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- Methods of Proof
2.1. Types of Proofs. Suppose we wish to prove an implication p → q. Here are some strategies we have available to try.
- Trivial Proof: If we know q is true then p → q is true regardless of the truth value of p.
- Vacuous Proof: If p is a conjunction of other hypotheses and we know one or more of these hypotheses is false, then p is false and so p → q is vacuously true regardless of the truth value of q.
- Direct Proof: Assume p, and then use the rules of inference, axioms, defi- nitions, and logical equivalences to prove q.
- Indirect Proof or Proof by Contradiction: Assume p and ¬q and derive a contradiction r ∧ ¬r.
- Proof by Contrapositive: (Special case of Proof by Contradiction.) Give a direct proof of ¬q → ¬p. Assume ¬q and then use the rules of inference, axioms, definitions, and logical equivalences to prove ¬p.(Can be thought of as a proof by contradiction in which you assume p and ¬q and arrive at the contradiction p ∧ ¬p.)
- Proof by Cases: If the hypothesis p can be separated into cases p 1 ∨ p 2 ∨ · · · ∨ pk, prove each of the propositions, p 1 → q, p 2 → q,... , pk → q, separately. (You may use different methods of proof for different cases.)
Discussion
We are now getting to the heart of this course: methods you can use to write proofs. Let’s investigate the strategies given above in some detail.
2.2. Trivial Proof/Vacuous Proof.
Example 2.2.1. Prove the statement: If there are 100 students enrolled in this course this semester, then 62 = 36.
Proof. The assertion is trivially true, since the conclusion is true, independent of the hypothesis (which, may or may not be true depending on the enrollment).
Example 2.2.2. Prove the statement. If 6 is a prime number, then 62 = 30.
Proof. The hypothesis is false, therefore the statement is vacuously true (even though the conclusion is also false).
Discussion
The first two methods of proof, the “Trivial Proof” and the “Vacuous Proof” are certainly the easiest when they work. Notice that the form of the “Trivial Proof”, q → (p → q), is, in fact, a tautology. This follows from disjunction introduction, since p → q is equivalent to ¬p ∨ q. Likewise, the “Vacuous Proof” is based on the tautology ¬p → (p → q).
Exercise 2.2.1. Fill in the reasons for the following proof of the tautology ¬p → (p → q).
[¬p → (p → q)] ⇔ [p ∨ (¬p ∨ q)] ⇔ [(p ∨ ¬p) ∨ q] ⇔ T ∨ q ⇔ T
Exercise 2.2.2. Let A = { 1 , 2 , 3 } and R = {(2, 3), (2, 1)}(⊆ A × A). Prove: if a, b, c ∈ A are such that (a, b) ∈ R and (b, c) ∈ R then (a, c) ∈ R.
Since it is a rare occasion when we are able to get by with one of these two methods of proof, we turn to some we are more likely to need. In most of the following examples the underlying “theorem” may be a fact that is well known to you. The purpose in presenting them, however, is not to surprise you with new mathematical facts, but to get you thinking about the correct way to set up and carry out a mathematical argument, and you should read them carefully with this in mind.
2.3. Direct Proof.
Example 2.3.1. Prove the statement: For all integers m and n, if m and n are odd integers, then m + n is an even integer.
Proof. Assume m and n are arbitrary odd integers. Then m and n can be written in the form
m = 2a + 1 and n = 2b + 1,
A common mistake of this type might arise as follows:
“Well, m is an odd integer, so I can write m = 2k + 1, where k is an integer. Since n is also an odd integer, I can write n = 2k + 1, where k is an integer.”
Do you see the mistake? By allowing the same letter k to represent what might be different integers, we have inadvertently added another assumption, namely, that m = n! Of course, we didn’t mean to do this, but, unfortunately, our intentions haven’t been carried out, and so our proof breaks down at this point. In order to maintain the “arbitrariness” of m and n, we must allow, at the least, that they be different. We accomplish this by choosing different letters a and b in our representations of m and n as “twice an integer plus one.” There is nothing sacred about a and b; we could have used k and `, or x and y, or α and β, or any pair of symbols that have not been appropriated for some other use.
