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Proof By Contradiction

Proof By Contradiction. Chapter 3 Indirect Argument Contradiction Theorems 3.6.1 and 3.6.2 pg. 171. Your Hosts. Robert Di Battista Introduction, slides Da’niel Rowan Theorem 3.6.1 Annette Stiller Theorem 3.6.2. Proof By Contradiction. Indirect Argument Prove the negation is false.

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Proof By Contradiction

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  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 Nis the greatest integer and Nis 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 – bis an integer and a – bis 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

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