A mathematical system consists of axioms, definitions, and undefined terms. Axioms are assumed to be true. Definitions are used to create new concepts in terms of existing ones. Some terms are not explicitly defined but rather are implicitly defined by the axioms.
Within a mathematical system we can derive theorems. A theorem is a proposition that has been proved to be true. Special kinds of theorems are referred to as lemmas and corollaries. A lemma is a theorem that is usually not too interesting in its own right but is useful in proving another theorem. A corollary is a theorem that follows easily from another theorem.
An argument that establishes the truth of a theorem is called a proof. Logic is a tool for the analysis of proofs. In this section and the next, we introduce some general methods of proof. In Sections 2.3–2.5, we discuss resolution and mathematical induction, which are special proof techniques. We begin by giving some examples of mathematical systems.
Example 2.1.1 Euclidean geometry furnishes an example of a mathematical system. Among the axioms are
■ Given two distinct points, there is exactly one line that contains them.
■ Given a line and a point not on the line, there is exactly one line parallel to the line through the point.
The terms point and line are undefined terms that are implicitly defined by the axioms that describe their properties.
Among the definitions are
■ Two triangles are congruent if their vertices can be paired so that the corresponding sides and corresponding angles are equal.
■ Two angles are supplementary if the sum of their measures is 180◦.
Example 2.1.2 The real numbers furnish another example of a mathematical system. Among the axioms are
■ For all real numbers x and y, x y= yx.
■ There is a subset P of real numbers satisfying (a) If x and y are in P, then x+ y and xy are in P.
(b) If x is a real number, then exactly one of the following statements is true:
x is in P, x= 0, −x is in P.
Multiplication is implicitly defined by the first axiom and others that describe the prop-erties multiplication is assumed to have.
Among the definitions are
■ The elements in P (of the preceding axiom) are called positive real numbers.
■ The absolute value|x| of a real number x is defined to be x if x is positive or 0 and−x otherwise.
We give several examples of theorems, corollaries, and lemmas in Euclidean geometry and in the system of real numbers.
Example 2.1.3 Examples of theorems in Euclidean geometry are
■ If two sides of a triangle are equal, then the angles opposite them are equal.
■ If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.
Example 2.1.4 An example of a corollary in Euclidean geometry is
■ If a triangle is equilateral, then it is equiangular.
This corollary follows immediately from the first theorem of Example 2.1.3.
Example 2.1.5 Examples of theorems about real numbers are
■ x· 0 = 0 for every real number x.
■ For all real numbers x, y, and z, if x ≤ y and y ≤ z, then x ≤ z.
Example 2.1.6 An example of a lemma about real numbers is
■ If n is a positive integer, then either n− 1 is a positive integer or n − 1 = 0.
Surely this result is not that interesting in its own right, but it can be used to prove other results.
Direct Proofs
Theorems are often of the form
For all x1, x2,. . . , xn, if p(x1, x2,. . . , xn), then q(x1, x2,. . . , xn).
This universally quantified statement is true provided that the conditional proposition if p(x1, x2,. . . , xn), then q(x1, x2,. . . , xn) (2.1.1) is true for all x1, x2,. . . , xnin the domain of discourse. To prove (2.1.1), we assume that x1, x2,. . . , xnare arbitrary members of the domain of discourse. If p(x1, x2,. . . , xn) is false, by Definition 1.3.3, (2.1.1) is vacuously true; thus, we need only consider the case that p(x1, x2,. . . , xn) is true. A direct proof assumes that p(x1, x2,. . . , xn) is true and then, using p(x1, x2,. . . , xn) as well as other axioms, definitions, previously derived theorems, and rules of inference, shows directly that q(x1, x2,. . . , xn) is true.
Everyone “knows” what an even or odd integer is, but the following definition makes these terms precise and provides a formal way to use the terms “even integer”
and “odd integer” in proofs.
Definition 2.1.7 An integer n is even if there exists an integer k such that n= 2k. An integer n is odd if there exists an integer k such that n= 2k + 1.
