I ban Kimo because..
Kimo- Kim = o...(1)
Kimo - Kim = o
o(Kim) - Kim = o
[squaring both sides]
[o(kim)]^2 - Kim^2 = o^2
{[o(kim) + Kim]} {[o(kim) - kim} = o^2
{[o(kim) - Kim]} = o^2 / {[o(kim) + kim}
Kimo - Kim = o^2 / {[o(kim) + kim}
As Kimo - Kim = o (1)
Substituting ..
o = o^2 / {[o(kim) + kim}
o [ o (Kim) + Kim] = o^2
Kimo^2 - o^2 + kimo = 0
o (Kimo - o + kim) = 0
Kimo - o + kim = 0
Kimo + kim = o ..(2)
Hence (wrongly) proved that Kimo + kim can be = o as well as Kimo - kim be = o.. therefore my efforts lie in vain and hope Kim and Kimo don't mind.
I ban Kimo, Kim and Ash for Warning - while you were typing 15 new replies have been posted. You may wish to review your post.