3.6. Disproving Quantified Statements Disproving Existential Statements

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Louise Weaver
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1 36 Dproving Quantified Statements 361 Dproving Extential Statements A statement of the form x D, P( if P ( false for all x D false if and only To dprove th kind of statement, we need to show the for all x D, P ( false That we need to prove it s negation: ~ ( x D, P( ) x D,~ P( Th equivalent to proving a universal statement and so the method of exhaustion or the generalized proof method used Example: Dprove: There exts a positive number n such that n + 3n + prime Proving the given statement false equivalent to proving its negation true That proving that for all numbers n, n + 3n + not prime Since th statement universal, its proof requires the generaled proof method WUCT11 Logic 108

2 36 Dproving Universal Statements A statement of the form x D, P( if P ( false for at least one x D false if and only To dprove th kind of statement, we need to find one x D such that P ( false That we need to prove it s negation: ~ ( x D, P( ) x D,~ P( Th known as finding a counterexample Example: Dprove: a, b,( a = b ) ( a = b) Let P ( a, b) : ( a = b ) ( a = b) We need to show a, b,~ P( a, b) Counterexample: Let a = 1, b = 1 Then a = b however a b Now true false false, thus P ( a, b) false, and ~ P ( a, b) true So, we have shown, by counterexample a, b,~ P( a, b) WUCT11 Logic 109

3 Exerces: Dprove: x,( x > 0 x < 0) z 1 Dprove z,( z odd ) ( odd) Prove or dprove: x, y,( x + y = 0) WUCT11 Logic 110

4 37 Generaled Proof Methods Before proving a statement, it of great use to write the statement using logic notation, including quantifiers, where appropriate Doing th means you have clearly written the assumptions you can make AND the conclusion you are aiming to reach Example: Prove: For all integers n, if n odd, then n odd Rewritten using logic notation: n, n odd n odd Here the domain given as,, and the predicate involves the statements: P(n) n odd, Q(n) The form of the predicate P( n) Q( n) n odd Thus the assumption that can be made P(n) and the conclusion to be reached Q(n) WUCT11 Logic 111

5 371 Direct Proof A direct proof one in which we begin with the assumptions and work in a straightforward fashion to the conclusion The steps in the final proof must proceed in the correct direction beginning with the initial assumption and following known laws, rules, definitions etc until the final conclusion reached The proof must not start with what you are trying to prove Example: Prove that if 3 x 9 = 15 then x = 8 Assuming the domain to be, then the statement of the form x, P( Q( Thus the assumption P ( : 3x 9 = 15, and the conclusion Q ( : x = 8 3x 9 = 15 3x x = 4 3x 4 = 3 3 x = 8 = WUCT11 Logic 11

6 Exerce: Prove: For all integers n, if n odd, then n odd Rewritten using logic notation: Here the domain given as,, and the predicate involves the statements: P(n) n odd, Q(n) The form of the predicate P( n) Q( n) n odd Thus the assumption that can be made P(n) and the conclusion to be reached Q(n) Proof: WUCT11 Logic 113

7 When the statement to be proven of the form: P( Q(, the assumption which begins the proof P ( If the form not P( Q(, or P ( not clear, it may be necessary to examine what you are aiming to prove and establh an assumption from which to begin Example: ( ) Prove that for, x + x + 1 (Do not start with th!) x In th case, the form not P( Q( By examining what we are aiming to prove, ie ( ) + x + 1 x a beginning to the proof can be found working : x + x + 1 x x x ( x 1) We can now put the proof together: We know that ( 1) 0 Thus, x 1 0 x for any x true WUCT11 Logic 114

8 ( x 1) x x x 0 x x 1 + x + 1 what known expanding subtracting from both multiplying by 1 sides Note In the example, we did NOT start with the statement ( ) x + x + 1, as we technically do not know whether it true or not We started our proof with a statement we know to be true Exerce: Prove the following: If the right angled triangle XYZ with sides of length x and y z and hypotenuse z has an area of 4, then the triangle osceles X z y Y x Z WUCT11 Logic 115

9 The form of statement P( Q( What known ie the assumptions that can be made are: What to be proven: That triangle XYZ osceles Thus we must show two sides have equal length Proof: So two sides are equal and thus triangle XZY osceles WUCT11 Logic 116

10 38 Indirect Proofs 381 Method of Proof by Contradiction The method of proof by contradiction can be used when the statement to be proven not of the form P( Q( The method as follows: 1 Suppose the statement to be proven false That, suppose that the negation of the statement true Show that th supposition leads to a contradiction 3 Conclude that the statement to be proven true Example: Prove there no greatest integer Suppose not, that suppose there a greatest integer N Then N n for every integer n Let M = N + 1 Now M an integer since it the sum of integers Also M = N +1 M > N since Thus M an integer that greater than N So N the greatest integer and N not the greatest integer, which a contradiction Thus the assumption that there a greatest integer false, hence there no greatest integer true WUCT11 Logic 117

