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CHAPTER 6 Proof by Contradiction, Study notes of Logic

City University of Seattle Logic

The basic idea is to assume that the statement we want to prove is false, and then show that this assumption leads to nonsense.

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CHAPTER 6

Proof by Contradiction

We now explore

a third method of proof:

proof by contradiction

.

This method is not limited to proving just conditional statements—

it can be used to prove any kind of statement whatsoever. The basic idea

is to assume that the statement we want to prove is false, and then show

that this assumption leads to nonsense. We are then led to conclude that

we were wrong to assume the statement was false, so the statement must

be true. As an example, consider the following proposition and its proof.

Proposition If a,b∈Z, then a2−4b6=2.

Proof. Suppose this proposition is false.

This conditional statement being false means there exist numbers

a

and

b

for which a,b∈Zis true, but a2−4b6=2is false.

In other words, there exist integers a,b∈Zfor which a2−4b=2.

From this equation we get a2=4b+2=2(2b+1), so a2is even.

Because a2is even, it follows that ais even, so a=2cfor some integer c.

Now plug a=2cback into the boxed equation to get (2c)2−4b=2,

so 4c2−4b=2. Dividing by 2, we get 2c2−2b=1.

Therefore 1=2(c2−b), and because c2−b∈Z, it follows that 1 is even.

We know 1 is not even, so something went wrong.

But all the logic after the first line of the proof is correct, so it must be

that the first line was incorrect. In other words, we were wrong to assume

the proposition was false. Thus the proposition is true. ■

You may be a bit suspicious of this line of reasoning, but in the next

section we will see that it is logically sound. For now, notice that at

the end of the proof we deduced that 1 is even, which conflicts with our

knowledge that

1

is odd. In essence, we have obtained the statement

(1 is odd)∧∼(1 is odd), which has the form C∧∼C. Notice that no matter

what statement

C

is, and whether or not it is true, the statement

C∧∼C

is false. A statement—like this one—that cannot be true is called a

contradiction. Contradictions play a key role in our new technique.

Partial preview of the text

Download CHAPTER 6 Proof by Contradiction and more Study notes Logic in PDF only on Docsity!

CHAPTER 6 Proof by Contradiction

W

e now explore a third method of proof: proof by contradiction.

This method is not limited to proving just conditional statements—

it can be used to prove any kind of statement whatsoever. The basic idea

is to assume that the statement we want to prove is false , and then show

that this assumption leads to nonsense. We are then led to conclude that

we were wrong to assume the statement was false, so the statement must

be true. As an example, consider the following proposition and its proof.

Proposition If a, b ∈ Z, then a^2 − 4 b 6 = 2.

Proof. Suppose this proposition is false.

This conditional statement being false means there exist numbers a and b

for which a, b ∈ Z is true, but a^2 − 4 b 6 = 2 is false.

In other words, there exist integers a, b ∈ Z for which a^2 − 4 b = 2.

From this equation we get a^2 = 4 b + 2 = 2(2b + 1), so a^2 is even.

Because a^2 is even, it follows that a is even, so a = 2 c for some integer c.

Now plug a = 2 c back into the boxed equation to get (2c)^2 − 4 b = 2 ,

so 4 c^2 − 4 b = 2. Dividing by 2, we get 2 c^2 − 2 b = 1.

Therefore 1 = 2(c^2 − b), and because c^2 − b ∈ Z, it follows that 1 is even.

We know 1 is not even, so something went wrong.

But all the logic after the first line of the proof is correct, so it must be

that the first line was incorrect. In other words, we were wrong to assume

the proposition was false. Thus the proposition is true. ■

You may be a bit suspicious of this line of reasoning, but in the next

section we will see that it is logically sound. For now, notice that at

the end of the proof we deduced that 1 is even, which conflicts with our

knowledge that 1 is odd. In essence, we have obtained the statement

(1 is odd)∧ ∼ (1 is odd), which has the form C∧ ∼ C. Notice that no matter

what statement C is, and whether or not it is true, the statement C∧ ∼ C

is false. A statement—like this one—that cannot be true is called a

contradiction. Contradictions play a key role in our new technique.

