- 262 Views
- Updated On :
- Presentation posted in: General

Indirect Argument: Contradiction and Contraposition. Method of Proof by Contradiction. Suppose the statement to be proved is false. Show that this supposition logically leads to a contradiction. Conclude that the statement to be proved is true. Method of Proof by Contradiction (Ex.).

Indirect Argument: Contradiction and Contraposition

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Indirect Argument: Contradiction and Contraposition

- Suppose the statement to be proved is false.
- Show that this supposition logically leads to a contradiction.
- Conclude that the statement to be proved is true.

Theorem: There is no least positive rational number.

Proof:Suppose the opposite:

least positive rational number x.

That is, xQ+ s.t. for yQ+, y≥x. (1)

Consider the number y*=x/2.

x>0 implies that y*=x/2>0.(2)

xQimplies thaty*=x/2Q.(3)

x>0 implies thaty*=x/2<x.(4)

Based on (2),(3),(4), y*Q+ and y<x.

This contradicts (1).

Thus, the supposition is false and

there is no least positive rational number.■

1. Express the statement to be proved in the form:

xD, if P(x) then Q(x) .

2. Rewrite in the contrapositive form:

xD, if Q(x) is false then P(x) is false.

3. Prove the contrapositive by a direct proof:

(a) Suppose x is an element of D

such that Q(x) is false.

(b) Show that P(x) is false.

Proposition 1:For any integer n,

if n2 is even then n is also even.

Proof: The contrapositive is:

For any integer n,

if n is not even then n2 is not even. (1)

Let’s show (1) by direct proof.

Supposen isnot even.

Then n is odd. So n=2k+1 for some kZ.

Hence n2 =(2k+1)2=4k2+4k+1

Thus, n2 isnot even. ■

- Advantage of contradiction method:
- Contrapositive method only for universal conditional statements.
- Contradiction method is more general.

- Advantage of contrapositive method:
- Easier structure: after the first step, Contrapositive method requires a direct proof.
- Contradiction method normally has more complicated structure.

- Statements starting with “There is no”.
(E.g.,“There is no greatest integer” ).

- If the negation of the statement deals with
sets which are easier to handle with.

(E.g.,“is irrational”; rational numbers are more structured and easier to handle with than irrational numbers).

- If the infinity of some set to be shown.
(E.g.,“The set of prime numbers is infinite” ).

Theorem: is irrational.

Proof:Assume the opposite:is rational.

Then by definition of rational numbers,

(1)

where m and n are integers with no common factors.

( by dividing m and n by any common factors if necessary)

Squaring both sides of (1),

Then m2=2n2 (by basic algebra) (2)

Proof (cont.):

(2) implies thatm2is even. (by definition)

Then m is even. (by Proposition 1)(3)

So m=2k for some integer k. (by definition)(4)

By substituting (4) into (2):

2n2 = m2 =(2k)2 = 4k2.

By dividing both sides by 2, n2 = 2k2.

Thus, n2 is even (by definition) and n is even (by Prop. 1).(5)

Based on (3) and (5),

m and n have a common factor of 2.

This contradicts (1). ■

Lemma 1: For any integer a and

any prime number p,

ifp|a then p doesn’t divide a+1.

Proof (by contradiction):

Assume the opposite: p|a and p|(a+1).

Then a=p·n and a+1=p·m for some n,m Z.

So 1=(a+1)-a=p·(m-n) which implies that p|1.

But the only integer divisors of 1 are 1 and -1.

Contradiction.■

- Theorem: The set of prime numbers is infinite.
- Proof (by contradiction): Assume the opposite:
The set of prime numbers is finite.

Then they can be listed as

p1=2, p2=3, …, pn in ascending order.

Consider M = p1· p2·…·pn+1.

p|M for some prime number p (1)

(based on the th-m from handout 9/23).

p is one of p1, p2, …, pn..

Thus, p | p1· p2·…·pn..(2)

By (2) and Lemma 1, p is not a divisor of M. (3)

(3) contradicts (1).■