Example 2.1.8 The integer n = 12 is even because there exists an integer k (namely k = 6) such that n= 2k; that is, 12 = 2 · 6.
Example 2.1.9 The integer n= −21 is odd because there exists an integer k (namely k = −11) such that n= 2k + 1; that is, −21 = 2 · − 11 + 1.
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Example 2.1.10 Give a direct proof of the following statement. For all integers m and n, if m is odd and n is even, then m+ n is odd.
Discussion In a direct proof, we assume the hypotheses and derive the conclusion. A good start is achieved by writing out the hypotheses and conclusion so that we are clear where we start and where we are headed. In the case at hand, we have
m is odd and n is even. (Hypotheses)
· · ·
m+ n is odd. (Conclusion)
The gap (· · ·) represents the part of the proof to be completed that leads from the hy-potheses to the conclusion.
We can begin to fill in the gap by using the definitions of “odd” and “even” to obtain
m is odd and n is even. (Hypotheses)
There exists an integer, say k1, such that m= 2k1+ 1. (Because m is odd) There exists an integer, say k2, such that n = 2k2. (Because n is even)
· · ·
m+ n is odd. (Conclusion)
(Notice that we cannot assume that k1 = k2. For example if m = 15 and n = 4, then k1 = 7 and k2 = 2. That k1is not necessarily equal to k2 is the reason that we must denote the two integers with different symbols.)
The missing part of our proof is the argument to show that m+ n is odd. How can we reach this conclusion? We can use the definition of “odd” again if we can show that m+ n is equal to
2× some integer + 1. (2.1.2)
We already know that m= 2k1+ 1 and n = 2k2. How can we use these facts to reach our goal (2.1.2)? Since the goal involves m+ n, we can add the equations m = 2k1+ 1 and n= 2k2to obtain a fact about m+ n, namely,
m+ n = (2k1+ 1) + 2k2.
Now this expression is supposed to be of the form (2.1.2). We can use a little algebra to show that it is of the desired form:
m+ n = (2k1+ 1) + 2k2= 2(k1+ k2)+ 1.
We have our proof.
Proof Let m and n be arbitrary integers, and suppose that m is odd and n is even. We prove that m+ n is odd. By definition, since m is odd, there exists an integer k1 such that m= 2k1+ 1. Also, by definition, since n is even, there exists an integer k2such that n= 2k2. Now the sum is
m+ n = (2k1+ 1) + (2k2)= 2(k1+ k2)+ 1.
Thus, there exists an integer k (namely k= k1+ k2) such that m+ n = 2k+1. Therefore, m+ n is odd.
Example 2.1.11 Give a direct proof of the following statement. For all sets X , Y , and Z , X∩ (Y − Z) = ( X∩ Y ) − (X ∩ Z).
Discussion The outline of the proof is X , Y , and Z are sets. (Hypothesis)
· · ·
X∩ (Y − Z) = (X ∩ Y ) − (X ∩ Z) (Conclusion)
The conclusion asserts that the two sets X∩ (Y − Z) and (X ∩ Y ) − (X ∩ Z) are equal.
Recall (see Section 1.1) that to prove from the definition of set equality that these sets are equal, we must show that for all x,
if x∈ X ∩ (Y − Z), then x ∈ (X ∩ Y ) − (X ∩ Z) and
if x∈ (X ∩ Y ) − (X ∩ Z), then x ∈ X ∩ (Y − Z).
Thus our proof outline becomes
X , Y , and Z are sets. (Hypothesis)
If x∈ X ∩ (Y − Z), then x ∈ (X ∩ Y ) − (X ∩ Z).
If x∈ (X ∩ Y ) − (X ∩ Z), then x ∈ X ∩ (Y − Z).
X∩ (Y − Z) = (X ∩ Y ) − (X ∩ Z) (Conclusion)
We should be able to use the definitions of intersection (∩) and set difference (−) to complete the proof.
