Download Proof by Contradiction

Proof by Contradiction
Outline:
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Proposition: P is true.
Proof : Suppose ∼ P.
Proof by Contradiction
Outline:
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Proposition: P is true.
Proof : Suppose ∼ P.
..
.
We conclude that something ridiculous happens.
For example, 3 is both even and odd.
Proof by Contradiction
Outline:
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Proposition: P is true.
Proof : Suppose ∼ P.
..
.
We conclude that something ridiculous happens.
For example, 3 is both even and odd.
Therefore, P is true.
A First Example: Proof by Contradiction
Proposition: There are no natural number solutions to the equation
x 2 − y 2 = 1.
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
A First Example: Proof by Contradiction
Proposition: There are no natural number solutions to the equation
x 2 − y 2 = 1.
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Proof : Suppose x, y ∈ N and x 2 − y 2 = 1.
Then (x + y )(x − y ) = 1, so x − y and x + y are divisors of 1. Then
(x − y ) = (x + y ) = ±1.
So, 0 = (x + y ) − (x − y ) = 2y . Therefore y = 0, contradicting that it is
positive.
Rational Numbers
Proposition: There is no smallest positive rational number.
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Rational Numbers
Proposition: There is no smallest positive rational number.
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Proof : Suppose q is the smallest positive number. Then 0 < q2 < q and
q ∈ Q. This contradicts the assumption that q is the smallest positive rational
number.
The square root of two is irrational.
For p, q ∈ Z, q 6= 0, we say the fraction
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
q > 0.
p
is reduced if gcd(p, q) = 1 and
q
The square root of two is irrational.
For p, q ∈ Z, q 6= 0, we say the fraction
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
p
is reduced if gcd(p, q) = 1 and
q
q > 0.
For any integer a, a2 is even if and only if a is even.
The square root of two is irrational.
For p, q ∈ Z, q 6= 0, we say the fraction
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
p
is reduced if gcd(p, q) = 1 and
q
q > 0.
For any integer a, a2 is even if and only if a is even.
Proposition:
√
2 6∈ Q.
The square root of two is irrational.
For p, q ∈ Z, q 6= 0, we say the fraction
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
p
is reduced if gcd(p, q) = 1 and
q
q > 0.
For any integer a, a2 is even if and only if a is even.
√
Proposition: 2√6∈ Q.
Proof : Suppose 2 ∈ Q.
The square root of two is irrational.
For p, q ∈ Z, q 6= 0, we say the fraction
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
p
is reduced if gcd(p, q) = 1 and
q
q > 0.
For any integer a, a2 is even if and only if a is even.
√
Proposition: 2√6∈ Q.
Proof : Suppose 2 ∈ Q.
Then
√
2 =
2
p
q
p
,
q2
2
for some p, q ∈ Z, and WLOG
2
2
p
q
is a reduced fraction.
Then 2 =
so p = 2q , so 2|p. Then p = 2n for some n ∈ Z. Then
(2n)2 = 2q , so q 2 = 2n2 , so 2|q. Therefore gcd(p, q) ≥ 2. This contradicts that qp is a reduced fraction.
Therefore,
√
2 6∈ Q.
Proving Conditional Statements by Contradiction
Outline:
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Proposition: P =⇒ Q
Proof : Suppose P∧ ∼ Q.
Proving Conditional Statements by Contradiction
Outline:
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Proposition: P =⇒ Q
Proof : Suppose P∧ ∼ Q.
..
.
We conclude that something ridiculous happens.
Proving Conditional Statements by Contradiction
Outline:
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Proposition: P =⇒ Q
Proof : Suppose P∧ ∼ Q.
..
.
We conclude that something ridiculous happens.
Therefore, ∼ P ∨ Q. That is, P =⇒ Q.
Proving Conditional Statements by Contradiction
Outline:
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Proposition: P =⇒ Q
Proof : Suppose P∧ ∼ Q.
..
.
We conclude that something ridiculous happens.
Therefore, ∼ P ∨ Q. That is, P =⇒ Q.
Lemma: Given integers a and b, with a > 1, if a|b then a 6 | (b + 1).
