SCHIZOPHRENIA IN CONTEMPORARY MATHEMATICS 17 routine method for proving or disproving Goldbach's conjecture ! Most of the usual theorems about 5 and remain true constructively, with the exception of trichotomy. Not only does the usual form "a b or a = b or a b" fail, but such weaker forms as "a b or a 2 b, I' or even "a g b or a 2 b" fail as well. For example, we are not entitled to assert "0 H or 0 = H or 0 H. " If we consider the closely related number we are not even entitled to assert that "H' 2 0 or H' g 0. Since trichotomy is so fundamental, we might expect constructive math- ematics to be hopelessly enfeebled because of its failure. The situation is saved, because trichotomy does have a constructive version, which of course is considerably weaker than the classical. THEOREM. For arbitrary real numbers a, b, and c, with a b, either c a or c b . 1 PROOF. Choose integers M and No such that an s bn - whenever n 2 No . Choose integers N , Nb , and N such that I an - a 1 6 ( 6 ~ ) - I a c m whenever n, m 2 Na . I bn - bm 1 5 (6M)-' whenever n. m 2 Nb , 1 cn - c I 5 m (6M)-' whenever n,m 2 Nc . Set N = max {N Na, Nb. Nc } . Since a 0' N' bN , and cN are all rational numbers, either 1 . Consider first the case c 2 (aN t bN). Since a N 5 bN - M-' , it follows that aN g cN - ( 2 ~ ) ~ ~ . For each n 2 N we therefore haye 1 Therefore, a c. In the other case, cN 2 (aN + b ), it follows similarly N that c b. This completes the proof of the theorem. Do not be deceived by the use of the word "choose" in the above proof, which is simply a carry-over from classical usage. No choice is involved,
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