112 Proof by Contradiction

6.1 Proving Statements with Contradiction

Let’s now see why the proof on the previous page is logically valid. In

that proof we needed to show that a statement P : (a, b ∈ Z) ⇒ (a^2 − 4 b 6 = 2)

was true. The proof began with the assumption that P was false, that is

that ∼ P was true, and from this we deduced C∧ ∼ C. In other words we

proved that ∼ P being true forces C∧ ∼ C to be true, and this means that

we proved that the conditional statement (∼ P) ⇒ (C ∧ ∼ C) is true. To see

that this is the same as proving P is true, look at the following truth table

for (∼ P) ⇒ (C ∧ ∼ C). Notice that the columns for P and (∼ P) ⇒ (C ∧ ∼ C)

are exactly the same, so P is logically equivalent to (∼ P) ⇒ (C ∧ ∼ C).

P C ∼ P C ∧ ∼ C (∼ P) ⇒ (C ∧ ∼ C)

T T F F T

T F F F T

F T T F F

F F T F F

Therefore to prove a statement P, it suffices to instead prove the conditional

statement (∼ P) ⇒ (C ∧ ∼ C). This can be done with direct proof: Assume

∼ P and deduce C ∧ ∼ C. Here is the outline:

Outline for Proof by Contradiction

Proposition P.

Proof. Suppose ∼ P.

Therefore C ∧ ∼ C. ■

One slightly unsettling feature of this method is that we may not know

at the beginning of the proof what the statement C is going to be. In

doing the scratch work for the proof, you assume that ∼ P is true, then

deduce new statements until you have deduced some statement C and its

negation ∼ C.

If this method seems confusing, look at it this way. In the first line of

the proof we suppose ∼ P is true, that is we assume P is false. But if P is

really true then this contradicts our assumption that P is false. But we

haven’t yet proved P to be true, so the contradiction is not obvious. We

use logic and reasoning to transform the non-obvious contradiction ∼ P to

an obvious contradiction C∧ ∼ C.

114 Proof by Contradiction

Proof by contradiction gives us a starting point: Assume

p

2 is rational,

and work from there.

In the above proof we got the contradiction (b is even) ∧ ∼(b is even)

which has the form C∧ ∼ C. In general, your contradiction need not

necessarily be of this form. Any statement that is clearly false is sufficient.

For example 2 6 = 2 would be a fine contradiction, as would be 4 | 2 , provided

that you could deduce them.

Here is another ancient example, dating back at least as far as Euclid:

Proposition There are infinitely many prime numbers.

Proof. For the sake of contradiction, suppose there are only finitely many

prime numbers. Then we can list all the prime numbers as p 1 , p 2 , p 3 ,... pn,

where p 1 = 2 , p 2 = 3 , p 3 = 5 , p 4 = 7 and so on. Thus pn is the nth and largest

prime number. Now consider the number a = (p 1 p 2 p 3 · · · pn)+ 1 , that is, a is

the product of all prime numbers, plus 1. Now a, like any natural number

greater than 1, has at least one prime divisor, and that means pk | a for at

least one of our n prime numbers pk. Thus there is an integer c for which

a = c pk, which is to say

(p 1 p 2 p 3 · · · pk− 1 pk pk+ 1 · · · pn) + 1 = c pk.

Dividing both sides of this by pk gives us

(p 1 p 2 p 3 · · · pk− 1 pk+ 1 · · · pn) +

pk

= c,

so

pk

= c − (p 1 p 2 p 3 · · · pk− 1 pk+ 1 · · · pn).

The expression on the right is an integer, while the expression on the left

is not an integer. This is a contradiction. ■

Proof by contradiction often works well in proving statements of the

form ∀x, P(x). The reason is that the proof set-up involves assuming

∼ ∀x, P(x), which as we know from Section 2.10 is equivalent to ∃ x, ∼ P(x).

This gives us a specific x for which ∼ P(x) is true, and often that is enough

to produce a contradiction. Here is an example:

Proposition For every real number x ∈ [0, π /2], we have sin x + cos x ≥ 1.