Proving Conditional Statements by Contradiction
Outline:
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Proposition: P =⇒ Q
Proof : Suppose P∧ ∼ Q.
..
.
We conclude that something ridiculous happens.
Therefore, ∼ P ∨ Q. That is, P =⇒ Q.
Lemma: Given integers a and b, with a > 1, if a|b then a 6 | (b + 1).
Proof : Suppose a|b and a|(b + 1). Then there exist integers x and y such
that ax = b and ay = b + 1. Then y = b+1
= ax+1
= x + 1a . Since a > 1, 1a is
a
a
not an integer, so y is not an integer. This contradicts the assumption that y
is an integer.
How Many Primes Are There?
Lemma: Given integers a and b, with a > 1, if a|b then a 6 | (b + 1).
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Theorem: Given any finite set P whose elements are prime numbers, there
exists a prime number that is not in P.
How Many Primes Are There?
Lemma: Given integers a and b, with a > 1, if a|b then a 6 | (b + 1).
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Theorem: Given any finite set P whose elements are prime numbers, there
exists a prime number that is not in P.
Proof : Let P be a finite set whose elements are prime numbers. If P = ∅,
then 3 is a prime not in P, so suppose P = {p1 , · · · , pn } and n ≥ 1. Let
n := p1 · . . . · pn + 1. Since n > 1, it has at least one prime factor p that is
greater than one. By the Lemma, no element of P divides n, so
p 6∈ P.
How Many Primes Are There?
Lemma: Given integers a and b, with a > 1, if a|b then a 6 | (b + 1).
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Theorem: Given any finite set P whose elements are prime numbers, there
exists a prime number that is not in P.
Proof : Let P be a finite set whose elements are prime numbers. If P = ∅,
then 3 is a prime not in P, so suppose P = {p1 , · · · , pn } and n ≥ 1. Let
n := p1 · . . . · pn + 1. Since n > 1, it has at least one prime factor p that is
greater than one. By the Lemma, no element of P divides n, so
p 6∈ P.
Example: P = {2, 3, 5, 7, 11, 13}; p = 30031 = 59 × 509;
59 and 509 are not in P.
How Many Primes Are There?
Lemma: Given integers a and b, with a > 1, if a|b then a 6 | (b + 1).
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Theorem: Given any finite set P whose elements are prime numbers, there
exists a prime number that is not in P.
Proof : Let P be a finite set whose elements are prime numbers. If P = ∅,
then 3 is a prime not in P, so suppose P = {p1 , · · · , pn } and n ≥ 1. Let
n := p1 · . . . · pn + 1. Since n > 1, it has at least one prime factor p that is
greater than one. By the Lemma, no element of P divides n, so
p 6∈ P.
Example: P = {2, 3, 5, 7, 11, 13}; p = 30031 = 59 × 509;
59 and 509 are not in P.
Corollary: There are infinitely many prime numbers.
How Many Primes Are There?
Lemma: Given integers a and b, with a > 1, if a|b then a 6 | (b + 1).
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Theorem: Given any finite set P whose elements are prime numbers, there
exists a prime number that is not in P.
Proof : Let P be a finite set whose elements are prime numbers. If P = ∅,
then 3 is a prime not in P, so suppose P = {p1 , · · · , pn } and n ≥ 1. Let
n := p1 · . . . · pn + 1. Since n > 1, it has at least one prime factor p that is
greater than one. By the Lemma, no element of P divides n, so
p 6∈ P.
Example: P = {2, 3, 5, 7, 11, 13}; p = 30031 = 59 × 509;
59 and 509 are not in P.
Corollary: There are infinitely many prime numbers.
Proof : We define P to be the set of all primes. If P is finite, then by the
Proposition, P is not the set of all primes. Therefore, P is infinite.
Prove the Following, Somehow
Proposition: The sum of a rational number and an irrational number is
irrational.
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Proposition: Every nonzero rational number can be expressed as the product
of two irrational numbers.
Prove the Following, Somehow
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Proposition: The sum of a rational number and an irrational number is
irrational.