To prove
if x ∈ X ∩ (Y − Z), then x ∈ (X ∩ Y ) − (X ∩ Z),
we begin by assuming that (the arbitrary element) x is in X∩ (Y − Z). Because this latter set is an intersection, we immediately deduce that x ∈ X and x ∈ Y − Z. The proof proceeds in this way. As one constructs the proof, it is essential to keep the goal in mind: x∈ (X ∩ Y ) − (X ∩ Z). To help guide the construction of the proof, it may be helpful to translate the goal using the definition of set difference: x ∈ (X ∩ Y ) −(X ∩ Z) means x∈ X ∩ Y and x /∈ X ∩ Z.
Proof Let X , Y , and Z be arbitrary sets. We prove
X∩ (Y − Z) = (X ∩ Y ) − (X ∩ Z) by proving
if x∈ X ∩ (Y − Z), then x ∈ (X ∩ Y ) − (X ∩ Z) (2.1.3) and
if x∈ (X ∩ Y ) − (X ∩ Z), then x ∈ X ∩ (Y − Z). (2.1.4) To prove equation (2.1.3), let x∈ X ∩ (Y − Z). By the definition of intersection, x ∈ X and x ∈ Y − Z. By the definition of set difference, since x ∈ Y − Z, x ∈ Y and x /∈ Z. By the definition of intersection, since x ∈ X and x ∈ Y , x ∈ X ∩ Y . Again by the definition of intersection, since x /∈ Z, x /∈ X ∩ Z. By the definition of set difference, since x ∈ X ∩ Y , but x /∈ X ∩ Z, x ∈ (X ∩ Y ) − (X ∩ Z). We have proved equation (2.1.3).
To prove equation (2.1.4), let x ∈ (X ∩ Y ) − (X ∩ Z). By the definition of set difference, x∈ X ∩ Y and x /∈ X ∩ Z. By the definition of intersection, since x ∈ X ∩ Y , x∈ X and x ∈ Y . Again, by the definition of intersection, since x /∈ X ∩ Z and x ∈ X,
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x /∈ Z. By the definition of set difference, since x ∈ Y and x /∈ Z, x ∈ Y − Z. Finally, by the definition of intersection, since x ∈ X and x ∈ Y − Z, x ∈ X ∩ (Y − Z). We have proved equation (2.1.4).
Since we have proved both equations (2.1.3) and (2.1.4), it follows that X∩ (Y − Z) = (X ∩ Y ) − (X ∩ Z).
Our next example shows that in constructing a proof, we may find that we need some auxiliary results, at which point we pause, go off and prove these auxiliary results, and then return to the main proof. We call the proofs of auxiliary results subproofs. (For those familiar with programming, a subproof is similar to a subroutine.)
Example 2.1.12 If a and b are real numbers, we define min{a, b} to be the minimum of a and b or the common value if they are equal. More precisely,
min{a, b} =
⎧⎨
⎩
a if a< b a if a= b b if b< a.
We will give a direct proof of the following statement. For all real numbers d, d1, d2, x, if d= min{d1, d2} and x ≤ d, then x ≤ d1and x ≤ d2.
Discussion The outline of the proof is d = min{d1, d2} and x ≤ d (Hypotheses)
· · ·
x≤ d1and x ≤ d2(Conclusion)
To help understand what is being asserted, let us look at a specific example. As we have remarked previously, when we are asked to prove a universally quantified statement, a specific example does not prove the statement. It may, however, help us to understand the statement.
Let us set d1 = 2 and d2 = 4. Then d = min{d1, d2} = 2. The statement to be proved says that if x≤ d (= 2), then x ≤ d1(= 2) and x ≤ d2(= 4). Why is this true in general? The minimum d of two numbers, d1and d2, is equal to one of the two numbers (namely, the smallest) and less than or equal to the other one (namely, the largest)—in symbols, d ≤ d1and d ≤ d2. If x ≤ d, then from x ≤ d and d ≤ d1, we may deduce x ≤ d1. Similarly, from x≤ d and d ≤ d2, we may deduce x ≤ d2. Thus the outline of our proof becomes
d = min{d1, d2} and x ≤ d (Hypotheses) Subproof: Show that d≤ d1and d≤ d2. From x≤ d and d ≤ d1, deduce x≤ d1. From x≤ d and d ≤ d2, deduce x≤ d2.
x≤ d1and x ≤ d2(Conclusion—uses the conjunction inference rule)
At this point, the only part of the proof that is missing is the subproof to show that d ≤ d1and d ≤ d2. Let us look at the definition of “minimum.” If d1 ≤ d2, then d = min{d1, d2} = d1and d = d1 ≤ d2. If d2 < d1, then d = min{d1, d2} = d2and d = d2< d1. In either case, d ≤ d1and d≤ d2.