Proof. Suppose for the sake of contradiction that this is not true.

Then there exists an x ∈ [0, π /2] for which sin x + cos x < 1.

Proving Conditional Statements by Contradiction 115

Since x ∈ [0, π /2], neither sin x nor cos x is negative, so 0 ≤ sin x + cos x < 1.

Thus 02 ≤ (sin x + cos x)^2 < 12 , which gives 02 ≤ sin^2 x + 2 sin x cos x + cos^2 x < 12.

As sin^2 x + cos^2 x = 1 , this becomes 0 ≤ 1 + 2 sin x cos x < 1 , so 1 + 2 sin x cos x < 1.

Subtracting 1 from both sides gives 2 sin x cos x < 0.

But this contradicts the fact that neither sin x nor cos x is negative. ■

6.2 Proving Conditional Statements by Contradiction

Since the previous two chapters dealt exclusively with proving conditional

statements, we now formalize the procedure in which contradiction is used

to prove a conditional statement. Suppose we want to prove a proposition

of the following form.

Proposition If P, then Q.

Thus we need to prove that P ⇒ Q is a true statement. Proof by

contradiction begins with the assumption that ∼ (P ⇒ Q) is true, that is,

that P ⇒ Q is false. But we know that P ⇒ Q being false means that it is

possible that P can be true while Q is false. Thus the first step in the

proof is to assume P and ∼ Q. Here is an outline:

Outline for Proving a Conditional

Statement with Contradiction

Proposition If P, then Q.

Proof. Suppose P and ∼ Q.

Therefore C ∧ ∼ C. ■

To illustrate this new technique, we revisit a familiar result: If a^2 is

even, then a is even. According to the outline, the first line of the proof

should be “For the sake of contradiction, suppose a^2 is even and a is not

even.”

Proposition Suppose a ∈ Z. If a^2 is even, then a is even.

Proof. For the sake of contradiction, suppose a^2 is even and a is not even.

Then a^2 is even, and a is odd.

Since a is odd, there is an integer c for which a = 2 c + 1.

Then a^2 = (2c + 1)^2 = 4 c^2 + 4 c + 1 = 2(2c^2 + 2 c) + 1 , so a^2 is odd.

Thus a^2 is even and a^2 is not even, a contradiction. ■

Some Words of Advice 117

for integers c and d, so

p 2 = r

d

c

But we know r = a/b, which combines with the above equation to give

p 2 = r

d

c

a

b

d

c

ad

bc

This means

p

2 is rational, which is a contradiction because we know it is

irrational. Therefore r/

p

2 is irrational.

Consequently r =

p 2 · r/

p

2 is a product of two irrational numbers. ■

For another example of a proof-within-a-proof, try Exercise 5 at the

end of this chapter (or see its solution). Exercise 5 asks you to prove that

p

3 is irrational. This turns out to be slightly trickier than proving that

p

2 is irrational.

6.4 Some Words of Advice

Despite the power of proof by contradiction, it’s best to use it only when the

direct and contrapositive approaches do not seem to work. The reason for

this is that a proof by contradiction can often have hidden in it a simpler

contrapositive proof, and if this is the case it’s better to go with the simpler

approach. Consider the following example.

Proposition Suppose a ∈ Z. If a^2 − 2 a + 7 is even, then a is odd.

Proof. To the contrary, suppose a^2 − 2 a + 7 is even and a is not odd.

That is, suppose a^2 − 2 a + 7 is even and a is even.

Since a is even, there is an integer c for which a = 2 c.

Then a^2 − 2 a + 7 = (2c)^2 − 2(2c) + 7 = 2(2c^2 − 2 c + 3) + 1 , so a^2 − 2 a + 7 is odd.

Thus a^2 − 2 a + 7 is both even and odd, a contradiction. ■

Though there is nothing really wrong with this proof, notice that part

of it assumes a is not odd and deduces that a^2 − 2 a + 7 is not even. That is

the contrapositive approach! Thus it would be more efficient to proceed as

follows, using contrapositive proof.

