Download § 3.3 Proof by Contradiction

§ 3.3 Proof by Contradiction
Idea Behind Proof by Contradiction
We establish P → Q by assuming the hypothesis P is true and that the
conclusion Q is false and then we use P, ¬Q and any pertinent
axioms, definitions, etc. to derive a contradiction.
Idea Behind Proof by Contradiction
We establish P → Q by assuming the hypothesis P is true and that the
conclusion Q is false and then we use P, ¬Q and any pertinent
axioms, definitions, etc. to derive a contradiction.
A proof by contradiction is sometimes called an indirect proof since
to establish P → Q using proof by contradiction, we follow an
indirect route: we derive R ∧ ¬R and then conclude Q is true.
Idea Behind Proof by Contradiction
We establish P → Q by assuming the hypothesis P is true and that the
conclusion Q is false and then we use P, ¬Q and any pertinent
axioms, definitions, etc. to derive a contradiction.
A proof by contradiction is sometimes called an indirect proof since
to establish P → Q using proof by contradiction, we follow an
indirect route: we derive R ∧ ¬R and then conclude Q is true.
The only difference between the assumptions made in a direct proof
and a proof by contradiction is that in the latter, we negate the
conclusion. That is, we have logical equivalence between
P→Q
and
(P ∧ ¬Q) → (R ∧ ¬R)
First Example
Example
Show that at least 4 of any 22 days must fall on the same day of the
week.
First Example
Example
Show that at least 4 of any 22 days must fall on the same day of the
week.
Proof.
Let P be the proposition ‘At least 4 of 22 chosen days fall on the same
day of the week’.
First Example
Example
Show that at least 4 of any 22 days must fall on the same day of the
week.
Proof.
Let P be the proposition ‘At least 4 of 22 chosen days fall on the same
day of the week’. Suppose that ¬P is true. This means that at most of
3 of the 22 days fall on the same day of the week.
First Example
Example
Show that at least 4 of any 22 days must fall on the same day of the
week.
Proof.
Let P be the proposition ‘At least 4 of 22 chosen days fall on the same
day of the week’. Suppose that ¬P is true. This means that at most of
3 of the 22 days fall on the same day of the week. Because there are 7
days in a week, this implies that at most 21 days could be chosen
because for each of the days of the week, at most 3 of the chosen days
could fall on that day.
First Example
Example
Show that at least 4 of any 22 days must fall on the same day of the
week.
Proof.
Let P be the proposition ‘At least 4 of 22 chosen days fall on the same
day of the week’. Suppose that ¬P is true. This means that at most of
3 of the 22 days fall on the same day of the week. Because there are 7
days in a week, this implies that at most 21 days could be chosen
because for each of the days of the week, at most 3 of the chosen days
could fall on that day. This contradicts that we have 22 days under
consideration.
First Example
Example
Show that at least 4 of any 22 days must fall on the same day of the
week.
Proof.
Let P be the proposition ‘At least 4 of 22 chosen days fall on the same
day of the week’. Suppose that ¬P is true. This means that at most of
3 of the 22 days fall on the same day of the week. Because there are 7
days in a week, this implies that at most 21 days could be chosen
because for each of the days of the week, at most 3 of the chosen days
could fall on that day. This contradicts that we have 22 days under
consideration. That is, if R is the statement that ‘22 days are chosen’,
then we have that ¬P → (R ∧ ¬R). So, we know P is true.
Contradictions with Odd and Even Numbers
Example
Prove by contradiction: If 3n + 2 is odd, then n is odd.
Contradictions with Odd and Even Numbers
Example
Prove by contradiction: If 3n + 2 is odd, then n is odd.
This looks like a candidate for proof by contrapositive ... and it would
be if we didn’t put the method in the statement of the problem.
Contradictions with Odd and Even Numbers
Example
Prove by contradiction: If 3n + 2 is odd, then n is odd.
This looks like a candidate for proof by contrapositive ... and it would
be if we didn’t put the method in the statement of the problem.
Proof.
Assume 3n + 2 is odd but that n is even.
Contradictions with Odd and Even Numbers
Example
Prove by contradiction: If 3n + 2 is odd, then n is odd.
This looks like a candidate for proof by contrapositive ... and it would
be if we didn’t put the method in the statement of the problem.
Proof.
Assume 3n + 2 is odd but that n is even. Since n is even there exists
k ∈ Z such that n = 2k.
Contradictions with Odd and Even Numbers
Example
Prove by contradiction: If 3n + 2 is odd, then n is odd.
This looks like a candidate for proof by contrapositive ... and it would
be if we didn’t put the method in the statement of the problem.
Proof.