Proof Let d, d1, d2, and x be arbitrary real numbers, and suppose that d = min{d1, d2} and x ≤ d.
We prove that
x≤ d1and x ≤ d2.
We first show that d≤ d1and d≤ d2. From the definition of “minimum,” if d1 ≤ d2, then d = min{d1, d2} = d1and d= d1≤ d2. If d2< d1, then d = min{d1, d2} = d2
and d= d2< d1. In either case, d ≤ d1and d≤ d2. From x ≤ d and d ≤ d1, it follows that x ≤ d1 from a previous theorem (the second theorem of Example 2.1.5). From x ≤ d and d ≤ d2, we may derive x ≤ d2from the same previous theorem. Therefore, x≤ d1and x ≤ d2.
Example 2.1.13 There are frequently many different ways to prove a statement. We illustrate by giving two proofs of the statement
X∪ (Y − X) = X ∪ Y for all sets X and Y .
Discussion We first give a direct proof like the proof in Example 2.1.11. We show that for all x, if x∈ X ∪ (Y − X), then x ∈ X ∪ Y , and if x ∈ X ∪ Y , then x ∈ X ∪ (Y − X).
Our second proof uses Theorem 1.1.21, which gives laws of sets. The idea is to begin with X ∪ (Y − X) and use the laws of sets, which here we think of as rules to manipulate set equations, to obtain X∪ Y .
Proof [First proof] We show that for all x, if x ∈ X ∪ (Y − X), then x ∈ X ∪ Y , and if x∈ X ∪ Y , then x ∈ X ∪ (Y − X).
Let x ∈ X ∪ (Y − X). Then x ∈ X or x ∈ Y − X. If x ∈ X, then x ∈ X ∪ Y . If x∈ Y − X, then x ∈ Y , so again x ∈ X ∪ Y . In either case, x ∈ X ∪ Y .
Let x ∈ X ∪ Y . Then x ∈ X or x ∈ Y . If x ∈ X, then x ∈ X ∪ (Y − X). If x /∈ X, then x ∈ Y . In this case, x ∈ Y − X. Therefore, x ∈ X ∪ (Y − X). In either case, x∈ X ∪ (Y − X). The proof is complete.
Proof [Second proof] We use Theorem 1.1.21, which gives laws of sets, and the fact that Y− X = Y ∩ X, which follows immediately from the definition of set difference.
Letting U denote the universal set, we obtain
X∪ (Y − X) = X ∪ (Y ∩ X) [Y − X = Y ∩ X]
= (X ∪ Y ) ∩ (X ∪ X) [Distributive law; Theorem 1.1.21, part (c)]
= (X ∪ Y ) ∩ U [Complement law; Theorem 1.1.21, part (e)]
= X ∪ Y [Identity law; Theorem 1.1.21, part (d)].
Disproving a Universally Quantified Statement Recall (see Section 1.5) that to disprove
∀x P(x)
we simply need to find one member x in the domain of discourse that makes P(x) false.
Such a value for x is called a counterexample.
Example 2.1.14 The statement
∀n ∈ Z+(2n+ 1 is prime)
is false. A counterexample is n= 3 since 23+ 1 = 9, which is not prime.
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Example 2.1.15 If the statement
( A∩ B) ∪ C = A ∩ (B ∪ C), for all sets A, B, and C is true, prove it; otherwise, give a counterexample.