11 Exerce: Prove there no integer that both even and odd WUCT11 Logic 118

12 38 Proof by Contraposition Recall the following logical equivalence: ( P Q) (~ Q ~ P) (~ Q ~ P) known as the contrapositive of ( P Q) Th equivalence indicates that if (~ Q ~ P) a true statement, then so too ( P Q) Thus, in order to prove ( P Q), we prove the contrapositive, that (~ Q ~ P), true The method of proof by contraposition can be used when the statement to be proven of the form P( Q( The method as follows: 1 Express the statement to be proven in the form: x D, P( Q( Rewrite the statement in the contrapositive form: x D, ~ Q( ~ P( 3 Prove the contrapositive by a direct proof a Suppose that x a particular but arbitrary element of D, such that Q( false b Show that P( false WUCT11 Logic 119

13 Example: Prove that for all integers n if n even, n even The statement can be expressed in the form: n, n even n even Thus the contrapositive n, n not even n not even That n, n odd n odd To prove the contrapositive: Let n be any odd integer Then n = k + 1, k K(1) Show n odd, ie show n = l + 1, l So n = (k = 4k = (k = l + 1) k by (1) k) + 1 l = k + k n odd, and the contrapositive true Hence the statement for all integers n if even also true n even, n WUCT11 Logic 10

14 Exerce: Prove that for all integers n if 5 / n then 5 / n The statement can be expressed in the form: Thus the contrapositive To prove the contrapositive: WUCT11 Logic 11

15 Exerce: Prove if y irrational, then y + 7 irrational The statement can be expressed in the form: Thus the contrapositive To prove the contrapositive: WUCT11 Logic 1

16 383 Proof by Cases When the statement to be proven of the form, or can be written in the form: cases can be used It relies on the logical equivalence ( P Q) R, the method of proof by (( P R) ( Q R)) (( P Q) R) The method as follows: 1 Prove P R Prove Q R 3 Conclude ( P Q) R If the statement not written in the form ( P Q) R, it necessary to establh the particular cases by exhaustion WUCT11 Logic 13

17 Example: Prove: If x 0 or y 0, then x + y > 0 The statement can be expressed in the form: ( x 0) ( y 0) x + y > 0 We assume x, y, thus x 0, y 0 Proof: Case 1: Prove x 0 x + y > 0 > Let x 0, then x 0 and y 0 Thus x + y > 0 Case : Prove y 0 x + y > 0 > Let y 0, then y 0 and x 0 Thus x + y > 0 Therefore If x 0 or y 0, then x + y > 0 WUCT11 Logic 14

18 Exerce Prove: If x or x, then x 4 0 The statement can be expressed in the form: Proof: Case 1: Prove Therefore x x 4 0 Case : Prove Therefore x x 4 0 Thus if x or x, then x 4 0 WUCT11 Logic 15

19 If the statement not written in the form ( P Q) R, it necessary to establh the particular cases by exhaustion Example Prove: m, m + m + 1 odd The statement not in the form ( P Q) R However by considering m ( m even) ( m odd) Then the statement can be expressed in the form: ( m even) ( m odd) m + m + 1 odd Case 1: Prove m even m + m + 1 odd m even m = p, p K(1) m + m + 1 = ( p) + p + 1 by(1) = = 4 p + p + 1 ( ) p + p + 1 = k + 1, where k = ( ) p + p Therefore, m even m + m + 1 odd WUCT11 Logic 16

20 Case : Prove m odd m + m + 1 odd m odd m = q + 1, q K() m + m + 1 = ( q + 1) = 4q = 4q + 4q + 1+ q q ( ) q + 3q = ( l 1, where l q + 3q + 1 ) = + = + q by() Therefore, m odd m + m + 1 odd Therefore, ( m even) ( m odd) m + m + 1 odd Therefore, m, m + m + 1 odd WUCT11 Logic 17

21 Exerce Prove: n, n n + 3 odd The statement not in the form ( P Q) R However by considering n ( n even) ( n odd) Then the statement can be expressed in the form: ( n even) ( n odd) n n + 3 odd Case 1: WUCT11 Logic 18

22 Case : WUCT11 Logic 19

Proof by Contradiction

Proof by Contradiction MAT231 Transition to Higher Mathematics Fall 2014 MAT231 (Transition to Higher Math) Proof by Contradiction Fall 2014 1 / 12 Outline 1 Proving Statements with Contradiction 2 Proving