1. Proof By Contradiction Chapter 3 Indirect Argument Contradiction Theorems 3.6.1 and 3.6.2 pg. 171

2. Your Hosts Robert Di Battista  Introduction, slides Da’niel Rowan  Theorem 3.6.1 Annette Stiller  Theorem 3.6.2

3. Proof By Contradiction Indirect Argument  Prove the negation is false.  reductio ad absurdum  Assume its negation is true, until it is reduced to an impossibility or absurdity.  This leaves only one possibility.

4. Method of Proof By Contradiction  Suppose the statement to be proved is false, i.e. suppose the negation of the statement is true.  Show that this supposition leads logically to a contradiction.  Conclude that the statement to be proved is true.

5. Analogous to… Shows truth by discounting the opposite.  Sarcasm  Reverse psychology

6. Theorem 3.6.1 There is no greatest integer.

7. Theorem 3.6.1 [We take the negation of the theorem and suppose it to be true] Suppose not. That is, suppose there is a greatest integer N. [We must deduce a contradiction] Then N > n for every integer n.

8. Theorem 3.6.1 Let M = N + 1. M is an integer since it is the sum of integers. Also M > N since M = N + 1.

9. Theorem 3.6.1 Thus M is an integer that is greater than N. So N is the greatest integer and N is not the greatest integer, which is a contradiction. [This contradiction shows that the supposition is false and, hence, that the theorem is true.]

10. Theorem 3.6.1 Bill Gates is disgustingly rich, but someone can always have $1 more than him.

11. Theorem 3.6.1 Exam question: Prove by contradiction “There is no greatest odd integer.”

12. WAIT!

13. THERES MORE

14. Theorem 3.6.2 There is no integer that is both even and odd.

15. Theorem 3.6.2 [We take the negation of the theorem and suppose it to be true] Suppose not. That is, suppose there is an integer N that is both even and odd. [We must deduce a contradiction.]

16. Theorem 3.6.2 By definition of even, n = 2a for some integer a, and by definition of odd, n = 2b + 1 for some integer b. Consequently, 2a = 2b +1 [By equating the two expressions for n.] And so 2a – 2b = 1 2(a – b) = 1 (a – b) = 1/2 [by algebra]

17. Theorem 3.6.2 Now since a and b are integers, the difference a – b must also be an integer. But a – b = 1/2, and 1/2 is not an integer. Thus a – b is an integer and a – b is not an integer, which is a contradiction. [This contradiction shows that the supposition is false and, hence, that the theorem is true.]

18. Analogous to… Mutually exclusive groups:  Male or female  Positive or negative  True or false

19. Theorem 3.6.2 Exam question: Prove by contradiction “There is no real number that is both positive and negative.”

20. Related Homework Section 3.6  3 -15  21 – 27  32