Upon closer scrutiny, this first step now starts to seem like a big one indeed! Especially if we may not be sure just where it will lead. The rest of the proof, however, proceeds fairly routinely. We add m and n and observe that the resulting expression has a factor of 2. We now only have to get past the recognition problem: observing that the resulting expression gives us what we were looking for. Since we have expressed m + n as twice another integer, m + n is, by definition, an even integer. By Universal Generalization we may now confidently declare “Q.E.D.” (the abbreviation of quod erat demonstrandum or “which was to be demonstrated”). Often a box at the end of a proof or the abbrviation “Q.E.D.” is used at the end of a proof to indicate it is finished.
Exercise 2.3.1. Give a careful proof of the statement: For all integers m and n, if m is odd and n is even, then m + n is odd.
2.4. Proof by Contrapositive.
Example 2.4.1. Prove the statement: For all integers m and n, if the product of m and n is even, then m is even or n is even.
We prove the contrapositive of the statement: If m and n are both odd integers, then mn is odd.
Proof. Suppose that m and n are arbitrary odd integers. Then m = 2a + 1 and n = 2b + 1, where a and b are integers. Then
mn = (2a + 1)(2b + 1) (substitution) = 4ab + 2a + 2b + 1 (associative, commutative, and distributive laws) = 2(2ab + a + b) + 1 (distributive law)
Since mn is twice an integer (namely, 2ab + a + b) plus 1, mn is odd.
Discussion
If a direct proof of an assertion appears problematic, the next most natural strat- egy to try is a proof of the contrapositive. In Example 2.4.1 we use this method to prove that if the product of two integers, m and n, is even, then m or n is even. This statement has the form p → (r ∨ s). If you take our advice above, you will first try to give a direct proof of this statement: assume mn is even and try to prove m is even or n is even. Next, you would use the definition of “even” to write mn = 2k, where k is an integer. You would now like to conclude that m or n has the factor 2. This can, in fact, be proved directly, but it requires more knowledge of number theory than we have available at this point. Thus, we seem to have reached a dead-end with the direct approach, and we decide to try an indirect approach instead.
The contrapositive of p → (r ∨ s) is ¬(r ∨ s) → ¬p, or, by De Morgan’s Law, (¬r ∧ ¬s) → ¬p.
This translates into the statement
“If m and n are odd, then mn is odd”
(where “not even” translates to “odd”). This is a good illustration of how the symbolic form of a proposition can be helpful in finding the correct statement we wish to prove. In this particular example, the necessity of De Morgan’s Law may be more evident in the symbolic form than in the “English version.”
Now we give a direct proof of the contrapositive: we assume m and n are arbitrary odd integers and deduce mn is odd. This proof is carried out in very much the same way as the direct proof in Example 2.3.1. The main difficulty we encounter with the problem of proving the original assertion is to realize that a direct proof should be abandoned in favor of some other strategy.
Exercise 2.4.1. The following statement is a special case of the proposition proved in Example 2.4.1. Give a careful proof of this statement without assuming the result in Example 2.4.1.
For every integer n, if n^2 is even, then n is even.
Proof. Suppose a, b, and c are positive real numbers such that ab = c, and suppose a >
c and b >
c. (Notice the use of De Morgan’s Law again. Also, recall that the symbol
c represents the positive square root of c, not ±
c.) By order properties of the real numbers,
b >
c ⇔ ab > a
c, since a > 0 ,
and
a >
c ⇔ a
c >
c ·
c = c, since
c > 0.
Thus, ab > a
c >
c ·
c = c implies
ab > c.
But ab = c; hence, ab is not greater than c, a contradiction.
This proves our assumption a >
c and b >
c cannot be true when a, b, and c are positive real numbers such that ab = c. Therefore a ≤
c or b ≤
c.
Exercise 2.5.2. Consider the statement: For all nonnegative real numbers a, b, and c, if a^2 + b^2 = c^2 , then a + b ≥ c.
(a) Give a proof by contradiction. (b) Give a direct proof. [Hint: The extra idea needed for a direct proof should emerge naturally from a proof by contradiction.]
Let’s step back and compare direct proof, proof by contrapositive, and proof by contradiction.
