Proof Methods & Strategy

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Presentation transcript:

Proof Methods & Strategy

Copyright © Peter Cappello Proof by Cases ( p1  p2  . . .  pn )  q  ( p1  q )  ( p2  q )  . . .  ( pn  q ). Prove the compound implication by proving each implication. Why do you have to prove all of them? Copyright © Peter Cappello

Copyright © Peter Cappello Example Let the domain be the integers. a ( ( 3 | a – 1  3 | a – 2 )  3 | a2 – 1 ). Proof: 1. Let a be an arbitrary integer. 2. Case 3 | a - 1: Let integer q = ( a – 1 ) / 3. (3 divides a – 1 with no remainder). a = 3q + 1. a2 = ( 3q + 1 )2 = 9q2 + 6q + 1 = 3( 3q2 + 2q ) + 1 a2 – 1 = 3(3q2 + 2q ) . 3 | a2 – 1 3. Case 3 | a - 2: (Proved previously). 4. a ( ( 3 | a – 1  3 | a – 2 )  3 | a2 – 1 ). (Universal Generalization) Copyright © Peter Cappello

Copyright © Peter Cappello 2011 Existence Proofs x P( x ) Constructive Produce a particular x such that P( x ) is true. Non-constructive Prove that there is an x such that P( x ) is true without actually producing it. Copyright © Peter Cappello 2011

A Constructive Existence Proof A constructive proof that constructive proofs exist  Let the domain be natural numbers Prove: There are distinct natural numbers a, b, c, d such that a3 + b3 = c3 + d3. Proof: 103 + 93 = 1000 + 729 = 1728 + 1 = 123 + 13. Copyright © Peter Cappello 2011

A Non-Constructive Existence Proof If 1, 2, …, 10 are placed randomly in a circle then the sum of some 3 adjacent numbers  17. Proof given previously. Copyright © Peter Cappello 2011

Copyright © Peter Cappello Uniqueness Proof There is exactly 1 x that makes P( x ) true. x ( P( x)  y ( x  y  P( y ) ) ). Alternatively (contrapositive of 2nd part), x ( P( x )  y ( P( y )  y = x ) ). Copyright © Peter Cappello

Copyright © Peter Cappello Example Domain: Integers Prove: There is a unique additive identity: x a ( a + x = a  y ( a + y = a  x = y ) ). Proof: Let a be an arbitrary integer. a + 0 = a (There is an additive identity for integers: x a a + x = a ) Let y be an arbitrary integer. Assume a + y = a. a + 0 = a + y  0 = y. (From steps 2 & 4.) x a ( a + x = a  y ( a + y = a  y = x ) ). Copyright © Peter Cappello