Proof :
p
Let r ∈ Q and n ∈ R. Then r = for some p, q ∈ Z. Suppose r + n ∈ Q.
q
Then r + n = ba for some a, b ∈ Z. Now:
a
b
a
p
aq − bp
a
n = −r = − =
b
b
q
bq
r +n =
Since bq ∈ Z and aq − bp ∈ Z, we conclude n ∈ Q.
Proposition: Every nonzero rational number can be expressed as the product
of two irrational numbers.
Prove the Following, Somehow
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Proposition: The sum of a rational number and an irrational number is
irrational.
Proof :
p
Let r ∈ Q and n ∈ R. Then r = for some p, q ∈ Z. Suppose r + n ∈ Q.
q
Then r + n = ba for some a, b ∈ Z. Now:
a
b
a
p
aq − bp
a
n = −r = − =
b
b
q
bq
r +n =
Since bq ∈ Z and aq − bp ∈ Z, we conclude n ∈ Q.
Proposition: Every nonzero rational number can be expressed as the product
of two irrational numbers.
Proof :
√
If r ∈ Q − {0}, then r = 2 · √r 2 , and √r 2 6= 0. We claim √r 2 6∈ Q. If √r 2 ∈ Q,
√
then √r 2 = pq for some p, q ∈ Z and p 6= 0. Then 2 = qrp . Since qr ∈ Z and
√
p ∈ Z − {0}, 2 ∈ Q. But, we know that’s not true. Therefore,
√r 6∈ Q.
2
Prove the Following, Somehow
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Proposition: There are no rational number solutions to the equation
x 3 + x + 1 = 0.
Prove the Following, Somehow
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Proposition: There are no rational number solutions to the equation
x 3 + x + 1 = 0.
Proof : Suppose x ∈ Q and x 3 + x + 1 = 0. Then x = qp for two integers p
and q with gcd(p, q) = 1.
3
Then qp3 + pq + 1 = 0. Multiplying both sides of the equation by q 3 , it follows
that p 3 + pq 2 + q 3 = 0.
Case 1: p and q are both odd.
Case 2: p is odd and q is even.
Case 3: p is even and q is odd.
Case 4: p and q are both even.
Prove the Following, Somehow
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Lemma: The product of two odd numbers is odd. The product of an even
and odd number is even. The sum of two integers is even if and only if they
have the same parity.
Proposition: There are no rational number solutions to the equation
x 3 + x + 1 = 0.
Proof : Suppose x ∈ Q and x 3 + x + 1 = 0. Then x = qp for two integers p
and q with gcd(p, q) = 1.
3
Then qp3 + pq + 1 = 0. Multiplying both sides of the equation by q 3 , it follows
that p 3 + pq 2 + q 3 = 0.
Case 1: p and q are both odd.
Case 2: p is odd and q is even.
Case 3: p is even and q is odd.
Case 4: p and q are both even.
Prove the Following, Somehow
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Lemma: The product of two odd numbers is odd. The product of an even
and odd number is even. The sum of two integers is even if and only if they
have the same parity.
Proposition: There are no rational number solutions to the equation
x 3 + x + 1 = 0.
Proof : Suppose x ∈ Q and x 3 + x + 1 = 0. Then x = qp for two integers p
and q with gcd(p, q) = 1.
3
Then qp3 + pq + 1 = 0. Multiplying both sides of the equation by q 3 , it follows
that p 3 + pq 2 + q 3 = 0.
Case 1: p and q are both odd.
Case 2: p is odd and q is even.
Case 3: p is even and q is odd.
Case 4: p and q are both even.
Prove the Following, Somehow
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Lemma: The product of two odd numbers is odd. The product of an even
and odd number is even. The sum of two integers is even if and only if they
have the same parity.
Proposition: There are no rational number solutions to the equation
x 3 + x + 1 = 0.
Proof : Suppose x ∈ Q and x 3 + x + 1 = 0. Then x = qp for two integers p
and q with gcd(p, q) = 1.
3
Then qp3 + pq + 1 = 0. Multiplying both sides of the equation by q 3 , it follows
that p 3 + pq 2 + q 3 = 0.