Let us begin by trying to prove the statement. We will first try to show that if x∈ ( A ∩ B) ∪ C, then x ∈ A ∩ (B ∪ C). If x ∈ ( A ∩ B) ∪ C, then
x∈ A ∩ B or x ∈ C. (2.1.5)
We have to show that x ∈ A ∩ (B ∪ C), that is,
x∈ A and x ∈ B ∪ C. (2.1.6)
Equation (2.1.5) is true if x is in C, and equation (2.1.6) is false if x /∈ A. Thus the given statement is false; there is no direct proof (or any other proof!). If we choose sets A and C so that there is an element that is in C, but not in A, we will have a counterexample.
Let
A= {1, 2, 3}, B= {2, 3, 4}, C = {3, 4, 5}
so that there is an element that is in C, but not in A. Then
( A∩ B) ∪ C = {2, 3, 4, 5} and A ∩ (B ∪ C) = {2, 3}
and
( A∩ B) ∪ C = A ∩ (B ∪ C).
Thus A, B, and C provide a counterexample that shows that the given statement is false.
Problem-Solving Tips
To construct a direct proof of a universally quantified statement, first write down the hypotheses (so you know what you are assuming), and then write down the conclusion (so you know what you must prove). The conclusion is what you will work toward—
something like the answer in the back of the book to an exercise, except here it is essential to know the goal before proceeding. You must now give an argument that begins with the hypotheses and ends with the conclusion. To construct the argument, remind yourself what you know about the terms (e.g., “even,” “odd”), symbols (e.g., X∩Y , min{d1, d2}), and so on. Look at relevant definitions and related results. For example, if a particular hypothesis refers to an even integer n, you know that n is of the form 2k for some integer k. If you are to prove that two sets X and Y are equal from the definition of set equality, you know you must show that for every x, if x∈ X then x ∈ Y , and if x ∈ Y then x ∈ X.
To understand what is to be proved, look at some specific values in the domain of discourse. When we are asked to prove a universally quantified statement, showing that the statement is true for specific values does not prove the statement; it may, however, help to understand the statement.
To disprove a universally quantified statement, find one element in the domain of discourse, called a counterexample, that makes the propositional function false. Here, your proof consists of presenting the counterexample together with justification that the propositional function is indeed false for your counterexample.
When you write up your proof, begin by writing out the statement to be proved.
Indicate clearly where your proof begins (e.g., by beginning a new paragraph or by
writing “Proof.”). Use complete sentences, which may include symbols. For example, it is perfectly acceptable to write
Thus x∈ X.
In words, this is the complete sentence: Thus x is in X . End a direct proof by clearly stating the conclusion, and, perhaps, giving a reason to justify the conclusion. For example, Example 2.1.10 ends with:
Thus, there exists an integer k (namely k= k1+ k2) such that m+ n = 2k + 1.
Therefore, m+ n is odd.
Here the conclusion (m+ n is odd) is clearly stated and justified by the statement m + n = 2k+ 1.
Alert the reader where you are headed. For example, if you are going to prove that X = Y , write “We will prove that X = Y ” before launching into this part of the proof.
Justify your steps. For example, if you conclude that x ∈ X or x ∈ Y because it is known that x ∈ X ∪ Y , write “Since x ∈ X ∪ Y , x ∈ X or x ∈ Y ,” or perhaps even
“Since x ∈ X ∪ Y , by the definition of union x ∈ X or x ∈ Y ” if, like Richard Nixon, you want to be perfectly clear.
If you are asked to prove or disprove a universally quantified statement, you can begin by trying to prove it. If you succeed, you are finished—the statement is true and you proved it! If your proof breaks down, look carefully at the point where it fails.
The given statement may be false and your failed proof may give insight into how to construct a counterexample (see Example 2.1.15.) On the other hand, if you have trouble constructing a counterexample, check where your proposed examples fail. This insight may show why the statement is true and guide construction of a proof.
