Download Indirect Argument: Contradiction and Contraposition

Indirect Argument:
Contradiction and Contraposition
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Method of Proof by Contradiction
1. Suppose the statement to be proved is
false.
2. Show that this supposition logically leads
to a contradiction.
3. Conclude that the statement to be
proved is true.
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Method of Proof by Contradiction (Ex.)
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Q implies that y*=x/2Q.
(3)
x>0 implies that y*=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.
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Method of Proof by
Contraposition
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.
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Method of Proof by
Contraposition (Example)
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.
Let’s show (1) by direct proof.
Suppose n is not even.
Then n is odd. So n=2k+1 for some kZ.
Hence n2 =(2k+1)2=4k2+4k+1
Thus, n2 is not even.
(1)
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Comparison of Contradiction and
Contrapositive methods
• 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.
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When to use indirect proof
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., “ 2 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” ).
Method of Proof by Contradiction
(Ex.)
Theorem:
2 is irrational.
Proof: Assume the opposite:
2 is rational.
m
Then by definition of rational numbers,
2
n (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
2
m
2 2
n
(by basic algebra)
(2)
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Method of Proof by Contradiction (Ex.)
Proof (cont.):
(2) implies that m2 is 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).
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Infinity of Prime Numbers
Lemma 1: For any integer a and
any prime number p,
if p|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.
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Infinity of Prime Numbers
• 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).
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