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Proof by Contradiction CS 270 Math Foundations of CS Jeremy Johnson.

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Presentation on theme: "Proof by Contradiction CS 270 Math Foundations of CS Jeremy Johnson."— Presentation transcript:

1 Proof by Contradiction CS 270 Math Foundations of CS Jeremy Johnson

2 Outline

3 Decimals and Fractions

4 Repeating Decimals

5 Theorem. A number r is rational iff it has a terminating or repeating decimal expansion Proof  If r = a/b perform long division to compute the decimal expansion. At each step divide what is left by b, m = qb + r, 0 ≤ r < b. There are b possible remainders. If r = 0 the expansion is terminating. If r has occurred previously the expansion is repeating. After at most b steps one of these must happen.

6 Repeating Decimals

7 Decimal Expansion of sqrt(2)

8 x0 := 1.0; x1 := 2.0; n := 20; for i from 1 to n do x := (x0+x1)/2; if x^2 > 2 then x1 := x; else x0 := x; end if; end do; 1.500000000 1.250000000 1.375000000 1.437500000 1.406250000 1.421875000 1.414062500 1.417968750 1.416015625 1.415039062 1.414550781 1.414306640 1.414184570 1.414245605 1.414215088 1.414199829 1.414207458 1.414211273 1.414213180 1.414214134

9 Decimal Expansion of sqrt(2)  Does the expansion terminate or repeat?  Maybe it doesn’t?  How long should I look?  Maybe it’s not rational?

10 Proof that sqrt(2) is not Rational

11 Negation Rules

12 Introduction and elimination rules Double negation     e 

13 Proof by Contradiction Negation elimination called proof by contradiction Assume  and derive a a contradiction

14 Exercise Prove that  A  A and A   A

15 Law of the Excluded Middle 1  (p  p) assumption 2Assumption 3 (p  p) 4 5  p 6 p  p 7 8

16 De Morgan’s Law  (P  Q)   P   Q 1 (P  Q)(P  Q) premise 2assumption 3 P  QP  Q  i1 2 4 5 PP 6Q 7 P  QP  Q  i2 6 8 9 10 P  QP  Q  i 5,9

17 De Morgan’s Law  (P  Q)   P   Q 1 P  QP  Q premise 2  e1 1 3  e2 1 4assumption 5P 6  e 2,5 7Q  i2 6 8  e 3,7 9  e 4,5-6, 7-8 10 (P  Q)(P  Q)  i 4-9

18 Exercise Prove  (P  Q)   P   Q

19 Exercise Prove  P   Q   (P  Q)


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