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# Indirect Argument: Contradiction and Contraposition - PowerPoint PPT Presentation

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.).

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### Indirect Argument: Contradiction and Contraposition

• Suppose the statement to be proved is false.

• 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, xQ+ s.t. for  yQ+, y≥x. (1)

Consider the number y*=x/2.

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

xQimplies thaty*=x/2Q. (3)

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

Based on (2),(3),(4), y*Q+ and y<x.

Thus, the supposition is false and

there is no least positive rational number. ■

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

 xD, if P(x) then Q(x) .

2. Rewrite in the contrapositive form:

 xD, 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 kZ.

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

Thus, n2 isnot even. ■

• Contrapositive method only for universal conditional statements.

• Contradiction method is more general.

• 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.

Lemma 1: For any integer a and

any prime number p,

ifp|a then p doesn’t divide a+1.

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.

• 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)