1. Proof Methods & Strategy

2. 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?
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3. 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 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)
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4. 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

5. 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:
= = =
Copyright © Peter Cappello 2011

6. A Non-Constructive Existence Proof
If , 2, …, 10 are placed randomly in a circle
then the sum of some 3 adjacent numbers  17.
Proof given previously.
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7. 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 ) ).
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8. 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