Assume 3n + 2 is odd but that n is even. Since n is even there exists
k ∈ Z such that n = 2k. Then
3n + 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1)
Contradictions with Odd and Even Numbers
Example
Prove by contradiction: If 3n + 2 is odd, then n is odd.
This looks like a candidate for proof by contrapositive ... and it would
be if we didn’t put the method in the statement of the problem.
Proof.
Assume 3n + 2 is odd but that n is even. Since n is even there exists
k ∈ Z such that n = 2k. Then
3n + 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1)
Since 3k + 1 ∈ Z, 3n + 2 is even, which contradicts our assumption
that 3n + 2 is odd when n is even.
Contradictions with Odd and Even Numbers
Example
Prove that no odd integer can be expressed as the sum of three even
integers.
Contradictions with Odd and Even Numbers
Example
Prove that no odd integer can be expressed as the sum of three even
integers.
Proof.
Assume, to the contrary that there exists an odd integer n that can be
expressed as the sum of three even integers x, y and z.
Contradictions with Odd and Even Numbers
Example
Prove that no odd integer can be expressed as the sum of three even
integers.
Proof.
Assume, to the contrary that there exists an odd integer n that can be
expressed as the sum of three even integers x, y and z. Since x, y and z
are even integers, there exists a, b, c ∈ Z such that x = 2a, y = 2b
and z = 2c.
Contradictions with Odd and Even Numbers
Example
Prove that no odd integer can be expressed as the sum of three even
integers.
Proof.
Assume, to the contrary that there exists an odd integer n that can be
expressed as the sum of three even integers x, y and z. Since x, y and z
are even integers, there exists a, b, c ∈ Z such that x = 2a, y = 2b
and z = 2c. Therefore
n = x + y + z = 2a + 2b + 2c = 2(a + b + c)
Contradictions with Odd and Even Numbers
Example
Prove that no odd integer can be expressed as the sum of three even
integers.
Proof.
Assume, to the contrary that there exists an odd integer n that can be
expressed as the sum of three even integers x, y and z. Since x, y and z
are even integers, there exists a, b, c ∈ Z such that x = 2a, y = 2b
and z = 2c. Therefore
n = x + y + z = 2a + 2b + 2c = 2(a + b + c)
Since a + b + c is an integer, n is even, contradicting our assumption
that n is odd.
Contradiction With DeMorgan
Example
Prove that for all real numbers x and y, if x + y ≥ 2, then either x ≥ 1
or y ≥ 1.
Contradiction With DeMorgan
Example
Prove that for all real numbers x and y, if x + y ≥ 2, then either x ≥ 1
or y ≥ 1.
Where we need DeMorgan’s Law is with the statement of the
negation of the conclusion. ¬[(x ≥ 1) ∨ (y ≥ 1)] ≡
Contradiction With DeMorgan
Example
Prove that for all real numbers x and y, if x + y ≥ 2, then either x ≥ 1
or y ≥ 1.
Where we need DeMorgan’s Law is with the statement of the
negation of the conclusion. ¬[(x ≥ 1) ∨ (y ≥ 1)] ≡(x < 1) ∧ (y < 1).
Contradiction With DeMorgan
Example
Prove that for all real numbers x and y, if x + y ≥ 2, then either x ≥ 1
or y ≥ 1.
Where we need DeMorgan’s Law is with the statement of the
negation of the conclusion. ¬[(x ≥ 1) ∨ (y ≥ 1)] ≡(x < 1) ∧ (y < 1).
Proof.
Let x, y ∈ R. Assume x + y ≥ 2 but that x < 1 and y < 1.
Contradiction With DeMorgan
Example
Prove that for all real numbers x and y, if x + y ≥ 2, then either x ≥ 1
or y ≥ 1.
Where we need DeMorgan’s Law is with the statement of the
negation of the conclusion. ¬[(x ≥ 1) ∨ (y ≥ 1)] ≡(x < 1) ∧ (y < 1).
Proof.
Let x, y ∈ R. Assume x + y ≥ 2 but that x < 1 and y < 1. Then,
x+y<1+1=2
a contradiction, since we have assumed x + y ≥ 2 but have shown that
x + y < 2.
Contradictions with Reals
Example
Prove there is no smallest positive real number.
Contradictions with Reals
Example
Prove there is no smallest positive real number.
How would you prove this directly?
Contradictions with Reals
Example
Prove there is no smallest positive real number.
How would you prove this directly?
Proof.
Assume to the contrary that there is a smallest positive real number,
say r.
Contradictions with Reals
Example
Prove there is no smallest positive real number.
How would you prove this directly?
Proof.
Assume to the contrary that there is a smallest positive real number,
say r. Since 0 < 2r < r, it follows that 2r is a positive real number that
is smaller than r, which contradicts our choice of r as the smallest
positive real number.
