Consider the following claim:

    Claim 1  Let x, y and z be real numbers.  If x > 0, y > 0, z > 0, and xy > z,
                    then x > z1/2 or y > z1/2
 

      You must first READ the claim and decide whether or not you think it is true (you may be wrong, but you have to practice this step; it is based on your prior experience and knowledge).   It is an inductive step; hence, there is no guarantee that you are right.

    Next, after considering claim 1, suppose we think it true.  Thinking it is true is not proving it is true.  Hence, we need to construct a proof.   We must announce it is a proof and frame it at the beginning (Proof:) and at the end (Q.E.D.[Quod Erat Demonstratum]).

Proof:

1.  Let x, y, and z be real numbers                                                  1. Premise
2.  x > 0, y > 0, z > 0, and xy > z                                                   2. Hypothesis
3.  Ø (x > z1/2   Ú   y > z1/2 )                                                           3. Negation of the conclusion
4. ( x £  z1/2 )  Ù  ( y £  z1/2 )                                                          4. DeMorgan (3)
5. [( x =  z1/2 )  Ú  ( x <  z1/2 )] Ù  [( y =  z1/2 )  Ú   ( y <  z1/2 )]      5. Def. of "£"

    Case A: [( x =  z1/2 )    Ù ( y =  z1/2 ) ]
    6A. Consider   xy                                            6A. Cases
    7A.     = (z1/2 )(z1/2 )                                        7A. Substitution
    8A.     =  (z1/2 )2                                              8A. Definition of square
    9A.     = z                                                       9A. Law of exponents
    10A. So, xy = z                                            10A. Transitivity of =
    11A. (xy = z) Ù (xy > z )                                11A. Adjunction (10A, 2)

    Case B: [( x <  z1/2 )    Ù ( y =  z1/2 ) ]
    6B. Consider    x <  z1/2                                  6B. Cases hypothesis
    7B.     y =  z1/2                                                7B. Cases hypothesis
    8B.    (x)(y)     <  (z1/2 )(y)                               8B. Preservation of order positive
                                                                                multiplier axiom (see axioms of IR)
    9B.     (x)(y )     < (z1/2 )(z1/2 )                          9B. Substitution
    10B.     (x)(y )     < (z1/2 )2                                10B. Definition of square
    11B.     (x)(y )     <  z                                         11B. Law of exponents
    12B. (xy < z) Ù (xy > z )                                 12B. Adjunction (11B, 2)

   Case C: [( x =  z1/2 )   Ù  ( y < z1/2 ) ]
    6C. Consider    y <  z1/2                                  6C. Cases hypothesis
    7C.     x =  z1/2                                                7C. Cases hypothesis
    8C.    (x)(y)     < (x)(z1/2 )                                8C. Preservation of order positive
                                                                                multiplier axiom  (see axioms of IR)
    9C.     (x)(y )     < (z1/2 )(z1/2 )                          9C. Substitution
    10C.     (x)(y )     < (z1/2 )2                                10C. Definition of square
    11C.     (x)(y )     <  z                                         11C. Law of exponents
    12C. (xy < z) Ù (xy > z )                                12C. Adjunction (11B, 2)

 Case D: [( x <  z1/2 )   Ù  ( y <  z1/2 ) ]
    6D. Consider    y <   z1/2                                 6D. Cases hypothesis
    7D.     x <  z1/2                                                7D. Cases hypothesis
    8D.     (x)(y )     <  (z1/2 )(z1/2 )                         8D. Practice Problem 1 page 38
    9D.     (x)(y )     < (z1/2 )2                                  9D. Definition of square
    10D.     (x)(y )     <  z                                        10D. Law of exponents
    11D. (xy < z) Ù (xy > z )                                11D. Adjunction (11B, 2)

13. x > z1/2 Ú  y > z1/2                                         13. Contradiction
                                                                            (of the trichotomy law (see axioms of IR))
 

Q. E. D.
 

Comment:  We proved the claim using an indirect proof [proof by contradiction - Reducto Ad Absurdum] note in each of the cases we conclude there is a logical contradiction - - breaking the law of the excluded middle.
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    Finally, as with all the discussions, examples, proofs, counterexamples, claims, etc. that we encounter; it is my opinion that few can do well in this class through just attending and watching others do the work.  I opine that only through doing can we understand and KNOW.  Hence, my advise is: "practice, practice, practice."
 
 

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 Last revised 10 February 2000