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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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. ■
(E.g.,“There is no greatest integer” ).
sets which are easier to handle with.
(E.g.,“is irrational”; rational numbers are more structured and easier to handle with than irrational numbers).
(E.g.,“The set of prime numbers is infinite” ).
Theorem: is irrational.
Proof:Assume the opposite: is rational.
Then by definition of rational numbers,
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)
(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.
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).■