Proof Methods & Strategy

Copyright © Peter Cappello 2

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 3

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 2011 4

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:

103 + 93 = 1000 + 729 = 1728 + 1 = 123 + 13.

Copyright © Peter Cappello 2011 6

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 7

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 8

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 ) ).