Proposition Suppose a ∈ Z. If a^2 − 2 a + 7 is even, then a is odd.

Proof. (Contrapositive) Suppose a is not odd.

Then a is even, so there is an integer c for which a = 2 c.

Then a^2 − 2 a + 7 = (2c)^2 − 2(2c) + 7 = 2(2c^2 − 2 c + 3) + 1 , so a^2 − 2 a + 7 is odd.

Thus a^2 − 2 a + 7 is not even. ■

118 Proof by Contradiction

Exercises for Chapter 6

A. Use the method of proof by contradiction to prove the following statements.

(In each case, you should also think about how a direct or contrapositive proof would work. You will find in most cases that proof by contradiction is easier.)

1. Suppose n ∈ Z. If n is odd, then n^2 is odd. 2. Suppose n ∈ Z. If n^2 is odd, then n is odd. 3. Prove that

p 3 2 is irrational.

4. Prove that

p 6 is irrational.

5. Prove that

p 3 is irrational.

6. If a, b ∈ Z, then a^2 − 4 b − 2 6 = 0. 7. If a, b ∈ Z, then a^2 − 4 b − 3 6 = 0. 8. Suppose a, b, c ∈ Z. If a^2 + b^2 = c^2 , then a or b is even. 9. Suppose a, b ∈ R. If a is rational and ab is irrational, then b is irrational. 10. There exist no integers a and b for which 21 a + 30 b = 1. 11. There exist no integers a and b for which 18 a + 6 b = 1. 12. For every positive x ∈ Q, there is a positive y ∈ Q for which y < x. 13. For every x ∈ [ π /2, π ], sin x − cos x ≥ 1. 14. If A and B are sets, then A ∩ (B − A) = ;. 15. If b ∈ Z and b - k for every k ∈ N, then b = 0. 16. If a and b are positive real numbers, then a + b ≥ 2

p ab.

17. For every n ∈ Z, 4 - (n^2 + 2). 18. Suppose a, b ∈ Z. If 4 | (a^2 + b^2 ), then a and b are not both odd.

B. Prove the following statements using any method from Chapters 4, 5 or 6.

19. The product of any five consecutive integers is divisible by 120. (For example, the product of 3,4,5,6 and 7 is 2520, and 2520 = 120 · 21 .) 20. We say that a point P = (x, y) in R^2 is rational if both x and y are rational. More precisely, P is rational if P = (x, y) ∈ Q^2. An equation F(x, y) = 0 is said to have a rational point if there exists x 0 , y 0 ∈ Q such that F(x 0 , y 0 ) = 0. For example, the curve x^2 + y^2 − 1 = 0 has rational point (x 0 , y 0 ) = (1, 0). Show that the curve x^2 + y^2 − 3 = 0 has no rational points. 21. Exercise 20 (above) involved showing that there are no rational points on the curve x^2 + y^2 − 3 = 0. Use this fact to show that

p 3 is irrational.

22. Explain why x^2 + y^2 − 3 = 0 not having any rational solutions (Exercise 20) implies x^2 + y^2 − 3 k^ = 0 has no rational solutions for k an odd, positive integer. 23. Use the above result to prove that

√ 3 k^ is irrational for all odd, positive k.

24. The number log 2 3 is irrational.

CHAPTER 6 Proof by Contradiction, Study notes of Logic

Related documents

Partial preview of the text

Download CHAPTER 6 Proof by Contradiction and more Study notes Logic in PDF only on Docsity!

CHAPTER 6

Proof by Contradiction

W

e now explore a third method of proof: proof by contradiction.

This method is not limited to proving just conditional statements—

it can be used to prove any kind of statement whatsoever. The basic idea

is to assume that the statement we want to prove is false , and then show

that this assumption leads to nonsense. We are then led to conclude that

we were wrong to assume the statement was false, so the statement must

be true. As an example, consider the following proposition and its proof.

Proposition If a, b ∈ Z, then a^2 − 4 b 6 = 2.

Proof. Suppose this proposition is false.

This conditional statement being false means there exist numbers a and b

for which a, b ∈ Z is true, but a^2 − 4 b 6 = 2 is false.