Some Common Errors
In Example 2.1.10, we pointed out that it is an error to use the same notation for two possibly distinct quantities. As an example, here is a faulty “proof” that for all m and n, if m and n are even integers then mn is a square (i.e., mn= a2for some integer a):
Since m and n are even, m = 2k and n = 2k. Now mn = (2k)(2k) = (2k)2. If we let a= 2k, then m = a2. The problem is that we cannot use k for two potentially different quantities. If m and n are even, all we can conclude is that m = 2k1 and n= 2k2for some integers k1and k2. The integers k1and k2need not be equal. (In fact, it is false that for all m and n, if m and n are even integers then mn is a square. A counterexample is m= 2 and n = 4.)
Given a universally quantified propositional function, showing that the proposi-tional function is true for specific values in the domain of discourse is not a proof that the propositional function is true for all values in the domain of discourse. (Such specific values may, however, suggest that the propositional function might be true for all values in the domain of discourse.) Example 2.1.10 is to prove that for all integers m and n,
if m is odd and n is even, then m+ n is odd. (2.1.7) Letting m= 11 and n = 4 and noting that m + n = 15 is odd does not constitute a proof that (2.1.7) is true for all integers m and n, it merely proves that (2.1.7) is true for the specific values m= 11 and n = 4.
In constructing a proof, you cannot assume what you are supposed to prove. As an example, consider the erroneous “proof” that for all integers m and n, if m and m+ n are even, then n is even: Let m= 2k1and n= 2k2. Then m+ n = 2k1+ 2k2. Therefore,
n= (m + n) − m = (2k1+ 2k2)− 2k2= 2(k1+ k2− k2).
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Thus n is even. The problem with the preceding “proof” is that we cannot write n= 2k2
since this is true if and only if n is even—which is what we are supposed to prove! This error is called begging the question or circular reasoning. [It is true that if m and m+n are even integers, then n is even (see Exercise 12).]
Section Review Exercises
1. What is a mathematical system?
2. What is an axiom?
3. What is a definition?
4. What is an undefined term?
5. What is a theorem?
6. What is a proof?
7. What is a lemma?
8. What is a direct proof?
9. What is the formal definition of “even integer”?
10. What is the formal definition of “odd integer”?
11. What is a subproof?
12. How do you disprove a universally quantified statement?
Exercises
1. Give an example (different from those of Example 2.1.1) of an axiom in Euclidean geometry.
2. Give an example (different from those of Example 2.1.2) of an axiom in the system of real numbers.
3. Give an example (different from those of Example 2.1.1) of a definition in Euclidean geometry.
4. Give an example (different from those of Example 2.1.2) of a definition in the system of real numbers.
5. Give an example (different from those of Example 2.1.3) of a theorem in Euclidean geometry.
6. Give an example (different from those of Example 2.1.5) of a theorem in the system of real numbers.
7. Prove that for all integers m and n, if m and n are even, then
13. Prove that for all rational numbers x and y, x+ y is rational.
14. Prove that for all rational numbers x and y, x y is rational.
15. Prove that for every rational number x, if x = 0, then 1/x is rational.
16. If a and b are real numbers, we define max{a, b} to be the maximum of a and b or the common value if they are equal.
Prove that for all real numbers d, d1, d2, x,
if d= max{d1, d2} and x ≥ d, then x ≥ d1and x≥ d2.
17. Justify each step of the following direct proof, which shows that if x is a real number, then x· 0 = 0. Assume that the fol-lowing are previous theorems: If a, b, and c are real numbers, then b+ 0 = b and a(b + c) = ab + ac. If a + b = a + c, then b= c.
Proof x· 0 + 0 = x · 0 = x · (0 + 0) = x · 0 + x · 0;
therefore, x· 0 = 0.
18. If X and Y are nonempty sets and X× Y = Y × X, what can we conclude about X and Y ? Prove your answer.
19. Prove that X∩ Y ⊆ X for all sets X and Y . 30. Give a direct proof along the lines of the second proof in
Ex-ample 2.1.13 of the statement
X∩ (Y − Z) = (X ∩ Y ) − (X ∩ Z) for all sets X, Y , and Z.
(In Example 2.1.11, we gave a direct proof of this statement using the definition of set equality.)
In each of Exercises 31–43, if the statement is true, prove it; other-wise, give a counterexample. The sets X , Y , and Z are subsets of a