Exercise 2.5.3. Fill in the blanks.
If we are proving the implication p → q we assume...
(1) p for a direct proof. (2) for a proof by contrapositive (3) for a proof by contradiction.
We are then allowed to use the truth of the assumption in 1, 2, or 3 in the proof.
After the initial assumption, we prove p → q by showing
(4) q must follow from the assumptions for a direct proof. (5) must follow the assumptions for a proof by contrapositive. (6) must follow the assumptions for a proof by contradiction.
2.6. Proof by Cases.
Example 2.6.1. If x is a real number such that
x^2 − 1 x + 2
0 , then either x > 1 or
− 2 < x < − 1.
Proof. Assume x is a real number for which the inequality x^2 − 1 x + 2
holds. Factor the numerator of the fraction to get the inequality
(x + 1)(x − 1) x + 2
For this combination of x + 1, x − 1, and x + 2 to be positive, either all are positive or two are negative and the other is positive. This gives four cases to consider:
Case 1. x + 1 > 0 , x − 1 > 0, and x + 2 > 0. In this case x > − 1 , x > 1, and x > −2, which implies x > 1. Case 2. x + 1 > 0 , x − 1 < 0, and x + 2 < 0. In this case x > − 1 , x < 1, and x < −2, and there is no x satisfying all three inequalities simultaneously. Case 3. x + 1 < 0 , x − 1 > 0, and x + 2 < 0. In this case x < − 1 , x > 1, and x < −2, and there is no x satisfying all three inequalities simultaneously. Case 4. x + 1 < 0 , x − 1 < 0, and x + 2 > 0. In this case x < − 1 , x < 1, and x > −2, which implies that − 2 < x < −1.
Thus, either x > 1 (Case 1) or − 2 < x < −1 (Case 4).
Discussion
Sometimes the hypothesis of a statement can be broken down into simpler cases that may be investigated separately. The validity of a proof by cases rests on the equivalence [(p 1 ∨ · · · ∨ pn) → q] ⇔ [(p 1 → q) ∨ · · · ∨ (pn → q)].
In Example 2.6.1 this method is used to verify the “solution” to the inequality, x^2 − 1 x + 2
Exercise 2.6.1. Prove: For every real number x,
x^2 = |x|. [Hint: Recall as above that
x^2 represents the positive square root of x^2 , and look at two cases: x ≥ 0 and x < 0 .]
A proof by cases can tend to be a little tedious. Here is an extreme example of such a proof.
In a constructive proof one finds an explicit example in the universe of discourse for which the statement is true.
Here is another example.
Example 2.8.2. Prove: If f (x) = x^3 + x − 5 , then there exists a positive real number c such that f ′(c) = 7.
Proof. Calculate the derivative of f : f ′(x) = 3x^2 + 1. Then we want to find a positive number c such that f ′(c) = 3c^2 + 1 = 7. Solving for c:
3 c^2 = 6 c^2 = 2 c = ±
Then c =
2 is a positive real number and f ′(
2)^2 + 1 = 7.
2.9. Nonconstructive Proof.
Example 2.9.1. Pigeon Hole Principle: If n + 1 objects (pigeons) are dis- tributed into n boxes (pigeon holes), then some box must contain at least 2 of the objects.
Proof. Assume n + 1 objects (pigeons) are distributed into n boxes. Suppose the boxes are labeled B 1 , B 2 , ..., Bn, and assume that no box contains more than 1 object. Let ki denote the number of objects placed in Bi. Then ki ≤ 1 for i = 1, ..., n, and so
k 1 + k 2 + · · · + kn ≤ 1 + 1 +︸ ︷︷ · · · + 1︸ n terms
≤ n.
But this contradicts the fact that k 1 + k 2 + · · · + kn = n + 1, the total number of objects we started with.
Discussion
Sometimes, constructing an example may be difficult, if not impossible, due to the nature of the problem. If you suspect this is the case, you should try a proof by contradiction: Assume there is no such example and show that this leads to a contradiction. If you are successful, you have established existence, but you have not exhibited a specific example. After you have studied the proof of the basic pigeon hole principal in Example 2.9.1, try your hand at the following variations.