In other words, there exist integers a, b ∈ Z for which a^2 − 4 b = 2.

From this equation we get a^2 = 4 b + 2 = 2(2b + 1), so a^2 is even.

Because a^2 is even, it follows that a is even, so a = 2 c for some integer c.

Now plug a = 2 c back into the boxed equation to get (2c)^2 − 4 b = 2 ,

so 4 c^2 − 4 b = 2. Dividing by 2, we get 2 c^2 − 2 b = 1.

Therefore 1 = 2(c^2 − b), and because c^2 − b ∈ Z, it follows that 1 is even.

We know 1 is not even, so something went wrong.

But all the logic after the first line of the proof is correct, so it must be

that the first line was incorrect. In other words, we were wrong to assume

the proposition was false. Thus the proposition is true. ■

You may be a bit suspicious of this line of reasoning, but in the next

section we will see that it is logically sound. For now, notice that at

the end of the proof we deduced that 1 is even, which conflicts with our

knowledge that 1 is odd. In essence, we have obtained the statement

(1 is odd)∧ ∼ (1 is odd), which has the form C∧ ∼ C. Notice that no matter

what statement C is, and whether or not it is true, the statement C∧ ∼ C

is false. A statement—like this one—that cannot be true is called a

contradiction. Contradictions play a key role in our new technique.

112 Proof by Contradiction

6.1 Proving Statements with Contradiction

Let’s now see why the proof on the previous page is logically valid. In

that proof we needed to show that a statement P : (a, b ∈ Z) ⇒ (a^2 − 4 b 6 = 2)

was true. The proof began with the assumption that P was false, that is

that ∼ P was true, and from this we deduced C∧ ∼ C. In other words we

proved that ∼ P being true forces C∧ ∼ C to be true, and this means that

we proved that the conditional statement (∼ P) ⇒ (C ∧ ∼ C) is true. To see

that this is the same as proving P is true, look at the following truth table

for (∼ P) ⇒ (C ∧ ∼ C). Notice that the columns for P and (∼ P) ⇒ (C ∧ ∼ C)

are exactly the same, so P is logically equivalent to (∼ P) ⇒ (C ∧ ∼ C).

Therefore to prove a statement P, it suffices to instead prove the conditional

statement (∼ P) ⇒ (C ∧ ∼ C). This can be done with direct proof: Assume

∼ P and deduce C ∧ ∼ C. Here is the outline:

Outline for Proof by Contradiction

Proposition P.

Proof. Suppose ∼ P.

Therefore C ∧ ∼ C. ■

One slightly unsettling feature of this method is that we may not know

at the beginning of the proof what the statement C is going to be. In

doing the scratch work for the proof, you assume that ∼ P is true, then

deduce new statements until you have deduced some statement C and its

negation ∼ C.

If this method seems confusing, look at it this way. In the first line of

the proof we suppose ∼ P is true, that is we assume P is false. But if P is

really true then this contradicts our assumption that P is false. But we

haven’t yet proved P to be true, so the contradiction is not obvious. We

use logic and reasoning to transform the non-obvious contradiction ∼ P to

an obvious contradiction C∧ ∼ C.

114 Proof by Contradiction

Proof by contradiction gives us a starting point: Assume

2 is rational,

and work from there.

In the above proof we got the contradiction (b is even) ∧ ∼(b is even)

which has the form C∧ ∼ C. In general, your contradiction need not

necessarily be of this form. Any statement that is clearly false is sufficient.

For example 2 6 = 2 would be a fine contradiction, as would be 4 | 2 , provided

that you could deduce them.

Here is another ancient example, dating back at least as far as Euclid:

Proposition There are infinitely many prime numbers.

Proof. For the sake of contradiction, suppose there are only finitely many

prime numbers. Then we can list all the prime numbers as p 1 , p 2 , p 3 ,... pn,

where p 1 = 2 , p 2 = 3 , p 3 = 5 , p 4 = 7 and so on. Thus pn is the nth and largest

prime number. Now consider the number a = (p 1 p 2 p 3 · · · pn)+ 1 , that is, a is