}}+5{l-f} 
          k + h = ({k-h}+{{Q}}+5{l-f})/(2{Q})


       k  l       [2]    h f

      [15+5[5]] = [2] + [7+3[5]]
        7 3              3 1
       ...
       27 9             23 3
       ...