Case 1: p and q are both odd. By repeated application of the lemma, p 3 ,
pq 2 , and q 3 are all odd, so p 3 + pq 2 + q 3 is odd, contradicting
p 3 + pq 2 + q 3 = 0.
Case 2: p is odd and q is even.
Case 3: p is even and q is odd.
Case 4: p and q are both even.
Prove the Following, Somehow
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Lemma: The product of two odd numbers is odd. The product of an even
and odd number is even. The sum of two integers is even if and only if they
have the same parity.
Proposition: There are no rational number solutions to the equation
x 3 + x + 1 = 0.
Proof : Suppose x ∈ Q and x 3 + x + 1 = 0. Then x = qp for two integers p
and q with gcd(p, q) = 1.
3
Then qp3 + pq + 1 = 0. Multiplying both sides of the equation by q 3 , it follows
that p 3 + pq 2 + q 3 = 0.
Case 1: p and q are both odd. By repeated application of the lemma, p 3 ,
pq 2 , and q 3 are all odd, so p 3 + pq 2 + q 3 is odd, contradicting
p 3 + pq 2 + q 3 = 0.
Case 2: p is odd and q is even. By repeated application of the lemma, p 3
is odd, while pq 2 and q 3 are even, so p 3 + pq 2 + q 3 is odd, contradicting
p 3 + pq 2 + q 3 = 0.
Case 3: p is even and q is odd.
Case 4: p and q are both even.
Prove the Following, Somehow
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Lemma: The product of two odd numbers is odd. The product of an even
and odd number is even. The sum of two integers is even if and only if they
have the same parity.
Proposition: There are no rational number solutions to the equation
x 3 + x + 1 = 0.
Proof : Suppose x ∈ Q and x 3 + x + 1 = 0. Then x = qp for two integers p
and q with gcd(p, q) = 1.
3
Then qp3 + pq + 1 = 0. Multiplying both sides of the equation by q 3 , it follows
that p 3 + pq 2 + q 3 = 0.
Case 1: p and q are both odd. By repeated application of the lemma, p 3 ,
pq 2 , and q 3 are all odd, so p 3 + pq 2 + q 3 is odd, contradicting
p 3 + pq 2 + q 3 = 0.
Case 2: p is odd and q is even. By repeated application of the lemma, p 3
is odd, while pq 2 and q 3 are even, so p 3 + pq 2 + q 3 is odd, contradicting
p 3 + pq 2 + q 3 = 0.
Case 3: p is even and q is odd. By repeated application of the lemma, p 3
and pq 2 are even, while q 3 is odd, so p 3 + pq 2 + q 3 is odd, contradicting
p 3 + pq 2 + q 3 = 0.
Case 4: p and q are both even.
Prove the Following, Somehow
6. Proof by
Contradiction
6.1
Proving
Statements
with Contradiction
6.2
Proving
Conditional
Statements
by Contradiction
6.3
Combining
Techniques
Lemma: The product of two odd numbers is odd. The product of an even
and odd number is even. The sum of two integers is even if and only if they
have the same parity.
Proposition: There are no rational number solutions to the equation
x 3 + x + 1 = 0.
Proof : Suppose x ∈ Q and x 3 + x + 1 = 0. Then x = qp for two integers p
and q with gcd(p, q) = 1.
3
Then qp3 + pq + 1 = 0. Multiplying both sides of the equation by q 3 , it follows
that p 3 + pq 2 + q 3 = 0.
Case 1: p and q are both odd. By repeated application of the lemma, p 3 ,
pq 2 , and q 3 are all odd, so p 3 + pq 2 + q 3 is odd, contradicting
p 3 + pq 2 + q 3 = 0.
Case 2: p is odd and q is even. By repeated application of the lemma, p 3
is odd, while pq 2 and q 3 are even, so p 3 + pq 2 + q 3 is odd, contradicting
p 3 + pq 2 + q 3 = 0.
Case 3: p is even and q is odd. By repeated application of the lemma, p 3
and pq 2 are even, while q 3 is odd, so p 3 + pq 2 + q 3 is odd, contradicting
p 3 + pq 2 + q 3 = 0.
Case 4: p and q are both even. Then 2 divides both p and q. This
contradicts that gcd(p, q) = 1.