Exercise 2.9.1. Prove: If 2 n + 1 objects are distributed into n boxes, then some box must contain at least 3 of the objects.
Exercise 2.9.2. Fill in the blank in the following statement and then give a proof.
Suppose k is a positive integer. If kn + 1 objects are distributed into n boxes, then some box must contain at least of the objects.
Exercise 2.9.3. Suppose that 88 chairs are arranged in a rectangular array of 8 rows and 11 columns, and suppose 50 students are seated in this array (1 student per chair).
(a) Prove that some row must have at least 7 students. (b) Prove that some column must have at most 4 students.
2.10. Nonexistence Proofs. Suppose we wish to establish the truth of the statement ¬∃xP (x), which is equivalent to ∀x¬P (x). One way is to assume there is a member, c, of the universe of discourse for which P (c) is true, and try to arrive at a contradiction.
Example 2.10.1. Prove there does not exist an integer k such that 4 k + 3 is a perfect square.
Proof. Proof by Contradiction: Assume there is an integer k such that 4k + 3 is a perfect square. That is, 4k + 3 = m^2 , where m is an integer. Since the square of an even integer is even and 4k + 3 is odd, m must be odd. Then m = 2a + 1 for some integer a. Thus,
4 k + 3 = m^2 4 k + 3 = (2a + 1)^2 4 k + 3 = 4a^2 + 4a + 1 4 k + 3 = 4(a^2 + a) + 1 3 − 1 = 4(a^2 + a) − 4 k 2 = 4(a^2 + a − k) 1 = 2(a^2 + a − k)
But this contradicts the fact that 1 is an odd integer.
Discussion
Discussion
From Example 2.6.1 one might be led to think that n^2 − n + 41 is a prime number for every natural number n. After all, it worked for the first 41 natural numbers. (Or so, you were led to believe. Did you finish the remaining 35 cases?) Showing that a predicate P (x) is true for a few, perhaps many millions of x’s in its universe of discourse, however, does not constitute a proof of ∀xP (x), unless you were able to exhaust all possibilities. This, of course, is not possible if the universe of discourse is an infinite set, such as the set of natural numbers or the set of real numbers. Since the negation of ∀xP (x) is ¬∀xP (x) ⇔ ∃x¬P (x), it only takes one x for which P (x) is false, a counterexample, to disprove ∀xP (x). The assertion “for every natural number n, n^2 − n + 41 is prime” is, in fact, false.
Exercise 2.12.1. Find a counterexample to the statement: For every natural number n, n^2 − n + 41 is prime.
2.13. Biconditional. In order to establish the truth of the statement p ↔ q, use the fact that (p ↔ q) is equivalent to (p → q) ∧ (q → p), and prove both implications using any of the previous methods.
Discussion
We conclude this module with a discussion on proving a biconditional or “if and only if” statement. As pointed out above, a proof of a biconditional requires two proofs: the proof of an implication and a proof of its converse. Our example below is very similar to theorems we have proved earlier. The point here is that the two implications may be proved independently of each other, and the decision on the best strategy to use should be made for each one separately.
Example 2.13.1. Prove: For any integer n, n is odd if and only if n^2 is odd.
In order to prove this statement, we must prove two implications:
(a) If n is odd, then n^2 is odd. (b) If n^2 is odd, then n is odd.
Proof of (a): We give a direct proof of this statement. Assume n is an odd integer. Then n = 2a + 1 for some integer a. Then n^2 = (2a + 1)^2 = 4a^2 + 4a + 1 = 2(2a^2 + 2a) + 1, which is twice an integer plus 1. Thus, n^2 is odd.
Proof of (b): We give a proof of the contrapositive of this statement: “If n is even (not odd), then n^2 is even (not odd). Assume n is an even integer. Then n = 2a for some integer a. Then n^2 = (2a)^2 = 4a^2 = 2(2a^2 ), which is an even integer.
Exercise 2.13.1. Prove the following statements are equivalent.
(1) n − 5 is odd. (2) 3 n + 2 is even. (3) n^2 − 1 is odd.
Hint: Prove the following implications:
(1) 1→ 2 (2) 2→ 1 (3) 1→ 3 (4) 3→ 1
