Proof By Contradiction

1
Proof By Contradiction

• Chapter 3
• Indirect Argument
• Contradiction
• Theorems 3.6.1 and 3.6.2
• pg. 171

2
Your Hosts

• Robert Di Battista
• Introduction, slides
• Daniel Rowan
• Theorem 3.6.1
• Annette Stiller
• Theorem 3.6.2

3
Proof By Contradiction

• Indirect Argument
• Prove the negation is false.
• reductio ad absurdum
• Assume its negation is true, until it is reduced
  to an impossibility or absurdity.
• This leaves only one possibility.

4
Method of Proof By Contradiction

• Suppose the statement to be proved is false, i.e.
  suppose the negation of the statement is true.
• Show that this supposition leads logically to a
  contradiction.
• Conclude that the statement to be proved is true.

5
Analogous to

• Shows truth by discounting the opposite.
• Sarcasm
• Reverse psychology

6
Theorem 3.6.1

• There is no greatest integer.

7
Theorem 3.6.1

• We take the negation of the theorem
• and suppose it to be true
• Suppose not. That is, suppose there is a greatest
  integer N.
• We must deduce
• a contradiction
• Then N gt n for every integer n.

8
Theorem 3.6.1

• Let M N 1.
• M is an integer since it is the sum of integers.
• Also M gt N since M N 1.

9
Theorem 3.6.1

• Thus M is an integer that is greater than N. So N
  is the greatest integer and N is not the greatest
  integer, which is a contradiction.
• This contradiction shows that the
• supposition is false and, hence,
• that the theorem is true.

10
Theorem 3.6.1

• Bill Gates is disgustingly rich, but someone can
  always have 1 more than him.

11
Theorem 3.6.1

• Exam question
• Prove by contradiction
• There is no greatest odd integer.

12
WAIT!
13
THERES MORE
14
Theorem 3.6.2

• There is no integer that
• is both even and odd.

15
Theorem 3.6.2

• We take the negation of the theorem
• and suppose it to be true
• Suppose not. That is, suppose there is an integer
  N that is both even and odd.
• We must deduce
• a contradiction.

16
Theorem 3.6.2

• By definition of even, n 2a for some integer a,
  and by definition of odd, n 2b 1 for some
  integer b. Consequently,
• 2a 2b 1 By equating the two
  expressions for n.
• And so
• 2a 2b 1
• 2(a b) 1
• (a b) 1/2 by algebra

17
Theorem 3.6.2

• Now since a and b are integers, the difference a
  b must also be an integer. But a b 1/2, and
  1/2 is not an integer. Thus a b is an integer
  and a b is not an integer, which is a
  contradiction.
• This contradiction shows that the
• supposition is false and, hence,
• that the theorem is true.

18
Analogous to

• Mutually exclusive groups
• Male or female
• Positive or negative
• True or false

19
Theorem 3.6.2

• Exam question
• Prove by contradiction
• There is no real number that is both positive
  and negative.

20
Related Homework

• Section 3.6
• 3 -15
• 21 27
• 32