Strategies for Proofs

Transcription

1 G. Carl Evans University of Illinois Summer 2013

2 Today Practice with proofs Become familiar with various strategies for proofs

3 Review: proving universal statements Claim: For any integer a, if a is odd, then a 2 is also odd. Definition: integer a is odd iff a = 2m + 1 for some integer m Suppose a Z is odd. Then by definition of odd, a = 2m + 1 for some m Z. So a 2 = (2m + 1) 2 = 4m 2 + 4m + 1 = 2(2m 2 + 2m) + 1 2m 2 + 2m is an integer since m Z. So a 2 is odd by definition QED.

4 Proving Existential Statements Claim: There exists a real number x, such that x 3 < x 2 Let x = 1 2. Then x 3 = 1 8 and x 2 = 1/4. Since 1 8 < 1 4 an x where x 3 < x 2 QED. there exists

5 Disproving Existential Statements Claim to disprove: There exists a real x, x 2 2x + 1 < 0

6 Disproving Existential Statements Claim to disprove: There exists a real x, x 2 2x + 1 < 0 In general, ( x, P(x)) x, P(x) ( x R, x 2 2x + 1 < 0) x R, (x 2 = 2x + 1 < 0) x R, x 2x Let x be a real number. x 2 2x + 1 = (x 1) 2. (x 1) 2 0 since x 1 is a real number and the square of a real number is non-negative QED.

7 Disproving Universal Statements Claim to disprove: For all real x, (x + 1) 2 > 0

8 Disproving Universal Statements Claim to disprove: For all real x, (x + 1) 2 > 0 In general, ( x, P(x)) x, P(x) ( x R, (x + 1) > 0) x R, (x + 1) 2 0) If x = 1, (x + 1) 2 = 0. So since 1 is a real there then exists a real that proves the negation of the claim and thus the original claim is disproved QED.

9 Proof By Cases Claim: for every real x, if x + 7 > 8, then x > 1 Suppose x R and x + 7 > 8. Then there are two cases. If x + 7 > 8, then x > 1, so x > 1. If x + 7 < 8, then x < 15, so x > 1. The conclusion holds for both cases QED.

10 Rephrasing Claims Claim: There is no integer k, such that k is odd and k 2 is even. ( k Z, odd(k) even(k 2 ) k Z, (odd(k) even(k 2 )) k Z, odd(k) even(k 2 ) k Z, even(k) odd(k 2 ) At this point there is a proof by cases.

11 Proof By Contrapositive Claim: For all integers a and b, (a + b 15) (a b 8) Contrapositive: a, b Z (a 8 b 8) (a + b 15) a, b Z, a < 8 b < 8 a + b < 15 To prove the claim we will prove the contrapositive, a, b Z, a < 8 b < 8 a + b < 15. Suppose a Z, a < 8 and a Z, b < 8. Then a 7 and b 7. So a + b 14 < 15 QED.

12 Proof strategies Does this proof require showing that the claim holds for all cases or just an example? Show all cases: prove universal, disprove existential Example: disprove universal, prove existential Can you figure a straightforward solution? If so, sketch it and then write it out clearly, and youre done If not, try to find an equivalent form that is easier Divide into subcases that combine to account for all cases OR in hypothesis is a hint that this may be a good idea Try the contrapositive OR in conclusion is a hint that this may be a good idea More generally rephrase the claim: convert to propositional logic and manipulate into something easier to solve

13 Claim: For integers j and k, if j is even or k is even, then jk is even. Definition: integer a is even iff a = 2m for some integer m

14 Claim: For all integers k, if 3k + 5 is even, then k is odd.

15 Disprove: For all real k, if k is rational, then k3 k is rational.

16 Alternate proof, by cases Claim: For any real k, if k is rational, then k 2 is rational. Definition: real k is rational iff k = m n for some integers m and n with n 0

17 Claim: For all integers x, if x is odd, then x = 4k + 1 or x = 4k 1 for some integer k.