Contradictions with Divisibility
Example
If a is an even integer and b is an odd integer, then 4 6 |(a2 + 2b2 ).
Contradictions with Divisibility
Example
If a is an even integer and b is an odd integer, then 4 6 |(a2 + 2b2 ).
Proof.
Assume, to the contrary, that there exists an even integer a and an odd
integer b such that 4|(a2 + 2b2 ).
Contradictions with Divisibility
Example
If a is an even integer and b is an odd integer, then 4 6 |(a2 + 2b2 ).
Proof.
Assume, to the contrary, that there exists an even integer a and an odd
integer b such that 4|(a2 + 2b2 ). Thus, there exists x, y, z ∈ Z such
that a = 2x, b = 2y + 1 and a2 + 2b2 = 4z.
Contradictions with Divisibility
Example
If a is an even integer and b is an odd integer, then 4 6 |(a2 + 2b2 ).
Proof.
Assume, to the contrary, that there exists an even integer a and an odd
integer b such that 4|(a2 + 2b2 ). Thus, there exists x, y, z ∈ Z such
that a = 2x, b = 2y + 1 and a2 + 2b2 = 4z. Consider
a2 + 2b2 = (2x)2 + 2(2y + 1)2 = 4z
Contradictions with Divisibility
Example
If a is an even integer and b is an odd integer, then 4 6 |(a2 + 2b2 ).
Proof.
Assume, to the contrary, that there exists an even integer a and an odd
integer b such that 4|(a2 + 2b2 ). Thus, there exists x, y, z ∈ Z such
that a = 2x, b = 2y + 1 and a2 + 2b2 = 4z. Consider
a2 + 2b2 = (2x)2 + 2(2y + 1)2 = 4z
When we simplify, we obtain
4x2 + 8y2 + 8y + 2 = 4z
Contradictions with Divisibility
Example
If a is an even integer and b is an odd integer, then 4 6 |(a2 + 2b2 ).
Proof.
Assume, to the contrary, that there exists an even integer a and an odd
integer b such that 4|(a2 + 2b2 ). Thus, there exists x, y, z ∈ Z such
that a = 2x, b = 2y + 1 and a2 + 2b2 = 4z. Consider
a2 + 2b2 = (2x)2 + 2(2y + 1)2 = 4z
When we simplify, we obtain
4x2 + 8y2 + 8y + 2 = 4z
That is,
2 = 4z − 4x2 − 8y2 − 8y = 4(z − x2 − 2y2 − 2y)
Contradictions with Divisibility
Example
If a is an even integer and b is an odd integer, then 4 6 |(a2 + 2b2 ).
Proof.
Assume, to the contrary, that there exists an even integer a and an odd
integer b such that 4|(a2 + 2b2 ). Thus, there exists x, y, z ∈ Z such
that a = 2x, b = 2y + 1 and a2 + 2b2 = 4z. Consider
a2 + 2b2 = (2x)2 + 2(2y + 1)2 = 4z
When we simplify, we obtain
4x2 + 8y2 + 8y + 2 = 4z
That is,
2 = 4z − 4x2 − 8y2 − 8y = 4(z − x2 − 2y2 − 2y)
Since z − x2 − 2y2 − 2y ∈ Z, we have 4|2, which is impossible.
Contradictions with Irrationality
Example
Prove that
√
2 is irrational.
Contradictions with Irrationality
Example
Prove that
Proof.
√
2 is irrational.
√
Suppose that 2 is a √
rational number. Then, there exists p, q ∈ Z
with q 6= 0 such that 2 = pq , where p and q have no common factors.
Contradictions with Irrationality
Example
Prove that
Proof.
√
2 is irrational.
√
Suppose that 2 is a √
rational number. Then, there exists p, q ∈ Z
with q 6= 0 such that 2 = pq , where p and q have no common factors.
2
Then, we can assume that qp = 2.
Contradictions with Irrationality
Example
Prove that
Proof.
√
2 is irrational.
√
Suppose that 2 is a √
rational number. Then, there exists p, q ∈ Z
with q 6= 0 such that 2 = pq , where p and q have no common factors.
2
Then, we can assume that qp = 2. Note that p2 = 2q2 , so p2 is
even.
Contradictions with Irrationality
Example
Prove that
Proof.
√
2 is irrational.
√
Suppose that 2 is a √
rational number. Then, there exists p, q ∈ Z
with q 6= 0 such that 2 = pq , where p and q have no common factors.
2
Then, we can assume that qp = 2. Note that p2 = 2q2 , so p2 is
even. This implies that p is even.
Contradictions with Irrationality
Example
Prove that
Proof.
√
2 is irrational.
√
Suppose that 2 is a √
rational number. Then, there exists p, q ∈ Z
with q 6= 0 such that 2 = pq , where p and q have no common factors.
