Page 114 - HOW TO PROVE IT: A Structured Approach, Second Edition
P. 114

P1: PIG/
                   0521861241c03  CB996/Velleman  October 20, 2005  2:42  0 521 86124 1  Char Count= 0






                                   100                         Proofs
                                   After using strategy:
                                                    Givens                 Goal
                                                      ¬P                    P
                                                      —
                                                      —

                                   Form of final proof:
                                       [Proof of P goes here.]
                                     Since we already know ¬P, this is a contradiction.

                                     Although we have recommended proof by contradiction for proving goals
                                   of the form ¬P, it can be used for any goal. Usually it’s best to try the other
                                   strategies first if any of them apply; but if you’re stuck, you can try proof by
                                   contradiction in any proof.
                                     The next example illustrates this and also another important rule of proof-
                                   writing: In many cases the logical form of a statement can be discovered
                                   by writing out the definition of some mathematical word or symbol that
                                   occurs in the statement. For this reason, knowing the precise statements of
                                   the definitions of all mathematical terms is extremely important when you’re
                                   writing a proof.


                                   Example 3.2.3. Suppose A, B, and C are sets, A \ B ⊆ C, and x is anything
                                   at all. Prove that if x ∈ A \ C then x ∈ B.

                                   Scratch work
                                   We’re given that A \ B ⊆ C, and our goal is x ∈ A \ C → x ∈ B. Because the
                                   goal is a conditional statement, our first step is to transform the problem by
                                   adding x ∈ A \ C as a second given and making x ∈ B our goal:

                                                    Givens                  Goal
                                                  A \ B ⊆ C                 x ∈ B
                                                  x ∈ A \ C
                                   The form of the final proof will therefore be as follows:

                                       Suppose x ∈ A \ C.
                                         [Proof of x ∈ B goes here.]
                                       Thus, if x ∈ A \ C then x ∈ B.

                                     The goal x ∈ B contains no logical connectives, so none of the techniques
                                   we have studied so far apply, and it is not obvious why the goal follows from
   109   110   111   112   113   114   115   116   117   118   119