2
Then, we can assume that qp = 2. Note that p2 = 2q2 , so p2 is
even. This implies that p is even. Thus, p = 2n for some integer n.
Contradictions with Irrationality
Example
Prove that
Proof.
√
2 is irrational.
√
Suppose that 2 is a √
rational number. Then, there exists p, q ∈ Z
with q 6= 0 such that 2 = pq , where p and q have no common factors.
2
Then, we can assume that qp = 2. Note that p2 = 2q2 , so p2 is
even. This implies that p is even. Thus, p = 2n for some integer n.
We see that 2q2 = p2 = (2n)2 = 4n2 , so q2 = 2n2 .
Contradictions with Irrationality
Example
Prove that
Proof.
√
2 is irrational.
√
Suppose that 2 is a √
rational number. Then, there exists p, q ∈ Z
with q 6= 0 such that 2 = pq , where p and q have no common factors.
2
Then, we can assume that qp = 2. Note that p2 = 2q2 , so p2 is
even. This implies that p is even. Thus, p = 2n for some integer n.
We see that 2q2 = p2 = (2n)2 = 4n2 , so q2 = 2n2 . Thus q2 is even
and so q is even as well.
Contradictions with Irrationality
Example
Prove that
Proof.
√
2 is irrational.
√
Suppose that 2 is a √
rational number. Then, there exists p, q ∈ Z
with q 6= 0 such that 2 = pq , where p and q have no common factors.
2
Then, we can assume that qp = 2. Note that p2 = 2q2 , so p2 is
even. This implies that p is even. Thus, p = 2n for some integer n.
We see that 2q2 = p2 = (2n)2 = 4n2 , so q2 = 2n2 . Thus q2 is even
and so q is even as well. Since p and q are both even, we have a
common factor of 2, contradicting our assumption. Therefore our
assumption must be false.
Another Proof with Irrational Numbers
Example
Prove that the sum of a rational number and an irrational number is
irrational.
Another Proof with Irrational Numbers
Example
Prove that the sum of a rational number and an irrational number is
irrational.
Proof.
Assume, to the contrary, that there exists a rational number x and an
irrational number y whose sum is a rational number z.
Another Proof with Irrational Numbers
Example
Prove that the sum of a rational number and an irrational number is
irrational.
Proof.
Assume, to the contrary, that there exists a rational number x and an
irrational number y whose sum is a rational number z. That is,
x + y = z, where x = ab and z = dc for a, b, c, d ∈ Z with b 6= 0 and
d 6= 0.
Another Proof with Irrational Numbers
Example
Prove that the sum of a rational number and an irrational number is
irrational.
Proof.
Assume, to the contrary, that there exists a rational number x and an
irrational number y whose sum is a rational number z. That is,
x + y = z, where x = ab and z = dc for a, b, c, d ∈ Z with b 6= 0 and
d 6= 0. Consider
x+y=z
Another Proof with Irrational Numbers
Example
Prove that the sum of a rational number and an irrational number is
irrational.
Proof.
Assume, to the contrary, that there exists a rational number x and an
irrational number y whose sum is a rational number z. That is,
x + y = z, where x = ab and z = dc for a, b, c, d ∈ Z with b 6= 0 and
d 6= 0. Consider
x+y=z
a
c
+y=
b
d
Another Proof with Irrational Numbers
Example
Prove that the sum of a rational number and an irrational number is
irrational.
Proof.
Assume, to the contrary, that there exists a rational number x and an
irrational number y whose sum is a rational number z. That is,
x + y = z, where x = ab and z = dc for a, b, c, d ∈ Z with b 6= 0 and
d 6= 0. Consider
x+y=z
a
c
+y=
b
d
c a
y= −
d b
Another Proof with Irrational Numbers
Example
Prove that the sum of a rational number and an irrational number is
irrational.
Proof.
Assume, to the contrary, that there exists a rational number x and an
irrational number y whose sum is a rational number z. That is,
x + y = z, where x = ab and z = dc for a, b, c, d ∈ Z with b 6= 0 and
d 6= 0. Consider
x+y=z
a
c
+y=
b
d
c a
y= −
d b
bc − ad
y=
bd
Another Proof with Irrational Numbers
Example
Prove that the sum of a rational number and an irrational number is
irrational.
Proof.
Assume, to the contrary, that there exists a rational number x and an
irrational number y whose sum is a rational number z. That is,
x + y = z, where x = ab and z = dc for a, b, c, d ∈ Z with b 6= 0 and
d 6= 0. Consider
x+y=z
a
c
+y=
b
d
c a
y= −
d b
bc − ad
y=
bd
Since bc − ad ∈ Z and bd 6= 0, it follows that y is rational, a
contradiction.
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